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47639b61-2491-48af-a060-e83958a1dabd | 6 | 3 | 0 | 14 | 4 | 2 | 13 | 0 | In the boundary layer equations (Eqns. 14.7), | What limiting value of Mach number is considered?\nWhat limiting value of Reynolds number is considered?\nWhat observation did we make about the boundary layer to start the approximation?\nWhat is the approximate value of $\partial p / \partial y$?\nHow many equations are there? (see worked solutions for a discussion)\nHow many of the equations are second-order? (see discussion in worked solutions)\nHow many unknown, dependent variables are in the equations? See discussion in the worked solutions.\nHow many parameters are in the equations?\nWhat are the physical dimensions ($[M][L][T]$) of the equations as derived in the Lecture? | 9 | 0.333333 | 2 | \n\n\n\n\n\n\n\n | \n\n\n\n\n\n\n\n | In the boundary layer equations (Eqns. 14.7),What limiting value of Mach number is considered?\nWhat limiting value of Reynolds number is considered?\nWhat observation did we make about the boundary layer to start the approximation?\nWhat is the approximate value of $\partial p / \partial y$?\nHow many equations are there? (see worked solutions for a discussion)\nHow many of the equations are second-order? (see discussion in worked solutions)\nHow many unknown, dependent variables are in the equations? See discussion in the worked solutions.\nHow many parameters are in the equations?\nWhat are the physical dimensions ($[M][L][T]$) of the equations as derived in the Lecture? | 96 | 2 | 0 | 0 | 1 | 0 | 90 | 2 | 0 | 14.7,What limiting value of Mach number is considered? What limiting value of Reynolds number is considered? What observation did we make about the boundary layer to start the approximation? What is the approximate value of $\partial p / \partial y$? How many equations are there? see worked solutions for a discussion How many of the equations are second-order? see discussion in worked solutions How many unknown, dependent variables are in the equations? See discussion in the worked solutions. How many parameters are in the equations? What are the physical dimensions $[M][L][T]$ of the equations as derived in the Lecture? | 10 |
49bd88a1-8d46-4ec3-a6c1-65636eed2f3d | 0 | 1 | 0 | 9 | 4 | 2 | 6 | 18 | Establish if by choosing a capacitor $C_1 = 1~\mathrm{mF}$, the circuit below is underdamped or overdamped:
  
  
| The other component values are as follows:
  
$R_1 = 1~\mathrm{k\Omega}$\
$R_2 = 3~\mathrm{k\Omega}$\
$R_3 = 1~\mathrm{k\Omega}$\
$C_2 = 0.5~\mathrm{mF}$\
$C_3 = 3~\mathrm{mF}$
| 1 | 0.666667 | 2 | null | The circuit in the question is a band-pass filter, comprised of an active low-pass filter followed by an active high-pass filter.
***
The total gain of the circuit is the product of the high-pass filter gain and low-pass filter gain (see the lecture notes section 6.5.5). This can be written as follows:
  
$|H| = \frac{R_2}{R_1}\frac{1}{\sqrt{1+(\omega R_2C_1)^2}}\times \frac{C_2}{C_3}\frac{\omega R_3C_3}{\sqrt{1+(\omega R_3C_3)^2}}$
***
This can be re-written in complex form as follows:
  
$|H| = \frac{R_2}{R_1}\frac{1}{1+j\omega R_2C_1}\times \frac{C_2}{C_3}\frac{j\omega R_3C_3}{1+j\omega R_3C_3}$
***
Substituting $s = j\omega$:
  
$|H| = \frac{R_2}{R_1}\frac{1}{1+R_2C_1s}\times\frac{C_2}{C_3}\frac{R_3C_3s}{1+R_3C_3s}$
***
Substituting in component values:
  
$|H| = \frac{3}{1+3s}\times\frac{0.5s}{1+3s}$
***
Simplifying:
  
$|H| = \frac{1.5s}{9s^2+6s+1}$
***
The above is already in canonical form. Find $\omega_\mathrm{n}$:
  
$\frac{1}{\omega_\mathrm{n}^2}=9$
  
$\omega_\mathrm{n} = \sqrt{1}{9} = \frac{1}{3}~\mathrm{rad/s}$
***
Find $\zeta$:
  
$\frac{2\zeta}{\omega_\mathrm{n}} = 6$
  
$\zeta = \frac{6\times\frac{1}{3}}{2} = 1$
***
Hence:
  
The circuit is critically damped.
| Establish if by choosing a capacitor $C_1 = 1~\mathrm{mF}$, the circuit below is underdamped or overdamped:
  
  
The other component values are as follows:
  
$R_1 = 1~\mathrm{k\Omega}$\
$R_2 = 3~\mathrm{k\Omega}$\
$R_3 = 1~\mathrm{k\Omega}$\
$C_2 = 0.5~\mathrm{mF}$\
$C_3 = 3~\mathrm{mF}$
| 35 | 6 | 12 | 12 | 114 | 0 | 17 | 5 | 1 | Establish if by choosing a capacitor $C_1 = 1~\mathrm{mF}$, the circuit below is underdamped or overdamped: The other component values are as follows: $R_1 = 1~\mathrm{k\Omega}$ $R_2 = 3~\mathrm{k\Omega}$ $R_3 = 1~\mathrm{k\Omega}$ $C_2 = 0.5~\mathrm{mF}$ $C_3 = 3~\mathrm{mF}$ | 1 |
49be2315-31e5-477f-aa2d-6b3f71ee2e17 | 4 | 0 | 1 | 18 | 6 | 1 | 1 | 4 | Five street lamps A, B, C, D, and E are located on a straight line along the $x$ axis at an equal distance apart as shown in the figure below. They turn on at times $t_A$, $t_B$, $t_C$, $t_D$, and $t_E$ respectively, in the frame at rest relative to the ground. These five events are indicated in the spacetime diagram below, and a person stands at $x=0$ to watch them turn on.
   
   
A car is moving at velocity $v$ relative to the ground. At $t^\prime = t = 0$, it is at $x^\prime = x = 0$. The space and time axes of the moving frame of the car are also shown in the spacetime diagram. Answer the questions below by drawing the relevant lines and events in the spacetime diagram. | What is the order in which the lamps turn on in the ground rest frame? Rank the events from 1 to 5, with 1 occurring the earliest. If two events are simultaneous give them the same ranking. \nWhat is the order in which the lamps turn on in the car's rest frame?\nWhat is the order in which the light of the lamps reach the person at $x=0$?\nWhat is the order in which the light from the lamps reach the person driving the car?\nWhere is the car relative to the street lights when the light from street lamp D reaches it? | 5 | 0.666667 | 3 | In the rest frame, you can just read the times off the graph like you normally would. \nIt might help to go back at look at the solution to 2.4b
***
   
In the stationary frame, simultaneous events are defined as those which lie on a given line parallel to the $x$ axis. How can you use this definition to identify simultaneous events in the car's rest frame?
***
   
Events which are simultaneous in the car's rest frame will lie a line parallel to the $x'$ axis.\nHow can you draw the world lines of the light from the streetlamps onto the spacetime diagram?
***
   
What points on the graph should the lines representing the light ray start from?
***
   
You are interested in the order in which the world lines of the light rays intersect the $t$ axis. \nYou can answer this question using the same diagram as the previous part.
***
   
The intersection of the light rays with which axis will tell you about the order of events for the person in the car?\nHow can you indicate the point at which the car receives the light from streetlamp on the spacetime diagram?
***
   
Once you know this point, you need to figure out what position this corresponds to on the $x$ (not $x'$) axis...
***
   
You can do this the same way you would noramlly read the $x$ coordinate of a point off a graph.  | \
\nSee part (e)\n\n\n | Five street lamps A, B, C, D, and E are located on a straight line along the $x$ axis at an equal distance apart as shown in the figure below. They turn on at times $t_A$, $t_B$, $t_C$, $t_D$, and $t_E$ respectively, in the frame at rest relative to the ground. These five events are indicated in the spacetime diagram below, and a person stands at $x=0$ to watch them turn on.
   
   
A car is moving at velocity $v$ relative to the ground. At $t^\prime = t = 0$, it is at $x^\prime = x = 0$. The space and time axes of the moving frame of the car are also shown in the spacetime diagram. Answer the questions below by drawing the relevant lines and events in the spacetime diagram.What is the order in which the lamps turn on in the ground rest frame? Rank the events from 1 to 5, with 1 occurring the earliest. If two events are simultaneous give them the same ranking. \nWhat is the order in which the lamps turn on in the car's rest frame?\nWhat is the order in which the light of the lamps reach the person at $x=0$?\nWhat is the order in which the light from the lamps reach the person driving the car?\nWhere is the car relative to the street lights when the light from street lamp D reaches it? | 233 | 11 | 0 | 0 | 4 | 6 | 100 | 1 | 1 | Answer the questions below by drawing the relevant lines and events in the spacetime diagram.What is the order in which the lamps turn on in the ground rest frame? Rank the events from 1 to 5, with 1 occurring the earliest. If two events are simultaneous give them the same ranking. What is the order in which the lamps turn on in the car's rest frame? What is the order in which the light of the lamps reach the person at $x=0$? What is the order in which the light from the lamps reach the person driving the car? Where is the car relative to the street lights when the light from street lamp D reaches it? | 7 |
4a51d339-91dc-4e12-842d-b4e69018c0d9 | 1 | 0 | 0 | 9 | 4 | 2 | 6 | 12 | Determine the time domain response of the closed loop system below when subject to a ramp input $u(t) = t$ (for $t>0$). The transfer function of the process is $H(s) = \frac{3s}{s+4}$.  
| 
| 1 | 0.666667 | 2 | null | Find an expression for $Y(s)$. It can be helpful to label the nodes as in question (5):
  
  
$Y(s) = aH(s)$
  
where $a = U(s)-b$.
***
Find an expression for $b$:
  
$b = aH(s)$
***
Substitute this back into the expression for $a$:
  
$a = U(s)-aH(s)$
***
Make $a$ the subject:
  
$a = \frac{U(s)}{1+H(s)}$
***
Substitute the expression for $a$ back into the expression for $Y(s)$:
  
$Y(s) = \frac{U(s)H(s)}{1+H(s)}$
***
***
Substitute in the expression for $H(s)$ as given in the question:
  
$Y(s)={U(s)}\dfrac{\frac{3s}{s+4}}{1+\frac{3s}{s+4}}$
***
Simplify the fraction:
  
$Y(s)=U(s) \frac{3s}{s+4+3s} = U(s)\frac{3s}{4s+4}$
***
Using the Laplace Transform tables in the Data and Formula book, find an expression for $U(s)$ given that $u(t) = t$:
  
$t^n\rightarrow \frac{n!}{s^{n+1}}$
  
$t \rightarrow \frac{1}{s^2}$
***
Substituting this into the expression for $Y(s)$:
  
$Y(s) = \frac{3s}{s^2(4s+4)}$
***
Simplifying:
 
$Y(s) = \frac{3}{4s(s+1)}= \frac{3}{4}\frac{1}{s(s+1)}$
***
Transforming back to the time domain using the transform tables:
  
$\frac{a}{s(s+a)}\rightarrow 1-e^{-at}$
  
where in this case, $a = 1$.
***
Hence:
  
$y(t) = \frac{3}{4}(1-e^{-t})$
| Determine the time domain response of the closed loop system below when subject to a ramp input $u(t) = t$ (for $t>0$). The transfer function of the process is $H(s) = \frac{3s}{s+4}$.  
| 32 | 3 | 25 | 25 | 154 | 0 | 1 | 0 | 1 | Determine the time domain response of the closed loop system below when subject to a ramp input $u(t) = t$ for $t>0$. | 1 |
4a58c3bc-e81b-4fd8-a811-ca9d3ece0aaa | 2 | 0 | 0 | 9 | 4 | 2 | 6 | 3 | By matching the transfer operators of the op-amp stage and its block diagram representation, derive expressions for A and B.
| 
| 1 | 0.333333 | 1 | null | First write an expression for the op-amp stage. In this case, the op-amp is a summing amp, for which the general expression is:
  
$ v_\mathrm{o} = -R_\mathrm{f}\sum\limits_{n = 1,2,..}^N\frac{v_{\mathrm{i}n}}{R_n} $
***
Applying the expression to this particular op-amp stage:
  
$y = -R_\mathrm{f}[-\frac{x_1}{R_1}+\frac{x_2}{R_2}]$
***
Now write an expression using the block diagram:
  
$y = A(x_1-Bx_2)$
***
The coefficients for $x_1$ and $x_2$ from the two expressions can now be equated.
***
For $x_1$:
  
$\frac{R_\mathrm{f}}{R_1} = A$
***
For $x_2$:
  
$-\frac{R_f}{R_2} = -AB$
***
Substituting in the expression for $A$ and rearranging to find $B$:
  
$B = \frac{R_1}{R_2}$
| By matching the transfer operators of the op-amp stage and its block diagram representation, derive expressions for A and B.
| 21 | 0 | 12 | 12 | 89 | 0 | 1 | 0 | 1 | By matching the transfer operators of the op-amp stage and its block diagram representation, derive expressions for A and B. | 1 |
4a8f7227-9e61-4234-8426-b9843d27761d | 3 | 0 | 0 | 15 | 5 | 0 | 3 | 4 | There is a relationship between the size of an island (or other defined area) and the number of species it contains. This relationship is called (surprisingly enough) the species/area relationship and it is modelled by the following equation:
$$
S=C\cdot A^{z}
$$
* $S$ Number of species
* $A$ Area of the island
* $C$ Data specific constant
* $z$ Data specific constant
| Using logs, convert this equation into a linear form showing clearly what the gradient and intercept represent, and what you would need to plot to calculate them. Write $\log(x)$ as 'log(x)'. n this question it doesn't matter what base you use at all.
\nCalculate the values of $C$ and $z$ from this data set using linear regression (use Excel if you want). 
| **Island** | **Area of island (**$\mathrm{km^2}$**)** | **Number of (non-bat) mammal species** |
| :--------- | :--------------------------------------- | :------------------------------------- |
| Jersey | 116.3 | 9 |
| Guernsey | 63.5 | 5 |
| Alderney | 7.9 | 3 |
| Sark | 5.2 | 2 |
| Herm | 1.3 | 2 |
\n(Continue from last part...) Calculate the values of $C$ and $z$ from this data set using linear regression (use Excel if you want).
| **Island** | **Area of island (km2)** | **Number of (non-bat) mammal species** |
| :--------- | :----------------------- | :------------------------------------- |
| Jersey | 116.3 | 9 |
| Guernsey | 63.5 | 5 |
| Alderney | 7.9 | 3 |
| Sark | 5.2 | 2 |
| Herm | 1.3 | 2 |
| 3 | 1 | 1 | \n\n | Gradient : $z$
Y-intercept : $\log{C}$
Plot $\log{S}$ (Y-axis) against $\log{A}$ (X-axis)
$$
\log S=\log C+z \log A \newline Y=c+mX
$$
\nThe $Y$-intercept is $\log_{10}(C)=0.178$, so $C=10^{0.178} = 1.5$ . If you use a different base, you'll need a different antilog, but the result will be the same, *e.g.* if you use natural logs, you'll get $Y$-intercept is $\ln(C)=0.410$, so $C=e^{0.410}=1.5$. 
\nThe slope is $z$, so $z=0.329$
| There is a relationship between the size of an island (or other defined area) and the number of species it contains. This relationship is called (surprisingly enough) the species/area relationship and it is modelled by the following equation:
$$
S=C\cdot A^{z}
$$
* $S$ Number of species
* $A$ Area of the island
* $C$ Data specific constant
* $z$ Data specific constant
Using logs, convert this equation into a linear form showing clearly what the gradient and intercept represent, and what you would need to plot to calculate them. Write $\log(x)$ as 'log(x)'. n this question it doesn't matter what base you use at all.
\nCalculate the values of $C$ and $z$ from this data set using linear regression (use Excel if you want). 
| **Island** | **Area of island (**$\mathrm{km^2}$**)** | **Number of (non-bat) mammal species** |
| :--------- | :--------------------------------------- | :------------------------------------- |
| Jersey | 116.3 | 9 |
| Guernsey | 63.5 | 5 |
| Alderney | 7.9 | 3 |
| Sark | 5.2 | 2 |
| Herm | 1.3 | 2 |
\n(Continue from last part...) Calculate the values of $C$ and $z$ from this data set using linear regression (use Excel if you want).
| **Island** | **Area of island (km2)** | **Number of (non-bat) mammal species** |
| :--------- | :----------------------- | :------------------------------------- |
| Jersey | 116.3 | 9 |
| Guernsey | 63.5 | 5 |
| Alderney | 7.9 | 3 |
| Sark | 5.2 | 2 |
| Herm | 1.3 | 2 |
| 265 | 11 | 13 | 13 | 64 | 0 | 199 | 6 | 0 | This relationship is called surprisingly enough the species/area relationship and it is modelled by the following equation: $ S=C\cdot A^{z} $ $S$ Number of species $A$ Area of the island $C$ Data specific constant $z$ Data specific constant Using logs, convert this equation into a linear form showing clearly what the gradient and intercept represent, and what you would need to plot to calculate them. Write $\log(x)$ as 'logx'. n this question it doesn't matter what base you use at all. Calculate the values of $C$ and $z$ from this data set using linear regression use Excel if you want. | 4 |
4b789b57-bdcc-4255-833e-d051b488dd5e | 2 | 0 | 0 | 15 | 5 | 0 | 3 | 0 | Most cells produce thousands of different types of mRNA. Some mRNAs are more stable than others. The mRNAs for most proteins possess a ‘life-preserving’ poly-A tail, but mRNAs for the proteins called histones lack this tail and are consequently much shorter-lived than most mRNAs. The total amount of mRNA in a culture of rapidly dividing cells increases exponentially:
$$
M=M_{0}e^{kt}
$$
* $M$ Mass of mRNA
<!---->
* $M_{0}$ Initial mass of mRNA
<!---->
* $k$ Cell growth rate
<!---->
* $t$ time
| Use (natural) logs to derive an equation of the form $y=m \cdot x+c$ . This allows you to calculate the cell growth rate, $k$ , from a graph. (In lambda, you can use $\ln$ or $\log$ to represent natural log: both will work and are treated as synonymous).
\nIf the mass of mRNA quadruples in half an hour, what is the cell growth rate (in units of $\mathrm{min^{-1}}$)?
| 2 | 1 | 0 | \n | Start with
$$
M=M_0 e^{kt}
$$
Take natural logs:
$$
\ln(M)=\ln(M_0 e^{kt})
$$
Expand product in RHS using equivalency $\log(xy)=\log(x)+\log(y)$:
$$
\ln(M)=\ln(M_0)+\ln(e^{kt})
$$
Expand power in RHS using equivalency $\log(x^y)=y\log(x)$:
$$
\ln(M)=\ln(M_0)+kt\ln(e)
$$
Simplify product in RHS using identity $\log_b(b)=1$ (*i.e.* $\ln(e)=\log_e(e)=1$):
$$
\ln{M}=\ln{M_0}+kt\times1
$$
Which is simply:
$$
\ln{M}=\ln{M_0}+kt
$$
\n$$
k=\frac{\ln{M}-\ln{M_0}}{t}=\frac{\ln{(M/M_0)}}{t}=\frac{\ln{4}}{t}=\frac{1.386}{30\,\mathrm{min}}=0.0462\,\mathrm{min^{-1}}
$$
| Most cells produce thousands of different types of mRNA. Some mRNAs are more stable than others. The mRNAs for most proteins possess a ‘life-preserving’ poly-A tail, but mRNAs for the proteins called histones lack this tail and are consequently much shorter-lived than most mRNAs. The total amount of mRNA in a culture of rapidly dividing cells increases exponentially:
$$
M=M_{0}e^{kt}
$$
* $M$ Mass of mRNA
<!---->
* $M_{0}$ Initial mass of mRNA
<!---->
* $k$ Cell growth rate
<!---->
* $t$ time
Use (natural) logs to derive an equation of the form $y=m \cdot x+c$ . This allows you to calculate the cell growth rate, $k$ , from a graph. (In lambda, you can use $\ln$ or $\log$ to represent natural log: both will work and are treated as synonymous).
\nIf the mass of mRNA quadruples in half an hour, what is the cell growth rate (in units of $\mathrm{min^{-1}}$)?
| 148 | 10 | 11 | 11 | 42 | 0 | 67 | 5 | 3 | The total amount of mRNA in a culture of rapidly dividing cells increases exponentially: $ M=M_{0}e^{kt} $ $M$ Mass of mRNA embedded $M_{0}$ Initial mass of mRNA embedded $k$ Cell growth rate embedded $t$ time Use natural logs to derive an equation of the form $y=m \cdot x+c$ . This allows you to calculate the cell growth rate, $k$ , from a graph. In lambda, you can use $\ln$ or $\log$ to represent natural log: both will work and are treated as synonymous. If the mass of mRNA quadruples in half an hour, what is the cell growth rate in units of $\mathrm{min^{-1}}$? | 4 |
4b9be5bc-9941-404b-a653-681eec1b5108 | 1 | 0 | 0 | 24 | 6 | 1 | 0 | 13 | Consider the integral
$$
\ {\cal I} = \iint_R 8xy\left(x^2-y^2\right) e^{x^2 + y^2}\,dx\,dy
$$
where $R$ is the wedge of the circle bounded by $y=0$, $y=x$, and $x^2 + y^2 = 1$. Change to new variables $(u,v)$ given by $u=x^2 - y^2$, $v=x^2 + y^2$ and show that the integral becomes:
$$
{\cal I} = \int_{u=0}^1 \int_{v=u}^1 ue^v\,dv\,du
$$
Hence evaluate ${\cal I}$. 
***
*Hint*: When finding the Jacobian, you might find it useful to remember that: 
$$
\frac{\partial(u,v)}{\partial(x,y)} = 1/ \frac{\partial(x,y)}{\partial(u,v)}
$$
| Consider the integral
$$
\ {\cal I} = \iint_R 8xy\left(x^2-y^2\right) e^{x^2 + y^2}\,dx\,dy
$$
where $R$ is the wedge of the circle bounded by $y=0$, $y=x$, and $x^2 + y^2 = 1$. Change to new variables $(u,v)$ given by $u=x^2 - y^2$, $v=x^2 + y^2$ and show that the integral becomes:
$$
{\cal I} = \int_{u=0}^1 \int_{v=u}^1 ue^v\,dv\,du
$$
Hence evaluate ${\cal I}$. 
***
*Hint*: When finding the Jacobian, you might find it useful to remember that: 
$$
\frac{\partial(u,v)}{\partial(x,y)} = 1/ \frac{\partial(x,y)}{\partial(u,v)}
$$
| 1 | 0.666667 | 2 | Changing variables in integration involves the following:
1. Changing limits
2. Changing the integrand
3. Changing the differential using the Jacobian. 
***
(Step 1) Sketch the region in the $xy$ plane and identify 3 boundaries that describe the region...
***
(Step 1) ... Change these boundaries into $(u,v)$ form. Hence sketch the new region of integration in the $uv$ plane and determine the new limits.
***
(Step 3) ... The Jacobian is given by:
$$
|J|=\begin{vmatrix}\dfrac{\partial x}{\partial u} & \dfrac{\partial y}{\partial u}\\[1em] \dfrac{\partial{x}}{\partial v}&\dfrac{\partial y}{\partial v} \end{vmatrix}
$$
But, you should use the hint in the question (i.e., take 1 over the reciprocal of the Jacobian and change it to be in terms of $u$ and $v$). This makes the Jacobian easier to analyse as $u=u(x,y)$ and $v=v(x,y)$. 
***
You should find that the Jacobian cancels with one part of the integrand. Convert the remaining $(x,y)$ parts of the integrand into $(u,v)$ form.
| The region of integration in the $xy$ pane is:
***
Changing each limit into the new variables, starting with $y=x$:
***
$$
\underline{y=x}\implies u=0
$$
Next, $y=0$:
***
$$
\underline{y=0}\implies u=v
$$
Finally, $x^2+y^2=1$:
***
$$
\underline{(x^2+y^2)=1}\implies v=1
$$
The region of integration in the $uv$ plane is therefore:
***
The limits of the integral are:
$$
\begin{aligned}
&u: 0\to 1\\
&v: u\to 1
\end{aligned}
$$
***
Next, the differential $dx \, dy$ becomes:
$$
dx\, dy= |J|du\,dv
$$
***
***
$$
|J|=\begin{vmatrix}\dfrac{\partial x}{\partial u} & \dfrac{\partial y}{\partial u}\\[1em] \dfrac{\partial{x}}{\partial v}&\dfrac{\partial y}{\partial v} \end{vmatrix}
$$
However, we are given $u=u(x,y)$ and $v=v(x,y)$, so it is easier to take the reciprocal (as indicated in the hint):
***
$$
|J|=\frac{1}{|J'|} , \quad |J'|=\begin{vmatrix}\dfrac{\partial u}{\partial x} & \dfrac{\partial v}{\partial x}\\[1em] \dfrac{\partial{u}}{\partial y}&\dfrac{\partial v}{\partial y} \end{vmatrix}
$$
Evaluating $|J'|$:
***
$$
|J'| = \begin{vmatrix}2x & 2x\\-2y&2y \end{vmatrix} = 4xy+4xy=8xy
$$
Therefore,
$$
|J| = \frac{1}{8xy}
$$
***
Changing the integration variables (but leaving $8xy$ unchanged since it cancels with the Jacobian) and changing limits:
***
$$
\begin{aligned}\mathcal{I} &= \int_{u=0}^{1}\int_{v=u}^{1}8xy(u)e^v\frac{1}{8xy}dv\,du \\
&=\int_{u=0}^{1}\int_{v=u}^{1}{ue^v dv\,du}
\end{aligned}
$$
***
$$
\begin{aligned}
&=\int_{u=0}^{1}{u[e^v]_{v=u}^{1}}\,du\\
&=\int_{u=0}^{1}{u(e-e^u)}\,du\\
&=\left[\frac{eu^2}{2}\right]_0^1 - \int_{u=0}^{1}{ue^u\,du}
\end{aligned}
$$
Performing the remaining integral by parts:
***
$$
\begin{aligned}
&\cal{I}= \frac{e}{2}-[ue^u]^1_0 + \int_{0}^{1}e^u\,du \\
&= \frac{e}{2}-e+e-1\\
& = \frac{e}{2}-1
\end{aligned}
$$
| Consider the integral
$$
\ {\cal I} = \iint_R 8xy\left(x^2-y^2\right) e^{x^2 + y^2}\,dx\,dy
$$
where $R$ is the wedge of the circle bounded by $y=0$, $y=x$, and $x^2 + y^2 = 1$. Change to new variables $(u,v)$ given by $u=x^2 - y^2$, $v=x^2 + y^2$ and show that the integral becomes:
$$
{\cal I} = \int_{u=0}^1 \int_{v=u}^1 ue^v\,dv\,du
$$
Hence evaluate ${\cal I}$. 
***
*Hint*: When finding the Jacobian, you might find it useful to remember that: 
$$
\frac{\partial(u,v)}{\partial(x,y)} = 1/ \frac{\partial(x,y)}{\partial(u,v)}
$$
| 57 | 11 | 22 | 22 | 121 | 10 | 57 | 11 | 0 | Consider the integral $ \ {\cal I} = \iint_R 8xy\left(x^2-y^2\right) e^{x^2 + y^2}\,dx\,dy $ where $R$ is the wedge of the circle bounded by $y=0$, $y=x$, and $x^2 + y^2 = 1$. Change to new variables $(u,v)$ given by $u=x^2 - y^2$, $v=x^2 + y^2$ and show that the integral becomes: $ {\cal I} = \int_{u=0}^1 \int_{v=u}^1 ue^v\,dv\,du $ Hence evaluate ${\cal I}$. Hint: When finding the Jacobian, you might find it useful to remember that: $R$0 | 3 |
4c553593-e1b6-4c58-8d58-789fb45a411e | 0 | 0 | 2 | 24 | 6 | 1 | 3 | 6 | For $\Omega = 1/r$ where $r$ is the distance from the origin:
| Show that $\nabla\Omega = -\dfrac{1}{r^2}\mathbf{\hat{r}}$.
\nShow that $\nabla^2 \Omega = 0$
| 2 | 0.333333 | 1 | The gradient in spherical coordinates, ignoring $\theta$ and $\phi$ components is:
$$
\nabla\Omega = \frac{\partial \Omega}{\partial r}\mathbf{\hat{r}} + \frac{1}{r}\frac{\partial \Omega}{\partial \theta}\mathbf{\hat{\theta}} + \frac{1}{r\sin\theta}\frac{\partial \Omega}{\partial \phi}\mathbf{\hat{\phi}}
$$
***
What components of the gradient can you ignore?
\n$$
\nabla^2\Omega = \nabla\cdot(\nabla\Omega)
$$
***
The divergence of a vector field $\vec{A}$ in spherical coordinates (considering the $r$ component only for this case) is:
$$
\nabla\cdot\vec{A} = \frac{1}{r^2}\frac{\partial(r^2 A_r)}{\partial r}
$$
***
Replace $\vec{A}$ with $\nabla\Omega$. 
| $$
\nabla\Omega = \frac{\partial \Omega}{\partial r}\mathbf{\hat{r}} + \frac{1}{r}\frac{\partial \Omega}{\partial \theta}\mathbf{\hat{\theta}} + \frac{1}{r\sin\theta}\frac{\partial \Omega}{\partial \phi}\mathbf{\hat{\phi}}
$$
***
The $\mathbf{\hat{\theta}}$ and $\mathbf{\hat{\phi}}$ are zero since $\Omega=\Omega(r)$.
***
$$
\frac{\partial\Omega}{\partial r} =-\frac{1}{r^2}
$$
So,
$$
\nabla\Omega = -\frac{1}{r^2}\mathbf{\hat{r}}
$$
\nThe Laplacian $\nabla^2\Omega$ is given by:
$$
\nabla^2\Omega = \nabla\cdot(\nabla\Omega)
$$
***
The divergence of a vector field $\vec{A}$ in spherical coordinates (considering the $r$ component only for this case) is:
$$
\nabla\cdot\vec{A} = \frac{1}{r^2}\frac{\partial(r^2 A_r)}{\partial r}
$$
***
Replacing $\vec{A}$ with $\nabla\Omega$: 
***
$$
\nabla\cdot{\nabla\Omega} = \frac{1}{r^2} \frac{\partial}{\partial r}\left(r^2\left(-\frac{1}{r^2}\right)\right) = \frac{1}{r^2}\frac{\partial}{\partial r}(-1)=0
$$
| For $\Omega = 1/r$ where $r$ is the distance from the origin:
Show that $\nabla\Omega = -\dfrac{1}{r^2}\mathbf{\hat{r}}$.
\nShow that $\nabla^2 \Omega = 0$
| 17 | 4 | 14 | 14 | 51 | 9 | 7 | 2 | 0 | Show that $ abla^2 \Omega = 0$ | 1 |
4c57560c-2216-42ec-bdce-938f4ac71be0 | 0 | 0 | 1 | 4 | 3 | 3 | 1 | 10 | Construct the integral solution to the 1-D diffusion equation of a contaminant with initial concentration
$$
C_0 =
\begin{cases}
1 & \text{if }x<0 \\
0 & \text{if }x\geq 0\,.
\end{cases}
$$
Starting from the fundamental solution for an instantaneous point release, show that the integral can be evaluated to give
$$
C = \dfrac{1}{2} \text{erfc}\hspace*{-1pt}\left( \dfrac{x}{\sqrt{2}\sigma} \right).
$$
Hint: make a change of variables $\xi=(x-x')/\sqrt{2}\sigma$ to solve the integral.
| Construct the integral solution to the 1-D diffusion equation of a contaminant with initial concentration
$$
C_0 =
\begin{cases}
1 & \text{if }x<0 \\
0 & \text{if }x\geq 0\,.
\end{cases}
$$
Starting from the fundamental solution for an instantaneous point release, show that the integral can be evaluated to give
$$
C = \dfrac{1}{2} \text{erfc}\hspace*{-1pt}\left( \dfrac{x}{\sqrt{2}\sigma} \right).
$$
Hint: make a change of variables $\xi=(x-x')/\sqrt{2}\sigma$ to solve the integral.
| 1 | 1 | 2 | null | We approach this problem by building on the fundamental solution for 1-D diffusion in an initially unpolluted environment (i.e. the scenario where $C(x, t = 0) = 0$). It is necessary to account for the non-zero concentration for $x < 0$. Given that we consider diffusion in an unbounded environment, the initial concentration of pollutant $(C_0 = 1)$ extends (from the source at $x = 0$) to negative infinity.
***
A thin slice of the initial contaminant of concentration $C_0 = 1$ and thickness $\delta x'$ at a position $x' (< x)$ from the source at $x = 0$ may be treated as a "source" of strength
$$
M = C_0\delta x'.
$$
***
From the 1-D diffusion solution for an instantaneous point release, the concentration distribution resulting from this thin source of contaminant at $x'$ of strength $M$ is given by
$$
\delta C(x, t) = \dfrac{M}{\sqrt{2\pi} \sigma} e^{-\dfrac{(x - x')^2}{2\sigma^2}} = \dfrac{C_0}{\sqrt{2\pi} \sigma} e^{-\dfrac{(x - x')^2}{2\sigma^2}} \delta x'.
$$
Note that we have used a coordinate shift $(x^* = x - x')$ in order to account for the contaminant source (i.e. the thin layer) being a distance $x'$ away from the origin.
***
In order to obtain the total concentration distribution, we sum the contributions from each thin source of contaminant of concentration $C_0 = 1$. Integrating over all the thin layers of simple sources from $x = 0$ to $x = -\infty$ (i.e. integrating both sides of $\delta C(x, t) = \frac{C_0}{\sqrt{2\pi} \sigma} e^{-\frac{(x - x')^2}{2\sigma^2}} \delta x'$) yields
$$
\displaystyle C(x, t) = \int^{0}_{-\infty}{\dfrac{C_0}{\sqrt{2\pi} \sigma} e^{-\dfrac{(x - x')^2}{2\sigma^2}} \, \mathrm{d}x'}.
$$
***
We invoke a change of variables from $x'$ to $\xi$ using the substitution $\xi = (x - x')/\sqrt{2} \sigma$. Accounting for the change in the upper and lower limits of the integration, the equation $C(x, t) = \int^{0}_{-\infty}{\frac{C_0}{\sqrt{2\pi} \sigma} e^{-\frac{(x - x')^2}{2\sigma^2}} \, \mathrm{d}x'}$ then becomes
$$
\displaystyle C = \dfrac{C_0}{\sqrt{\pi}}\int^{\infty}_{x/\sqrt{2}\sigma}{e^{-\xi^2} \, \mathrm{d}\xi}.
$$
***
The error function is defined as
$$
\displaystyle \mathrm{erf}(x) = \dfrac{2}{\sqrt{\pi}}\int^{x}_{0}{e^{-r^2} \, \mathrm{d}r},
$$
where $r$ is a dummy variable. Noting that
$$
\displaystyle \int_{a}^{\infty}f(\xi)\mathrm{d}\xi =\int_{0}^{\infty}f(\xi)\mathrm{d}\xi-\int_{0}^{a}f(\xi)\mathrm{d}\xi,
$$
***
the solution to $C = \frac{C_0}{\sqrt{\pi}}\int^{\infty}_{x/\sqrt{2}\sigma}{e^{-\xi^2} \, \mathrm{d}\xi}$ is
$$
C(x, t) = \dfrac{C_0}{2} \Bigg\lbrace{\mathrm{erf}\left(\infty\right) - \mathrm{erf}\left(\dfrac{x}{\sqrt{2} \sigma}\right)\Bigg\rbrace}.
$$
***
Given that $\mathrm{erf}(\theta) \rightarrow 1$ as $\theta \rightarrow \infty$, and noting that $C_0 = 1$ in this problem,
$$
C(x, t) = \dfrac{1}{2}\mathrm{erfc}\left(\dfrac{x}{\sqrt{2} \sigma}\right),
$$
where we used in the last step that $\mathrm{erfc}(\theta) = 1 - \mathrm{erf}(\theta)$.
| Construct the integral solution to the 1-D diffusion equation of a contaminant with initial concentration
$$
C_0 =
\begin{cases}
1 & \text{if }x<0 \\
0 & \text{if }x\geq 0\,.
\end{cases}
$$
Starting from the fundamental solution for an instantaneous point release, show that the integral can be evaluated to give
$$
C = \dfrac{1}{2} \text{erfc}\hspace*{-1pt}\left( \dfrac{x}{\sqrt{2}\sigma} \right).
$$
Hint: make a change of variables $\xi=(x-x')/\sqrt{2}\sigma$ to solve the integral.
| 47 | 3 | 34 | 34 | 281 | 0 | 47 | 3 | 0 | Construct the integral solution to the 1-D diffusion equation of a contaminant with initial concentration $ C_0 = \begin{cases} 1 & \text{if }x<0 \\ 0 & \text{if }x\geq 0\,. \end{cases} $ Starting from the fundamental solution for an instantaneous point release, show that the integral can be evaluated to give $ C = \dfrac{1}{2} \text{erfc}\hspace*{-1pt}\left( \dfrac{x}{\sqrt{2}\sigma} \right). $ Hint: make a change of variables $\xi=(x-x')/\sqrt{2}\sigma$ to solve the integral. | 3 |
4c93a2e7-0473-454e-bd2f-9deaa7b505b0 | 2 | 0 | 3 | 14 | 4 | 2 | 11 | 1 | Couette flows are the archetype of lubrication-related issues. This problem investigates wall-temperature effects on the friction force. Two infinitely-long and infinitely-wide parallel impermeable walls are separated by a distance $h$ (see the image below). The bottom wall (located at $y = 0$ in the Cartesian frame of reference shown is stationary whilst the top wall (located at $y=h$) moves with constant and uniform velocity $U_0 \hat{e}_x$. A liquid bath assumed to be Newtonian with constant and uniform density $\rho_0$ separates both walls. No pressure gradient is applied in the $x$ direction. The flow is assumed to be steady, fully-developed in $x$ and $z$ with no flow in $z$, i.e. $w=0$. No body force is applied to this flow (it is simply driven by the motion of the wall). The temperature of the sliding wall can be made colder or warmer than the bottom wall, thus inducing a temperature gradient in the bath. The dynamic viscosity $\mu$ of the liquid depends on the temperature and is assumed to vary linearly across the bath:
$$
\mu(y)=\mu_0\left(1+\beta \frac{y}{h} \right)
$$
where $\beta$ is a constant parameter.
| Choose the appropriate set of equations to solve the problem and show that the component of the velocity normal to the wall is strictly zero, i.e. $v=0$.\nLet $\underline{\underline{\tau}}$ be the viscous stress tensor. Show that $\tau_{xy}$ is the only component of the viscous stress tensor different from zero and that $\tau_{xy}=\mu \mathrm{d} u/\mathrm{d} y$.\nShow that the pressure gradient is equal to the null vector throughout the flow.\nFind an expression for the $x$ component of the velocity.\nFind an expression for the shear stress on the top wall ($y=h$) and discuss how the shear stress changes as a function of $\beta$. | 5 | 1 | 4 | \n\n\n\n | \n\n\n\n | Couette flows are the archetype of lubrication-related issues. This problem investigates wall-temperature effects on the friction force. Two infinitely-long and infinitely-wide parallel impermeable walls are separated by a distance $h$ (see the image below). The bottom wall (located at $y = 0$ in the Cartesian frame of reference shown is stationary whilst the top wall (located at $y=h$) moves with constant and uniform velocity $U_0 \hat{e}_x$. A liquid bath assumed to be Newtonian with constant and uniform density $\rho_0$ separates both walls. No pressure gradient is applied in the $x$ direction. The flow is assumed to be steady, fully-developed in $x$ and $z$ with no flow in $z$, i.e. $w=0$. No body force is applied to this flow (it is simply driven by the motion of the wall). The temperature of the sliding wall can be made colder or warmer than the bottom wall, thus inducing a temperature gradient in the bath. The dynamic viscosity $\mu$ of the liquid depends on the temperature and is assumed to vary linearly across the bath:
$$
\mu(y)=\mu_0\left(1+\beta \frac{y}{h} \right)
$$
where $\beta$ is a constant parameter.
Choose the appropriate set of equations to solve the problem and show that the component of the velocity normal to the wall is strictly zero, i.e. $v=0$.\nLet $\underline{\underline{\tau}}$ be the viscous stress tensor. Show that $\tau_{xy}$ is the only component of the viscous stress tensor different from zero and that $\tau_{xy}=\mu \mathrm{d} u/\mathrm{d} y$.\nShow that the pressure gradient is equal to the null vector throughout the flow.\nFind an expression for the $x$ component of the velocity.\nFind an expression for the shear stress on the top wall ($y=h$) and discuss how the shear stress changes as a function of $\beta$. | 282 | 20 | 0 | 0 | 1 | 0 | 101 | 7 | 1 | A liquid bath assumed to be Newtonian with constant and uniform density $\rho_0$ separates both walls. No pressure gradient is applied in the $x$ direction. The flow is assumed to be steady, fully-developed in $x$ and $z$ with no flow in $z$, i.e. No body force is applied to this flow it is simply driven by the motion of the wall. The dynamic viscosity $y = 0$0 of the liquid depends on the temperature and is assumed to vary linearly across the bath: $y = 0$1 where $y = 0$2 is a constant parameter. Choose the appropriate set of equations to solve the problem and show that the component of the velocity normal to the wall is strictly zero, i.e. Let $y = 0$4 be the viscous stress tensor. Show that $y = 0$5 is the only component of the viscous stress tensor different from zero and that $y = 0$6. Show that the pressure gradient is equal to the null vector throughout the flow. Find an expression for the $x$ component of the velocity. Find an expression for the shear stress on the top wall $y=h$ and discuss how the shear stress changes as a function of $y = 0$2. | 11 |
4cc77538-a3b5-4773-80ca-bf9ede89dc4e | 6 | 0 | 1 | 21 | 6 | 1 | 1 | 1 | Here we will look at how the smoothness of a function affects its Fourier series.
| Find the trigonometric Fourier series for the following three functions.
**Note:** The last function is constructed from two parabolas stuck together. They join at $\pi/2+n\pi$, where $n$ is an integer.
The functions are referred to as (**Square**), (**Triangle**) and (**Parabola**) in the solutions.
\nHow quickly a Fourier series converges depends on the properties of the function. In general the terms in the series will be proportional to:
$$
\frac{1}{n^p}
$$
in the limit of large $n$. For the functions in part (a), identify $p$.
\nTry to identify the relationship between $p$ and which derivative of the function is discontinuous.
| 3 | 0.666667 | 4 | For all of the functions, start by considering whether $f(x)$ is odd or even about $x=0$. This will allow you to set either $a_n$ or $b_n$ to 0 (see **section 2.2**). 
***
Then, for the non-zero term, you may be able to find that it is zero for odd or even $n$. Start by reducing the integration interval to $[0,\pi]$.
***
... To do this, consider if $f(x)$ is odd/even about $x=\pi/2$ for odd/even $n$. 
***
After performing these checks, it will be much easier to evaluate the Fourier coefficients, as less integrals will be required. 
***
(**Triangle Function**): By using the properties of integrals of odd/even functions, you should be able to reduce your integration limits to the interval $[0,\pi]$. This is useful because then you will not have to define $f(x)$ in piecewise terms. 
***
(**Triangle Function**): Evaluate the integral by parts.
***
(**Parabola Function**): By using the properties of integral of odd/even functions, you should be able to reduce your integration limits to the interval $[0,\pi/2]$. This is useful because you will not have to define $f(x)$ in piecewise terms. 
***
(**Parabola Function**): Evaluate the integral by parts twice (see *Problem Sheet 1 Question 4* for a detailed breakdown of this). 
\nLook at each Fourier series in part (a). What is the power of $n$?
\nLook at each of the functions. What derivative of the function is discontinuous? Link this to the value of $p$ found in part (b).
| Worked solutions for each function. 
, First, we consider the even/odd nature of $f(x)$ to determine if $a_n$ and $b_n$ can be set to 0 and for which values of $n$.
***
$f(x)$ is even, so $f(x)\sin(nx)$ is odd. 
***
This means that $b_n=0$ for all $n$. We also see that the average is zero, so $a_0=0$ (you can check this by integrating). Now, we find $a_n$. 
***
Around $x=\pi/2$:
* The function is odd. 
* $\cos(nx)$ is even for even $n$, and odd for even $n$. 
***
This means that $a_n=0$ for even $n$, since $f(x)\cos(nx)$ is odd. It remains to evaluate $a_n$ for odd $n$.
***
Firstly, we can reduce the interval of the integral: 
$$
a_n = \frac{1}{\pi}\int^{\pi}_{-\pi}f(x)\cos(nx)\,\text{d}x = \frac{2}{\pi}\int^{\pi}_{0}f(x)\cos(nx)\,\text{d}x = \frac{4}{\pi}\int_{0}^{\pi/2}f(x)\cos(nx)\,\text{d}x
$$
The final step is acceptable because $f(x)\cos(nx)$ is even about $x=\pi/2$ for odd $n$. The advantage of this is that we do not need to find $f(x)$ as a *piecewise function*: we only need to consider its form in the region $[0,\pi/2]$. Looking at the graph, we can find $f(x)$ in this interval:
***
$$
f(x) = \frac{4}{\pi^2}{x^2-1}
$$
Therefore, evaluating $a_n$:
***
$$
a_n = \frac{1}{\pi}\int_{-\pi}^\pi{f(x) \cos (nx) \text{d}x}
= \frac{4}{\pi}\int_{0}^{\pi/2}{\left(\frac{4}{\pi^2}x^2-1\right)\cos (nx) \text{d}x}
$$
To evaluate this integral, we perform integration by parts *twice*. For a step-by-step process for integrating by parts twice, refer to *Problem Sheet 1, Question 4*. We obtain:
***
$$
a_n=\frac{32}{\pi^3n^3}(-1)^{(n+1)/2}
$$
***
Therefore, 
$$
f(x) = \sum_{n=1,3,5\cdots}^{\infty}{\frac{32}{\pi^3n^3}(-1)^{(n+1)/2}\cos(nx)} = \frac{32}{\pi^3}\left( - \frac{\cos x}{1} + \frac{\cos 3x}{3^3} - \frac{\cos 5x}{5^3} + \cdots \right)
$$
, First, we consider the even/odd nature of $f(x)$ to determine if $a_n$ and $b_n$ can be set to 0, and for which values of $n$.
***
Firstly, the function is odd about $x=0$. 
***
This means that $f(x)\cos(nx)$ is odd, and so $a_n=0$ (integral of an odd function is $0$). $a_0$ and $b_n$ are to be determined:
***
$$
a_0 = \frac{1}{\pi}\int^{\pi}_{-\pi}\text{d}x = 1
$$
Notice that $a_n=0$, but $a_0$ is non-zero. This is why it is important to evaluate $a_0$ independently.
***
Next, consider $b_n$. Since $f(x)\sin(nx)$ is even, we can reduce the interval of integration to $[0,\pi]$:
***
$$
b_n = \frac{1}{\pi}\int^{\pi}_{-\pi}{f(x)\sin(nx)~\text{d}x} = \frac{2}{\pi}\int_{0}^{\pi}{f(x)\sin(nx)~dx}
$$
***
Before evaluating the integral, it is best to see if odd/even values of $n$ cause the integral to vanish. 
***
About the half-way point of the interval, $x=\pi/2$, 
* $f(x)$ is even.
*  $\sin(nx)$ is odd for even $n$, and even for odd $n$ (see graphic below).
  
***
This means that $f(x)\sin(nx)$ is odd for even $n$, and so the integral evaluates to 0 for even $n$. This leaves us to find $b_n$ for odd $n$ only:
***
$$
b_n = \frac{1}{\pi}\int_{-\pi}^\pi{f(x) \sin (nx) \text{d}x}
= \frac{1}{\pi}\int_{0}^\pi{\sin (nx) \text{d}x} = \frac{2}{n\pi} \; .
$$
***
And so for $f(x)$ we have:
$$
f(x)= \frac{1}{2} +\frac{2}{\pi}\sum_{n=1,3,5,\cdots}^{\infty}\frac{\sin(nx)}{n}= \frac{1}{2} + \frac{2}{\pi}\left( \frac{\sin x}{1} + \frac{\sin 3x}{3} + \frac{\sin 5x}{5} + \cdots \right)
$$
, First, we consider the even/odd nature of $f(x)$ to determine if $a_n$ and $b_n$ can be set to 0 and for which values of $n$.
***
The function is even about $x=0$, so $f(x)\sin(nx)$ is odd. 
***
Therefore, all $b_n$ terms are zero.
***
We also see that the average is zero, so $a_0=0$.
***
Now considering $a_n$. Since $f(x)\cos(nx)$ is even, we can halve the interval and double the integral due to symmetry:
***
$$
a_n = \frac{2}{\pi}\int^{\pi}_{0}f(x)\cos(nx)dx
$$
***
About $x=\pi/2$:
* The function is odd.
*  $\cos(nx)$ is even for even $n$, and odd for odd $n$.
***
This means that $a_n$ is 0 for even $n$, and non-zero for odd $n$. Evaluating the integral for odd terms, we first need to determine $f(x)$ in the interval $[0,\pi]$. Looking at the graph, we see that $f(x)$ is given by:
***
$$
f(x) = \frac{2x}{\pi} -1
$$
***
$$
a_n
= \frac{2}{\pi}\int_{0}^\pi{\left(\frac{2x}{\pi}-1\right)\cos (nx) ~\text{d}x}
$$
Evaluating this integral by parts:
***
$$
\begin{aligned}
&\int\left(\frac{2x}{\pi}-1\right)\cos(nx)\,\text{d}x=\\
&= \frac{1}{n}\left(\frac{2x}{\pi}-1\right)\sin(nx) - \frac{2}{\pi n}\int{\sin(nx)~\text{d}x}\\
&= \frac{1}{n}\left(\frac{2x}{\pi}-1\right)\sin(nx) + \frac{2}{\pi n^2}\cos(nx)
\end{aligned}
$$
***
Inserting the limits, the $\sin$ term cancels ($\sin(0)$ and $\sin(n\pi)$ are 0). This leaves us with:
***
$$
a_n = \frac{4}{\pi^2 n^2}\left[(-1)^n - 1 \right]
$$
We see that for even $n$, $a_n=0$. For odd $n$:
$$
a_n = -\frac{8}{\pi^2 n^2}
$$
and hence bringing this together into $f(x)$:
***
$$
f(x) = -\sum^{\infty}_{n=1,3,5\cdots}\left(\frac{8}{\pi^2 n^2}\cos(nx) \right) =- \frac{8}{\pi^2}\left( \frac{\cos x}{1} + \frac{\cos 3x}{3^2} + \frac{\cos 5x}{5^2} + \cdots \right)
$$
\n
, The below animations demonstrate that the convergence speed is highest for the parabola function, and lowest for the square function.
We can see from above how convergence speed links to the Fourier coefficient like to $p$: the higher $p$, the higher the rate of convergence. A higher value of $p$ corresponds to a higher smoothness in the function.
, Taking the functions from part (a), and looking at the power of $n$, we see:
***
(**Square**): 
$$
f(x)= \frac{1}{2} +\frac{2}{\pi}\sum_{n=1,3,5,\cdots}^{\infty}\frac{\sin(nx)}{n}
$$
* $p=1$. 
***
(**Triangle**):
$$
f(x) = -\sum^{\infty}_{n=1,3,5\cdots}\left(\frac{8}{\pi^2 n^2}\cos(nx) \right)
$$
* $p=2$
***
(**Parabola**):
$$
f(x) = \sum_{n=1,3,5\cdots}^{\infty}{\frac{32}{\pi^3n^3}(-1)^{(n+1)/2}\cos(nx)}
$$
* $p=3$
The next branch demonstrates how the value of $p$ links to the rate of convergence.
\nThe convergence of a Fourier series depends on the highest derivative that has a discontinuity.
***
***
If a function discontinuities in the $k^{\rm th}$ derivative, it will have coefficients in the Fourier series that are proportional to $1/n^{k+1}$. 
***
Looking at the square function, we see that $f(x)$ itself is discontinuous, i.e., $k=0$. Indeed, its coefficients are proportional to $1/n$, i.e. $k+1=1$. 
***
For the triangle function, the first derivative is discontinuous:
Hence, $k=1$. Indeed, as we found in part (b), the Fourier coefficients are proportional to $1/n^2$ ($k+1=2$).
***
***
For the parabola function, the 2nd derivative is discontinuous, i.e. $k=2$:
Indeed, the Fourier series coefficients of the parabola function are proportional to $1/n^3$ ($k+1=3$).
| Here we will look at how the smoothness of a function affects its Fourier series.
Find the trigonometric Fourier series for the following three functions.
**Note:** The last function is constructed from two parabolas stuck together. They join at $\pi/2+n\pi$, where $n$ is an integer.
The functions are referred to as (**Square**), (**Triangle**) and (**Parabola**) in the solutions.
\nHow quickly a Fourier series converges depends on the properties of the function. In general the terms in the series will be proportional to:
$$
\frac{1}{n^p}
$$
in the limit of large $n$. For the functions in part (a), identify $p$.
\nTry to identify the relationship between $p$ and which derivative of the function is discontinuous.
| 115 | 6 | 47 | 47 | 647 | 15 | 100 | 6 | 0 | Find the trigonometric Fourier series for the following three functions. Note: The last function is constructed from two parabolas stuck together. How quickly a Fourier series converges depends on the properties of the function. Try to identify the relationship between $p$ and which derivative of the function is discontinuous. | 4 |
4ce870e1-9bbd-4bd9-8c6d-b2afec54d038 | 1 | 0 | 1 | 16 | 6 | 1 | 0 | 3 | For the function $f(x) = x^2(1 - x)^3$ ,
| Find the stationary points of $f(x)$ and determine their nature
\nSketch the graph of $y=f(x)$.
| 2 | 0.666667 | 2 | The stationary points are when $\displaystyle \frac{\mathrm{d}f}{\mathrm{d}x} = 0$. Now can you try finding them?
***
You can use the product rule to find $\displaystyle \frac{\mathrm{d}f}{\mathrm{d}x}$. Then set this to $0$ and factorise to find $x$.
Can you find the stationary points now? How can you determine their nature?
***
To determine their nature, look at higher derivatives at these points.
***
To find $\displaystyle \frac{\mathrm{d}^2f}{\mathrm{d}x^2}$, you can either do this manually from $\displaystyle \frac{\mathrm{d}f}{\mathrm{d}x}$, or you can use the product rule where
$$
\displaystyle
\frac{\mathrm{d}^2}{\mathrm{d}x^2}(fg) = \frac{\mathrm{d}^2f}{\mathrm{d}x^2}g + 2{\frac{\mathrm{d}f}{\mathrm{d}x}}{\frac{\mathrm{d}g}{\mathrm{d}x}} + f{\mathrm{d}^2g\over \mathrm{d}x^2}
$$
***
Remember that if the second derivative at the stationary point is $>0$, it is a minimum, if it is $=0$ it is a point of inflection, and if it is $<0$ it is a maximum point.
Are you able to find the nature of the three points now?
\nFind where the curve crosses the axes. Then by plotting the stationary points you can draw the curve.
| The stationary points are when $\displaystyle \frac{\mathrm{d}f}{\mathrm{d}x} = 0$. Now can you try finding them?
***
You can use the product rule to find $\displaystyle \frac{\mathrm{d}f}{\mathrm{d}x}$. Then set this to $0$ and factorise to find $x$.
***
$$
\displaystyle \frac{\mathrm{d}f}{\mathrm{d}x} = 2x(1-x)^3 + x^2\cdot 3(1-x)^2(-1)
$$
Setting $\displaystyle \frac{\mathrm{d}f}{\mathrm{d}x}=0$:
$$
\begin{aligned}
2x(1-x)^3 -3x^2(1-x)^2&=0
\\
(1-x)^2(2x(1-x) - 3x^2) &= 0
\\
x(1-x)^2(2-5x) &= 0
\end{aligned}
$$
Can you find the stationary points now? How can you determine their nature?
***
So the stationary points are at $\displaystyle x=0, x=1, x={2\over5}$.
To determine their nature, look at higher derivatives at these points.
***
To find $\displaystyle\frac{\mathrm{d}^2f}{\mathrm{d}x^2}$, you can either do this manually from $\displaystyle \frac{\mathrm{d}f}{\mathrm{d}x}$, or you can use the product rule where
$$
\displaystyle
\frac{\mathrm{d}^2}{\mathrm{dx^2}(fg)
= \frac{\mathrm{d}^2f}{\mathrm{d}x^2}g + 2{\frac{\mathrm{d}f}{\mathrm{d}x}}{\frac{\mathrm{d}g}{\mathrm{d}x}} + f{\mathrm{d}^2g\over \mathrm{d}x^2}
$$
***
Using the product rule (but you will get the same answer manually),
$$
\begin{aligned}
\frac{\mathrm{d}^2f}{\mathrm{d}x^2}
&= 2(1-x)^3 + 2(2x)[(-3)(1-x)^2] + x^2[6(1-x)]
\\
&=2(1-x)^3 - 12x(1-x)^2 + 6x^2(1-x)
\end{aligned}
$$
Now substitute in the stationary points to find their nature. 
***
Remember that if the second derivative at the stationary point is $>0$, it is a minimum, if it is $=0$ it is a point of inflection, and if it is $<0$ it is a maximum point.
Are you able to find the nature of the three points now?
***
$x=0$: $\displaystyle \frac{\mathrm{d}^2f}{\mathrm{d}x^2}\Big|_{x=0} = 2 > 0$ so a minimum point
$x=1:$ $\displaystyle \frac{\mathrm{d}^2f}{\mathrm{d}x^2} \Bigl|_{x=1} = 0$ so a point of inflection
$x=2$ : $\displaystyle\frac{\mathrm{d}^2f}{\mathrm{d}x^2}\Big|_{x={2\over5}} = -{18\over 25} < 0$ so a maximum point
Minimum $(0,0)$
Point of inflection $(1, 0)$
Maximum $ \displaystyle \left({2\over5}, {108\over3125}\right) $
\nFind where the curve crosses the axes. Then by plotting the stationary points you can draw the curve.
***
$y = x^2(1 - x)^3$
$$
\begin{aligned}
x=0 &: y= 0 \\
y=0 &: x^2(1-x)^3 =0 \quad \Rightarrow x=0, 1
\end{aligned}
$$
So crosses axes at $(0,0), (1,0)$.
***
| For the function $f(x) = x^2(1 - x)^3$ ,
Find the stationary points of $f(x)$ and determine their nature
\nSketch the graph of $y=f(x)$.
| 21 | 3 | 11 | 11 | 117 | 10 | 16 | 2 | 0 | For the function $f(x) = x^2(1 - x)^3$ , Find the stationary points of $f(x)$ and determine their nature Sketch the graph of $y=f(x)$. | 1 |
4ceb0f0d-4247-470d-b85d-766f50491227 | 0 | 0 | 1 | 16 | 6 | 1 | 7 | 6 | Locate the stationary points of $f(x, y) = xy(x + y -1)$ and deduce their nature
(i) from a contour sketch \[Sketch contours of the function and indicate regions where $f$ is respectively zero, positive and negative.]
(ii) from the criterion for the second partial derivatives of $f$ with respect to $x$ and $y$.
| Locate the stationary points of $f(x, y) = xy(x + y -1)$ and deduce their nature
(i) from a contour sketch \[Sketch contours of the function and indicate regions where $f$ is respectively zero, positive and negative.]
(ii) from the criterion for the second partial derivatives of $f$ with respect to $x$ and $y$.
| 1 | 1 | 3 | From section 5.5 of the notes, stationary points of a function $f(x,y)$ are where
$$
\begin{aligned}
{\partial f\over\partial x} = 0 = {\partial f\over\partial y}
\end{aligned}
$$
***
You should have found 4 stationary points, which you can find by solving the simultaneous equations that arise from the above equation.
***
**Part (i)** Now try sketching the contour graph by plotting the stationary points and zero contour lines (these are where $f(x,y)=0$).
You can find if $f(x,y)$ is positive or negative in the regions separated by the zero contour lines by substituting the relevant values for $x$ and $y$.
***
**Part (i)** The zero contours (where $f(x,y)=0$) are $x=0, y=0$ and $x+y-1=0$ which are the axes, and the line $y=-x+1$. The contour lines are the boundary between regions of positive and negative $f(x,y)$ values.
Can you finish the sketch now and find the nature of the stationary points?
***
**Part (ii)** You can determine their nature using
$$
\begin{aligned}
E_0
= \left[\left({\partial^2u\over \partial x\partial y}\right)^2 - \left({\partial^2u\over\partial x^2}\right)\left({\partial^2u\over\partial y^2}\right)\right]_{x_0, y_0}
\end{aligned} = B^2-AC
$$
where $(x_0, y_0)$ is the position of the stationary point.
If $E_0>0$ it is a saddle point.
If $E_0<0$ then if $\displaystyle \left({\partial^2u\over\partial x^2}\right)_{x_0, y_0} < 0$ it is a local maximum or if $ \displaystyle \left({\partial^2u\over\partial x^2}\right)_{x_0, y_0} > 0 $ it is a local minimum.
If $E_0 = 0$ then you need to look at higher derivatives.
You can substitute the saddle points to find $E_0$.
| From section 5.5 of the notes, stationary points of a function $f(x,y)$ are where
$$
\begin{aligned}
{\partial f\over\partial x} = 0 = {\partial f\over\partial y}
\end{aligned}
$$
***
Finding the derivatives,
$$
\begin{aligned}
{\partial f\over\partial x} &= 0 \\
2xy+y^2-y &= 0 \\
y(2x+y-1) &=0
\end{aligned}
$$
$$
\begin{aligned}
{\partial f\over\partial y} &= 0 \\
x^2+2xy-x &= 0 \\
x(2y+x-1) &=0
\end{aligned}
$$
Can you find the stationary points now?
***
$y =0, 1-2x$ and $x =0, 1-2y$
Solving the simultaneous equations gives the stationary points as,
$$
P_1(0,0),\quad P_2(0,1),\quad P_3(1,0),\quad P_4({1\over3}, {1\over3})
$$
, Now try sketching the contour graph by plotting the stationary points and zero contour lines (these are where $f(x,y)=0$).
You can find if $f(x,y)$ is positive or negative in the regions separated by the zero contour lines by substituting the relevant values for $x$ and $y$.
***
The zero contours (where $f(x,y)=0$) are $x=0, y=0$ and $x+y-1=0$ which are the axes, and the line $y=-x+1$. The contour lines are the boundary between regions of positive and negative $f(x,y)$ values.
Can you finish the sketch now?
***
The contour sketch is shown here, with the stationary points, zero contour lines, and the positive and negative regions shown.
From the sketch, can you see what the natures of the stationary points are now?
***
$P_4$ is a local minimum as $f(x,y)$ is increasing away from the point (it is going from negative to $0$ at the boundaries).
$P_1, P_2, P_3$ are all saddle points as $f(x,y)$ is increasing/decreasing depending on the direction you go.
, You can determine their nature using
$$
\begin{aligned}
E_0
= \left[\left({\partial^2u\over \partial x\partial y}\right)^2 - \left({\partial^2u\over\partial x^2}\right)\left({\partial^2u\over\partial y^2}\right)\right]_{x_0, y_0}
\end{aligned} = B^2-AC
$$
where $(x_0, y_0)$ is the position of the stationary point.
If $E_0>0$ it is a saddle point.
If $E_0<0$ then if $\displaystyle \left({\partial^2u\over\partial x^2}\right)_{x_0, y_0} < 0$ it is a local maximum or if $ \displaystyle \left({\partial^2u\over\partial x^2}\right)_{x_0, y_0} > 0 $ it is a local minimum.
If $E_0 = 0$ then you need to look at higher derivatives.
You can substitute the saddle points to find $E_0$.
***
$$
\begin{aligned}
{\partial ^2f\over\partial x^2} &= 2y \\
{\partial ^2f\over\partial y^2} &= 2x \\
{\partial ^2f\over\partial x\partial y} &= 2x+2y-1
\end{aligned}
$$
Can you find $E_0$ for each of the saddle points now, and then determine their nature?
***
| Stationary Point | A | B | C | $E_0$ |
| :-------------------------------------------------- | :------------------------ | :------------------------ | :------------------------ | :------------------------- |
| $P_1(0,0)$ | 0 | -1 | 0 | 1 |
| $P_2(0,1)$ | 2 | 1 | 0 | 1 |
| $P_3(1,0)$ | 0 | 1 | 2 | 1 |
| $\displaystyle P_4\left({1\over3},{1\over3}\right)$ | $\displaystyle {2\over3}$ | $\displaystyle {1\over3}$ | $\displaystyle {2\over3}$ | $\displaystyle -{1\over3}$ |
Can you determine their nature now?
***
$$
P_1(0,0)\Rightarrow \mathrm{saddle} \\
P_2(0,1)\Rightarrow \mathrm{saddle}\\
P_3(1,0) \Rightarrow \mathrm{saddle}\\
P_4({1\over3}, {1\over3})\Rightarrow \mathrm{local\,minimum}\\
$$
| Locate the stationary points of $f(x, y) = xy(x + y -1)$ and deduce their nature
(i) from a contour sketch \[Sketch contours of the function and indicate regions where $f$ is respectively zero, positive and negative.]
(ii) from the criterion for the second partial derivatives of $f$ with respect to $x$ and $y$.
| 49 | 5 | 38 | 38 | 354 | 16 | 49 | 5 | 0 | Locate the stationary points of $f(x, y) = xy(x + y -1)$ and deduce their nature i from a contour sketch Sketch contours of the function and indicate regions where $f$ is respectively zero, positive and negative. | 1 |
4d004959-68eb-4c1e-9665-d7cf0c0fe704 | 2 | 0 | 0 | 14 | 4 | 2 | 0 | 2 | A submerged submarine is towed horizontally at a steady speed $U$ in deep still water. An axially-symmetrical wake is formed behind the submarine in which the water velocity may be assumed to vary linearly from $U$ on the axis to zero at a radius of $R$. The variation of the water pressure with depth may be assumed to be unaffected by the presence of the submarine. The density of the water is $\rho$. Using a control-volume analysis, we want to find the required power to tow the submarine. For both choices of control volumes (A and B as shown above), derive an expression for:
| The drag force $F$ of the submarine.\nThe power $P$ required to tow the submarine. | 2 | 0.5 | 2 | \n | \n | A submerged submarine is towed horizontally at a steady speed $U$ in deep still water. An axially-symmetrical wake is formed behind the submarine in which the water velocity may be assumed to vary linearly from $U$ on the axis to zero at a radius of $R$. The variation of the water pressure with depth may be assumed to be unaffected by the presence of the submarine. The density of the water is $\rho$. Using a control-volume analysis, we want to find the required power to tow the submarine. For both choices of control volumes (A and B as shown above), derive an expression for:
The drag force $F$ of the submarine.\nThe power $P$ required to tow the submarine. | 121 | 6 | 0 | 0 | 1 | 0 | 14 | 2 | 1 | An axially-symmetrical wake is formed behind the submarine in which the water velocity may be assumed to vary linearly from $U$ on the axis to zero at a radius of $R$. The variation of the water pressure with depth may be assumed to be unaffected by the presence of the submarine. Using a control-volume analysis, we want to find the required power to tow the submarine. For both choices of control volumes A and B as shown above, derive an expression for: The drag force $F$ of the submarine. | 4 |
4db11d69-3efb-44cc-a481-ee3120658046 | 3 | 0 | 0 | 24 | 6 | 1 | 2 | 7 | Find the work done by the force $\vec{F} = (2xy -3)\mathbf{\hat{i}} +x^2\mathbf{\hat{j}}$ in moving an object \[in the $x-y$ plane] from $(1,0)$ to \$(0,1) along each of the following paths:
| The circular arc of radius 1, centre at the origin, from (1,0) to (0,1) \[**Hint:** parameterise this arc in terms of the plane polar angle $\phi$]
\nFrom $(1,0)$ to $(1,1)$ to $(0,1)$, i.e., along two line segments parallel to the axes.
\nShow that in fact the force $\vec{F}$ is conservative. Find the function $\Omega(x,y)$ such that:
$$
\vec{F} = \frac{\partial\Omega}{\partial x}\mathbf{\hat{i}} + \frac{\partial\Omega}{\partial y}\mathbf{\hat{j}}
$$
| 3 | 0.666667 | 2 | First, write down $\vec{F}\cdot d\vec{r}$. 
***
Convert $\vec{F}\cdot d\vec{r}$ to polar coordinates ($\rho=1$)...
***
... This will require you to express $x$ and $y$ as a function of $\phi$. What are $dx$ and $dy$?
***
Convert to a more integrable form and evaluate over the limits of $\phi$.
\nEvaluate the line integral separately along each section of the path...
***
... You will be able to simplify the integral by considering the parameters of the path...
***
... e.g., $x=1$, $dy=0$.
***
The total line integral is the sum of the integral over each part. How does your result compare to (a)? Before moving onto (c), think about why this is the case.
\nThe vector field $\vec{F}$ is conservative if the differential $\vec{F}\cdot d\vec{r}$ is *exact*.
***
For $\vec{F}\cdot d\vec{r}$ to be exact, it is required that:
$$
\frac{\partial F_x}{\partial y}=\frac{\partial F_y}{\partial x}
$$
***
Given that $\vec{F}$ is conservative, it has a 'parent' function $\Omega$. You know that:
$$
F_x = \frac{\partial\Omega}{\partial x}
$$
Integrate this with respect to $x$ to find $\Omega$....
***
... Remember that this gives a function of $y$ (let this be $g(y)$), as the 'constant term'...
***
Solve for $g(y)$ by finding $\partial \Omega/\partial y$ and setting this equal to $F_y$.
***
Can you think of how to verify that $\Omega$ is correct? Consider your answers to part (a) and (b).
| 
***
$$
\vec{F}\cdot d\vec{r} = F_x\,dx + F_y\,dy = (2xy-3)dx+x^2dy
$$
Converting to polar coordinates, with $\rho=1$:
***
$$
x=\cos\phi \implies dx = -\sin\phi\,d\phi
$$
***
$$
y=\sin\phi \implies dy = \cos\phi \,d\phi
$$
***
$$
\begin{aligned}
\implies \vec{F}\cdot d\vec{r} &= (2\cos\phi\sin\phi - 3)(-\sin\phi\,d\phi)+\cos^2\phi(\cos\phi)d\phi\\
&=(-2\sin^2\phi\cos\phi+3\sin\phi+(1-\sin^2\phi)\cos\phi)d\phi\\
& = (-3\sin^2\phi\cos\phi + 3\sin\phi+\cos\phi)d\phi
\end{aligned}
$$
***
Evaluating over the limits $\phi: 0\to \pi/2$:
***
$$
\begin{aligned}
\int_{C}\vec{F}\cdot d\vec{r} &= \int_{0}^{\pi/2}{(-3\sin^2\phi\cos\phi + 3\sin\phi+\cos\phi)d\phi} \\
& = \left[-\frac{3\sin^3\phi}{3}-3\cos\phi+\sin\phi\right]_{0}^{\pi/2}\\
& = 3
\end{aligned}
$$
\nWe will separate the path into two parts $(b_1)$ and $(b_2)$ and evaluate the line integral over each part.
***
Along $(b_1)$, $x=1$ and $dx=0$, with $y:0\to 1$. 
***
$$
\int_{b_1}{\vec{F}\cdot d\vec{r}} = \int_{y=0}^{1}F_y\,dy = \int_{y=0}^{1}x^2\,dy = \int_{y=0}^{1}\,dy = 1
$$
***
Along $(b_2)$, $y=1$ and $dy=0$, with $x:1\to 0$. 
***
$$
\int_{b_2}\vec{F}\cdot d\vec{r} = \int_{x=1}^{0}{(2xy-3)dx} = \int_{x=1}^{0}(2x-3)\,dx=[x^2-3x]_{1}^{0}=2
$$
***
Therefore, summing the line integral over each part:
***
$$
\int_{C}\vec{F}\cdot d\vec{r} = 1+2=3
$$
This is equivalent to the result from part (a). 
\nThe vector field $\vec{F}$ is conservative if the differential $\vec{F}\cdot d\vec{r}$ is *exact*. 
***
$$
F_x = 2xy-3 \implies \frac{\partial F_x}{\partial y} = 2x
$$
***
$$
F_y = x^2 \implies \frac{\partial F_y}{\partial x} = 2x
$$
***
Since:
$$
\frac{\partial F_x}{\partial y}=\frac{\partial F_y}{\partial x}
$$
The differential is exact. So, $ \vec{F} $ is conservative. 
***
Setting:
$$
F_x = \frac{\partial \Omega}{\partial x}
$$
and integrating with respect to $x$:
***
$$
\Omega = \int{(2xy-3)dx} = x^2y-3x+g(y)
$$
Partially differentiating this with respect to $y$:
***
$$
\frac{\partial \Omega}{\partial y} = x^2 + g'(y)
$$
This is equal to $F_y$, so:
***
$$
x^2 = x^2 + g'(y) \implies g'(y)=0\implies g(y) = C
$$
So, $\Omega$ is given by:
$$
\Omega = x^2y- 3x + C
$$
***
To verify that this is correct, the line integrals in part (a) and part (b) should be equivalent to the difference in $\Omega$ from $(1,0)$ to $(0,1)$:
***
$$
\int_{C}\vec{F}\cdot d\vec{r} = \Omega(0,1)-\Omega(1,0) = 0-(-3)=3
$$
| Find the work done by the force $\vec{F} = (2xy -3)\mathbf{\hat{i}} +x^2\mathbf{\hat{j}}$ in moving an object \[in the $x-y$ plane] from $(1,0)$ to \$(0,1) along each of the following paths:
The circular arc of radius 1, centre at the origin, from (1,0) to (0,1) \[**Hint:** parameterise this arc in terms of the plane polar angle $\phi$]
\nFrom $(1,0)$ to $(1,1)$ to $(0,1)$, i.e., along two line segments parallel to the axes.
\nShow that in fact the force $\vec{F}$ is conservative. Find the function $\Omega(x,y)$ such that:
$$
\vec{F} = \frac{\partial\Omega}{\partial x}\mathbf{\hat{i}} + \frac{\partial\Omega}{\partial y}\mathbf{\hat{j}}
$$
| 35 | 11 | 6 | 6 | 284 | 26 | 59 | 7 | 0 | Find the work done by the force $\vec{F} = (2xy -3)\mathbf{\hat{i}} +x^2\mathbf{\hat{j}}$ in moving an object in the $x-y$ plane from $(1,0)$ to $(0,1) along each of the following paths: The circular arc of radius 1, centre at the origin, from (1,0) to (0,1) \[**Hint:** parameterise this arc in terms of the plane polar angle $phi$ From $(1,0)$ to $1,1$ to $0,1$, i.e., along two line segments parallel to the axes. Show that in fact the force $vec{F}$ is conservative. Find the function $Omegax,y$ such that: $ vec{F} = frac{partialOmega}{partial x}mathbf{hat{i}} + frac{partialOmega}{partial y}mathbf{hat{j}} $ | 3 |
4de7d660-9a1c-486a-9cc3-9f8d98511e4e | 1 | 0 | 0 | 2 | 1 | 2 | 2 | 5 | Acceleration due to gravity on the moon is 1.62 m/s$^2,$and atmospheric pressure is $3\times10^{-15}$ bar (at night). What is the pressure at the bottom of a (hypothetical) $1$ mm high column of mercury, $\rho_{mercury}=13600$ kg/$m^3$?
Note that $1\,$bar $=100,000\,$Pa.
| Acceleration due to gravity on the moon is 1.62 m/s$^2,$and atmospheric pressure is $3\times10^{-15}$ bar (at night). What is the pressure at the bottom of a (hypothetical) $1$ mm high column of mercury, $\rho_{mercury}=13600$ kg/$m^3$?
Note that $1\,$bar $=100,000\,$Pa.
| 1 | 0.333333 | 1 | We have that the pressure in the column obeys
$\frac{\delta p_{\sf mercury}}{\delta z}=-\rho_{\sf mercury} g_{\sf moon}.$
***
Solving the above equation to get an expression for $p_{mercury}$
***
We get $p_{\sf mercury}=-\rho_{\sf mercury} g_{\sf moon}z+p_0$. Setting $z=0$ at the bottom of the column, at a height of $h=1$ mm what is the pressure in Pa?
Using this find an expression for $p_o$.
***
We get $p_0=p_{\sf atmosphere}+\rho_{\sf mercury} g_{\sf moon}h.$ Using $p_o$ find an expression for $p_{\sf mercury}$.
***
We get $p_{\sf mercury}=p_{\sf atmosphere}+\rho_{\sf mercury} g_{\sf moon}(h-z)$. Now substitute in $z=0$ to find the pressure at the bottom of the column.
***
Substitute in the values given in the question to get a numeric value for $p_{mercury}$.
| We have that the pressure in the column obeys
$\frac{\delta p_{\sf mercury}}{\delta z}=-\rho_{\sf mercury} g_{\sf moon}.$
Solving this, we get
$p_{\sf mercury}=-\rho_{\sf mercury} g_{\sf moon}z+p_0.$
Setting $z=0$ at the bottom of the column, at a height of $h=1$ mm, we have $p=p_{\sf atmosphere}=3\times10^{-15}$ bar $=3\times10^{-10}$ Pa, and hence
$p_0=p_{\sf atmosphere}+\rho_{\sf mercury} g_{\sf moon}h.$
Hence the pressure in the column is
$p_{\sf mercury}=p_{\sf atmosphere}+\rho_{\sf mercury} g_{\sf moon}(h-z),$
and thus at the bottom of the column ($z=0$),
$p_{\sf mercury}=p_{\sf atmosphere}+\rho_{\sf mercury} g_{\sf moon}h$
$=3\times10^{-10}+13,600\times1.6\times0.001$
$=3\times10^{-10}+22$
$\approx22$ Pa
Note that atmospheric pressure is very insignificant in comparison to the pressure from the column (unlike the case on the surface of the earth, for which the pressure from an identical column of fluid would be only about 0.1% of earth's atmospheric pressure).
| Acceleration due to gravity on the moon is 1.62 m/s$^2,$and atmospheric pressure is $3\times10^{-15}$ bar (at night). What is the pressure at the bottom of a (hypothetical) $1$ mm high column of mercury, $\rho_{mercury}=13600$ kg/$m^3$?
Note that $1\,$bar $=100,000\,$Pa.
| 45 | 7 | 13 | 13 | 105 | 12 | 45 | 7 | 0 | What is the pressure at the bottom of a hypothetical $1$ mm high column of mercury, $\rho_{mercury}=13600$ kg/$m^3$? Note that $1\,$bar $=100,000\,$Pa. | 2 |
4dec19fe-441b-402d-b868-7aaa3b5bb185 | 4 | 0 | 1 | 9 | 4 | 2 | 5 | 5 | Rotational speed can be measured using a small generator producing a voltage which is proportional the rotational speed (as seen in MECH50004 - Mechatronics 2 - Lab 4). Due to the characteristics of the generator, there will always be a “ripple” on the voltage signal caused by construction of the coils (stator and rotor rubbing against each other cause a high frequency noise which disrupts the main voltage). This ripple can be removed using a low pass filter. The speed sensor used in the lab generates $15\%$ ripple voltage (peak to peak) at $4$ times the frequency of the speed measured. For a signal at $1000~\mathrm{rpm}$, the output of the sensor is $4~\mathrm{V}$ plus ripple (i.e. $4~\mathrm{V}$ DC plus ripple voltage).
| Select a resistor which, in combination with a $36~\mu\mathrm{F}$ capacitor, will reduce the ripple by $20~\mathrm{dB}$. What will the peak-to-peak amplitude of the ripple voltage be after filtering?
\nSelect a resistor which, in combination with a $36~\mu\mathrm{F}$ capacitor, will reduce the ripple by $40~\mathrm{dB}$.
\nSelect a resistor which, in combination with a $36~\mu\mathrm{F}$ capacitor, will reduce the ripple by $60~\mathrm{dB}$.
\nWhat impact could this filter have on the control of the motor speed?
| 4 | 0.666667 | 3 | \n\n\n | Using the Bode plot for a normalised Low-Pass filter, one can estimate that when $|H|=-20~\mathrm{dB}$, $\frac{\omega}{\omega_\mathrm{c}}=10$ as shown:
  
***
Hence, since $\omega=418.9~\mathrm{rad/s}$:
  
$\omega_c=418.9/10=41.89$ and $R=\frac{1}{41.89\times36\times 10^{-6}}=663.1~\Omega$
***
Calculate the ripple voltage after filtering, noting that the ripple voltage generated is $15\%$ of the output:  
  
$V_\mathrm{peak-to-peak} = 0.15V|H| = 0.15\times 4\times0.1 = 60~\mathrm{mV}$
, $|H|_\mathrm{DB} = 20\log_{10}(|H|)$  
***
Substituting numbers and rearranging for $|H|$:
  
$|H| = 10^{-\frac{20}{20}} = 0.1$
***
For a passive low-pass filter:
  
$|H| = \frac{1}{\sqrt{1+(\omega RC)^2}}$
***
Substituting in numbers and rearranging for $R$:
  
$R = \dfrac{1}{418.9\times 36\times 10^{-6}}\sqrt{(\frac{1}{0.1})-1} = 659.8~\mathrm{\Omega}$
***
Calculate the ripple voltage after filtering, noting that the ripple voltage generated is $15\%$ of the output:
  
$V_\mathrm{peak-to-peak} = 0.15V|H| = 0.15\times 4\times0.1 = 60~\mathrm{mV}$
, Convert $\omega$ to $\mathrm{rad/s}$, noting that the frequency is $4$ times the frequency of the speed measured:
  
$\omega=1000\times \frac{2\pi\times 4}{60}=418.9~\mathrm{rad/sec}$
\n
, $|H|_\mathrm{DB} = 20\log_{10}(|H|)$  
***
Substituting numbers and rearranging for $|H|$:
  
$|H| = 10^{-\frac{40}{20}} = 0.1$
***
For a passive low-pass filter:
  
$|H| = \frac{1}{\sqrt{1+(\omega RC)^2}}$
***
Substituting in numbers and rearranging for $R$:
  
$R = \dfrac{1}{418.9\times 36\times 10^{-6}}\sqrt{(\frac{1}{0.01})-1} = 6631.1~\mathrm{\Omega}$
, Using the Bode plot for a normalised Low-Pass filter, one can estimate that when $|H|=-40~\mathrm{dB}$, $\frac{\omega}{\omega_\mathrm{c}}=100$ as shown:
  
  
***
Hence, since $\omega=418.9~\mathrm{rad/s}$:
  
$\omega_c=418.9/100=4.189$ and $R=\frac{1}{4.189\times36\times 10^{-6}}=6631.1~\Omega$
\n$|H|_\mathrm{DB} = 20\log_{10}(|H|)$
***
Substituting in numbers and rearranging for $|H|$:
  
$|H| = 10^{-\frac{60}{20}} = 0.001$
***
For a passive low-pass filter:
  
$|H| = \frac{1}{\sqrt{1+(\omega RC)^2}}$
***
Substituting in numbers and rearranging for $R$, noting that the frequency is $4$ times the frequency of the speed measured:
  
$R = \dfrac{1}{418.9\times 36\times 10^{-6}}\sqrt{(\frac{1}{0.001})^2-1} = 66311~\mathrm{\Omega}$
,
, Using the Bode plot for a normalised Low-Pass filter, one can estimate that when $|H|=-60~\mathrm{dB}$, $\frac{\omega}{\omega_\mathrm{c}}=1000$ as shown:
  
  
***
Hence, since $\omega=418.9~\mathrm{rad/s}$:
  
$\omega_c=418.9/1000=0.4189$ and $R=\frac{1}{0.4189\times36\times 10^{-6}}=66311~\Omega$
\n | Rotational speed can be measured using a small generator producing a voltage which is proportional the rotational speed (as seen in MECH50004 - Mechatronics 2 - Lab 4). Due to the characteristics of the generator, there will always be a “ripple” on the voltage signal caused by construction of the coils (stator and rotor rubbing against each other cause a high frequency noise which disrupts the main voltage). This ripple can be removed using a low pass filter. The speed sensor used in the lab generates $15\%$ ripple voltage (peak to peak) at $4$ times the frequency of the speed measured. For a signal at $1000~\mathrm{rpm}$, the output of the sensor is $4~\mathrm{V}$ plus ripple (i.e. $4~\mathrm{V}$ DC plus ripple voltage).
Select a resistor which, in combination with a $36~\mu\mathrm{F}$ capacitor, will reduce the ripple by $20~\mathrm{dB}$. What will the peak-to-peak amplitude of the ripple voltage be after filtering?
\nSelect a resistor which, in combination with a $36~\mu\mathrm{F}$ capacitor, will reduce the ripple by $40~\mathrm{dB}$.
\nSelect a resistor which, in combination with a $36~\mu\mathrm{F}$ capacitor, will reduce the ripple by $60~\mathrm{dB}$.
\nWhat impact could this filter have on the control of the motor speed?
| 198 | 11 | 42 | 42 | 268 | 0 | 76 | 6 | 0 | Rotational speed can be measured using a small generator producing a voltage which is proportional the rotational speed as seen in MECH50004 - Mechatronics 2 - Lab 4. This ripple can be removed using a low pass filter. The speed sensor used in the lab generates $15\%$ ripple voltage peak to peak at $4$ times the frequency of the speed measured. Select a resistor which, in combination with a $36~\mu\mathrm{F}$ capacitor, will reduce the ripple by $20~\mathrm{dB}$. What will the peak-to-peak amplitude of the ripple voltage be after filtering? Select a resistor which, in combination with a $36~\mu\mathrm{F}$ capacitor, will reduce the ripple by $40~\mathrm{dB}$. Select a resistor which, in combination with a $36~\mu\mathrm{F}$ capacitor, will reduce the ripple by $4$0. What impact could this filter have on the control of the motor speed? | 8 |
4e2e1aa7-7115-46b3-a00a-33c3be6adb8f | 0 | 1 | 0 | 17 | 6 | 1 | 3 | 1 | A ping-pong (table tennis) ball makes a head-on elastic collision with a much heavier basketball, which is initially stationary. Which of the following statements are true after the collision?
| A ping-pong (table tennis) ball makes a head-on elastic collision with a much heavier basketball, which is initially stationary. Which of the following statements are true after the collision?
| 1 | 0.666667 | 1 | The ping-pong ball of mass $m$ is so much lighter than the basketball of mass $M$ that its momentum almost reverses when it collides: $m\vec{\boldsymbol{u}}_m \rightarrow -m\vec{\boldsymbol{u}}_m$. 
***
What is the momentum of the basketball after the collision? ...
***
... Use conservation of momentum. This should allow you to distinguish between options (a), (b), (c) and (d).
***
Find the velocity of the basketball after collision, and hence find the kinetic energy of both balls after the collision.
***
Considering $m\ll M$, which ball has more kinetic energy? This allows you to distinguish between options (e), (f), (g) and (h).
| The ping-pong ball of mass $m$ is so much lighter than the basketball of mass $M$ that its momentum almost reverses when it collides: $m\vec{\boldsymbol{u}}_m \rightarrow -m\vec{\boldsymbol{u}}_m$. 
***
Since the total momentum $m\vec{\boldsymbol{u}}_m$ is conserved, the momentum of the basketball after the collision must be approximately $2m\vec{\boldsymbol{u}}_m$. The absolute value of the momentum of the basketball after the collision is larger than that of the ping-pong ball. Hence option (a) is true.
***
The momentum of the basketball after the collision is approximately $2m\vec{\boldsymbol{u}}_m$, so its velocity $\vec{\boldsymbol{v}}_M \approx \frac{2m}{M} \vec{\boldsymbol{u}}_m$. The final kinetic energies of the basketball and ping-pong ball are:
***
$$
\begin{aligned}
K_M
&= \frac{1}{2}M |\vec{\boldsymbol{v}}_M|^2 \approx \frac{1}{2} M
\left |\frac{2m\vec{\boldsymbol{u}}_m}{M}\right |^2 = \frac{2m^2}{M}
|\vec{\boldsymbol{u}}_m|^2 ,\\
K_m
&\approx \frac{1}{2} m |-\vec{\boldsymbol{u}}_m|^2 .
\end{aligned}
$$
The ratio $K_m/K_M = M/(4m)$ is much bigger than 1 (the mass of one basketball is much larger than the mass of four ping-pong balls), so the ping-pong ball has much more kinetic energy than the basketball.
| A ping-pong (table tennis) ball makes a head-on elastic collision with a much heavier basketball, which is initially stationary. Which of the following statements are true after the collision?
| 29 | 0 | 9 | 9 | 138 | 4 | 29 | 0 | 0 | Which of the following statements are true after the collision? | 1 |
4e6a7b17-240f-482b-aea1-e38b59c303d4 | 15 | 0 | 0 | 19 | 6 | 1 | 8 | 0 | Find the *real* eigenvalues and *normalised* eigenvectors of the following transformation matrices in $\mathbb{R}^{2}$, and use the information to identify the transformations.
***
**Note:** In the response area, type the same answer twice for repeated solutions. Type $\begin{pmatrix}n\\n\end{pmatrix}$ for no eigenvector.
| $$
\text{A}=\left(
\begin{array}{cc}
0&\hskip6pt 1\\
1&\hskip6pt 0
\end{array}
\right)\quad
$$
\n$$
\text{B}=\left(
\begin{array}{cc}
\frac{{1}}{2}& \frac{\sqrt{3}}{2} \\
\frac{\sqrt{3}}{2}&\hskip4pt -\frac{{1}}{2}
\end{array}
\right)\quad
$$
\n$$
\text{C}=\left(
\begin{array}{cc}
-1&\hskip7pt 0\\
\hskip7pt 0& -1
\end{array}
\right)\quad
$$
\n$$
\text{D}=\left(
\begin{array}{rr}
4&\hskip3pt 1\\
-1&\hskip6pt 4
\end{array}
\right)
$$
\n$$
\text{E}=\left(
\begin{array}{rr}
1&\hskip6pt 1\\
1&\hskip6pt 1
\end{array}\right)
$$
| 5 | 0.333333 | 3 | Set-up the characteristic equation $p(\lambda) = \det(\text{A}-\lambda\mathbb{I}_2)=0$ (**section 3.19**)
***
Hence solve for the eigenvalues $\lambda$ (you may find that none exist). 
***
Given $(\text{A}-\lambda\mathbb{I})\mathbf{\underline{x}} = 0$ and $\mathbf{\underline{x}}=(x,y)$, solve for $x$ and $y$ for each value of $\lambda$. 
***
Normalise your derived $\mathbf{\underline{x}}$ to give the normalised eigenvectors.
***
By definition, eigenvectors are those vectors that are invariant under a transformation matrix. Therefore, your eigenvectors will tell you about the nature of the transformation matrix. 
\nSet-up the characteristic equation $p(\lambda) = \det(\text{B}-\lambda\mathbb{I}_2)=0$ (**section 3.19**)
***
Hence solve for the eigenvalues $\lambda$ (you may find that none exist). 
***
Given $(\text{B}-\lambda\mathbb{I})\mathbf{\underline{x}} = 0$ and $\mathbf{\underline{x}}=(x,y)$, solve for $x$ and $y$ for each value of $\lambda$. 
***
Normalise your derived $\mathbf{\underline{x}}$ to give the normalised eigenvectors.
***
By definition, eigenvectors are those vectors that are invariant under a transformation matrix. Therefore, your eigenvectors will tell you about the nature of the transformation matrix. 
\nSet-up the characteristic equation $p(\lambda) = \det(\text{C}-\lambda\mathbb{I}_2)=0$ (**section 3.19**)
***
Hence solve for the eigenvalues $\lambda$ (you may find that none exist). 
***
Given $(\text{C}-\lambda\mathbb{I})\mathbf{\underline{x}} = 0$ and $\mathbf{\underline{x}}=(x,y)$, solve for $x$ and $y$ for each value of $\lambda$. 
***
Normalise your derived $\mathbf{\underline{x}}$ to give the normalised eigenvectors.
***
By definition, eigenvectors are those vectors that are invariant under a transformation matrix. Therefore, your eigenvectors will tell you about the nature of the transformation matrix. 
\nSet-up the characteristic equation $p(\lambda) = \det(\text{D}-\lambda\mathbb{I}_2)=0$ (**section 3.19**).
***
Hence solve for the eigenvalues $\lambda$ (you may find that none exist). 
***
Given $(\text{D}-\lambda\mathbb{I})\mathbf{\underline{x}} = 0$ and $\mathbf{\underline{x}}=(x,y)$, solve for $x$ and $y$ for each value of $\lambda$. 
***
Normalise your derived $\mathbf{\underline{x}}$ to give the normalised eigenvectors.
***
By definition, eigenvectors are those vectors that are invariant under a transformation matrix. Therefore, your eigenvectors will tell you about the nature of the transformation matrix. 
\nSet-up the characteristic equation $p(\lambda) = \det(\text{E}-\lambda\mathbb{I}_2)=0$ (**section 3.19**)
***
Hence solve for the eigenvalues $\lambda$ (you may find that none exist).
***
Given $(\text{B}-\lambda\mathbb{I})\mathbf{\underline{x}} = 0$ and $\mathbf{\underline{x}}=(x,y)$, solve for $x$ and $y$ for each value of $\lambda$. 
***
Normalise your derived $\mathbf{\underline{x}}$ to give the normalised eigenvectors.
***
By definition, eigenvectors are those vectors that are invariant under a transformation matrix. Therefore, your eigenvectors will tell you about the nature of the transformation matrix. 
| Setting up and solving the characteristic equation $p(\lambda) = \det(\text{A}-\lambda\mathbb{I}_2)=0$ (**section 3.19**):
***
$$
p(\lambda) = \det\left(\begin{pmatrix}0&1\\1&0\end{pmatrix}-\lambda\begin{pmatrix}1&0\\0&1\end{pmatrix}\right)
$$
***
$$
=\left| \begin{array}{cc}
-\lambda & 1\\
1 & -\lambda
\end{array}\right| =\lambda^{2} -1=0 \qquad \rightarrow \qquad \lambda =\pm 1
$$
***
***
Given $(\text{A}-\lambda\mathbb{I})\mathbf{\underline{x}} = 0$, where $\mathbf{\underline{x}}=(x,y)$ is the eigenvector, solving for $x,y$ for each value of $\lambda$:
***
$$
\lambda =1, \qquad \left(\begin{array}{ c c }
-1 & 1\\
1 & -1
\end{array}\right)\left(\begin{array}{ c }
x\\ y
\end{array}\right) =\left(\begin{array}{ c }
0\\ 0
\end{array}\right) \quad \rightarrow \quad -x+y=0
$$
***
$$
\lambda =-1,\qquad \left(\begin{array}{ c c }
1 & 1\\
1 & 1
\end{array}\right) \left(\begin{array}{ c }
x\\ y
\end{array}\right) =\left(\begin{array}{ c }
0\\ 0
\end{array}\right) \qquad \rightarrow \qquad x+y=0\qquad \rightarrow \qquad y=-x
$$
***
Therefore, the two normalised eigenvectors are given by:
***
$$
\quad \rightarrow \quad y=x \qquad \rightarrow \qquad \vec{v}_{1}= \left(\begin{array}{ c }
1\\ 1
\end{array}\right) \qquad \rightarrow \qquad \mathbf{\hat{v}_1} =\frac{1}{\sqrt{2}}\left(\begin{array}{ c }
1\\ 1
\end{array}\right)
$$
$$
\rightarrow \quad y=-x \quad \rightarrow \quad\vec{v}_{2} = \left(\begin{array}{ c }
1\\ -1
\end{array}\right) \qquad \rightarrow \qquad \mathbf{\hat{v}}_{2} =\frac{1}{\sqrt{2}}\left(\begin{array}{ c }
1\\ -1
\end{array}\right)
$$
The matrix operation is invariant along $\vec{v}_{1}= (1,1)$ which is the line $y=x$ but negates everything normal to that. Though it may not look like it in the diagram, the points have been flipped either side of $y=x$.
\nSetting up and solving the characteristic equation $p(\lambda) = \det(\text{B}-\lambda\mathbb{I}_2)=0$ (**section 3.19**):
***
$$
p( \lambda ) \ =\ \left| \begin{array}{ c c }
\frac{1}{2}-\lambda & \frac{\sqrt{3}}{2}\\
\frac{\sqrt{3}}{2} & -\frac{1}{2}-\lambda
\end{array}\right| = \lambda^{2} - \frac{1}{4} - \frac{3}{4} =0 \qquad \rightarrow \qquad \lambda =\pm 1
$$
***
***
***
Given $(\text{B}-\lambda\mathbb{I})\mathbf{\underline{x}} = 0$, where $\mathbf{\underline{x}}=(x,y)$ is the eigenvector, solving for $x,y$ for each value of $\lambda$:
***
$$
\lambda =1, \qquad \left(\begin{array}{ c c }
-\frac{1}{2} & \frac{\sqrt{3}}{2}\\
\frac{\sqrt{3}}{2} & -\frac{3}{2}
\end{array}\right)\left(\begin{array}{ c }
x\\ y
\end{array}\right) =\left(\begin{array}{ c }
0\\ 0
\end{array}\right) \quad \rightarrow \quad -x+\sqrt{3}y=0
$$
***
$$
\lambda =-1, \qquad \left(\begin{array}{ c c }
\frac{3}{2} & \frac{\sqrt{3}}{2}\\
\frac{\sqrt{3}}{2} & \frac{1}{2}
\end{array}\right)\left(\begin{array}{ c }
x\\ y
\end{array}\right) =\left(\begin{array}{ c }
0\\ 0
\end{array}\right) \quad \rightarrow \quad -\sqrt{3} x+y=0
$$
***
Therefore, the two normalised eigenvectors are given by:
***
$$
\quad \rightarrow \quad x=\sqrt{3}y \quad \rightarrow \quad \vec{v}_{1}= \left(\begin{array}{ c }
\sqrt{3} \\ 1
\end{array}\right) \quad \rightarrow \quad \mathbf{\hat{v}}_{1} =\frac{1}{{2}}\left(\begin{array}{ c }
\sqrt{3}\\ 1
\end{array}\right)
$$
$$
\quad \rightarrow \quad y=-\sqrt{3}x \quad \rightarrow \quad \vec{v}_{2}= \left(\begin{array}{ c }
1 \\ -\sqrt{3}
\end{array}\right) \quad \rightarrow \quad \mathbf{\hat{v}}_{2} =\frac{1}{{2}}\left(\begin{array}{ c }
1 \\ -\sqrt{3}
\end{array}\right)
$$
The matrix operation is invariant along $\vec{v}_{1}= (\frac{\sqrt{3}}{2},\frac{1}{2})$ which is a line at $30^\circ$ to the $x$-axis but reflects everything normal to that.
\nSetting up and solving the characteristic equation $p(\lambda) = \det(\text{C}-\lambda\mathbb{I}_2)=0$ (**section 3.19**):
***
$$
p( \lambda ) \ =\ \left| \begin{array}{ c c }
-1-\lambda & 0\\
0 & -1-\lambda
\end{array}\right| =( \lambda +1)^{2} =0\ \quad \rightarrow \quad \lambda =-1
$$
***
***
***
Given $(\text{C}-\lambda\mathbb{I})\mathbf{\underline{x}} = 0$, where $\mathbf{\underline{x}}=(x,y)$ is the eigenvector, solving for $x,y$ for each value of $\lambda$:
***
$$
\lambda =-1,\ \quad \left(\begin{array}{ c c }
0 & 0\\
0 & 0
\end{array}\right)\left(\begin{array}{ c }
x\\
y
\end{array}\right) =\left(\begin{array}{ c }
0\\
0
\end{array}\right) \qquad
\rightarrow \quad 0x+0y=0
$$
***
Only trivial solutions exist.
***
Note that though this matrix can also be thought of as pair of reflections, it is also a rotation of $\theta=\pi$ and so has no eigenvectors.
\nSetting up and solving the characteristic equation $p(\lambda) = \det(\text{D}-\lambda\mathbb{I}_2)=0$ (**section 3.19**):
***
$$
p( \lambda ) \ =\ \left| \begin{array}{ c c }
4-\lambda & 1\\
-1 & 4-\lambda
\end{array}\right| =16-8\lambda +\lambda ^{2} +1=0
$$
***
***
***
This is a quadratic with $ b^{2} -4ac\ =\ 64-68\ =-4< 0 $, so there are no solutions. This is because the matrix is a rotation (angle $ \tan^{-1}\left(-\tfrac{1}{4}\right) $) with a scaling of $ \sqrt{17} $ in all directions.
\nSetting up and solving the characteristic equation $p(\lambda) = \det(\text{E}-\lambda\mathbb{I}_2)=0$ (**section 3.19**):
***
$$
p( \lambda ) \ =\ \left| \begin{array}{ c c }
1-\lambda & 1\\
1 & 1-\lambda
\end{array}\right| =\lambda ^{2} -2\lambda +1-1=0\qquad \rightarrow \qquad \lambda ^{2} -2\lambda =0\ \quad \rightarrow \quad \lambda =0,2
$$
***
***
***
Given $(\text{E}-\lambda\mathbb{I})\mathbf{\underline{x}} = 0$, where $\mathbf{\underline{x}}=(x,y)$ is the eigenvector, solving for $x,y$ for each value of $\lambda$:
***
$$
\lambda =0,\qquad \left(\begin{array}{ c c }
1 & 1\\
1 & 1
\end{array}\right)\left(\begin{array}{ c }
x\\
y
\end{array}\right) =\left(\begin{array}{ c }
0\\
0
\end{array}\right) \qquad \rightarrow \qquad x+y=0\qquad \rightarrow\qquad y=-x
$$
***
$$
\lambda =2,\qquad \left(\begin{array}{ c c }
-1 & 1\\
1 & -1
\end{array}\right)\left(\begin{array}{ c }
x\\ y
\end{array}\right) =\left(\begin{array}{ c }
0\\ 0
\end{array}\right)
\qquad \rightarrow \qquad x-y=0\qquad \rightarrow \qquad y=x
$$
***
Therefore, the two normalised eigenvectors are given by:
***
$$
\vec{v}_{1}=\qquad \left(\begin{array}{ c }
1\\-1
\end{array}\right) \qquad \rightarrow \qquad \mathbf{\hat{v}}_{1} =\frac{1}{\sqrt{2}}\left(\begin{array}{ c }
1\\
-1
\end{array}\right)
$$
$$
\vec{v}_{2} = \left(\begin{array}{ c }
1\\ 1
\end{array}\right)\qquad \rightarrow \qquad \mathbf{\hat{v}}_{2} =\frac{1}{\sqrt{2}}\left(\begin{array}{ c }
1\\ 1
\end{array}\right)
$$
***
This collapses all points onto the line $ x=y\ $ ($\lambda_{1}=0$), and scales by a factor of 2 along that line ($\lambda_{1}=2$). **Note**:This is a singular matrix as one can attest by noting that $\det(D)=0$
| Find the *real* eigenvalues and *normalised* eigenvectors of the following transformation matrices in $\mathbb{R}^{2}$, and use the information to identify the transformations.
***
**Note:** In the response area, type the same answer twice for repeated solutions. Type $\begin{pmatrix}n\\n\end{pmatrix}$ for no eigenvector.
$$
\text{A}=\left(
\begin{array}{cc}
0&\hskip6pt 1\\
1&\hskip6pt 0
\end{array}
\right)\quad
$$
\n$$
\text{B}=\left(
\begin{array}{cc}
\frac{{1}}{2}& \frac{\sqrt{3}}{2} \\
\frac{\sqrt{3}}{2}&\hskip4pt -\frac{{1}}{2}
\end{array}
\right)\quad
$$
\n$$
\text{C}=\left(
\begin{array}{cc}
-1&\hskip7pt 0\\
\hskip7pt 0& -1
\end{array}
\right)\quad
$$
\n$$
\text{D}=\left(
\begin{array}{rr}
4&\hskip3pt 1\\
-1&\hskip6pt 4
\end{array}
\right)
$$
\n$$
\text{E}=\left(
\begin{array}{rr}
1&\hskip6pt 1\\
1&\hskip6pt 1
\end{array}\right)
$$
| 51 | 7 | 54 | 54 | 360 | 40 | 9 | 5 | 0 | Find the real eigenvalues and normalised eigenvectors of the following transformation matrices in $\mathbb{R}^{2}$, and use the information to identify the transformations. | 1 |
4e709bc9-f407-49dc-8e88-2f20a9da288b | 1 | 0 | 0 | 16 | 6 | 1 | 5 | 5 | The horizontal range $R$ of a projectile is given by $\displaystyle R = \left({{U^2} {\sin{2\alpha}}} \over {g}\right)$, 
where $U$ is the projection speed, $\alpha$ is the angle of elevation and $g$ is the gravitational acceleration.
If $U, \alpha$ are each known to an accuracy of $\pm 0.1\%$ and $g$ is exact, find an expression for the resulting accuracy bounds for $R$ when $\displaystyle \alpha= {{\pi}\over3}$.
| The horizontal range $R$ of a projectile is given by $\displaystyle R = \left({{U^2} {\sin{2\alpha}}} \over {g}\right)$, 
where $U$ is the projection speed, $\alpha$ is the angle of elevation and $g$ is the gravitational acceleration.
If $U, \alpha$ are each known to an accuracy of $\pm 0.1\%$ and $g$ is exact, find an expression for the resulting accuracy bounds for $R$ when $\displaystyle \alpha= {{\pi}\over3}$.
| 1 | 0.666667 | 1 | Find $\delta R$ to first order in $\delta U, \delta\alpha$. Use the total differential of $R$.
***
Using the total differential of $R$,
$$
\begin{aligned}
\delta R &\simeq {\partial R\over\partial U}\delta U + {\partial R\over\partial\alpha}\delta\alpha
\end{aligned}
$$
Can you find $\delta R$ now to first order?
***
Find $\displaystyle {\delta R\over R}$ and substitute the appropriate values to find the worst case.
What values will $\delta U$ and $\delta \alpha$ take for the worst case?
***
The bounds requested are for the worst case, when $\delta U, \delta\alpha$ have opposite signs. Can you find the accuracy bounds for $R$ now?
| Find $\delta R$ to first order in $\delta U, \delta\alpha$. Use the total differential of $R$.
***
$$
\displaystyle R = {U^2 \sin(2\alpha) \over g}
$$
Using the total differential of $R$,
$$
\begin{aligned}
\delta R &\simeq {\partial R\over\partial U}\delta U + {\partial R\over\partial\alpha}\delta\alpha
\end{aligned}
$$
Can you find $\delta R$ now to first order?
***
$$
\begin{aligned}
\delta R
&\simeq {2U\delta U\over g}\sin(2\alpha) + {2U^2\over g}\cos(2\alpha)\delta \alpha
\end{aligned}
$$
***
Find $\displaystyle {\delta R\over R}$ and substitute the appropriate values to find the worst case.
***
$$
\begin{aligned}
{\delta R\over R} \simeq 2\left({\delta U\over U}\right) + 2\alpha \cot(2\alpha) \left({\delta\alpha\over\alpha}\right)
\end{aligned}
$$
Substitute in $\displaystyle \alpha = {\pi\over 3}$:
$$
\begin{aligned}
{\delta R\over R} \simeq 2\left({\delta U\over U}\right) -{2\pi\over3\sqrt{3}}\left({\delta\alpha\over\alpha}\right)
\end{aligned}
$$
What values will $\delta U$ and $\delta \alpha$ take for the worst case?
***
The bounds requested are for the worst case, when $\delta U, \delta\alpha$ have opposite signs. Can you find the accuracy bounds for $R$ now?
***
$$
\begin{aligned}
\left|{\delta R\over R }\right|_\mathrm{max\,estimate}
&\simeq 2(0.001) - {2\pi\over3\sqrt{3}}(-0.001)
\\
&\simeq \left(2+{2\pi\over3\sqrt{3}}\right)(0.1)\%
\\
{\delta R\over R} &\leq 0.321 \%
\end{aligned}
$$
How likely is this worst case?
This rather depends on our assessment of the underlying **distributions** of $U$, $\alpha$ uncertainties... Certainly if we expect them to be separately **Gaussian** and **uncorrelated**, then we should **expect** the uncertainty in $\displaystyle{\delta R\over R}$ to be (say) $\left[2^2 +({2\pi\over3\sqrt{3}})^2\right]^{1\over2}(0.1)\% = 0.234\%$ (see measurement and uncertainties lectures).
Note: Similarly, adding up a column of $N$ figures, each rounded to the $d$th decimal place $\Rightarrow$ **Worst case** total error $\simeq$ $N(0.5)\times 10^{-d}$ whereas **Expected** error $\simeq N^{1\over2}(0.5)\times 10^{-d}$ in a typical case.
| The horizontal range $R$ of a projectile is given by $\displaystyle R = \left({{U^2} {\sin{2\alpha}}} \over {g}\right)$, 
where $U$ is the projection speed, $\alpha$ is the angle of elevation and $g$ is the gravitational acceleration.
If $U, \alpha$ are each known to an accuracy of $\pm 0.1\%$ and $g$ is exact, find an expression for the resulting accuracy bounds for $R$ when $\displaystyle \alpha= {{\pi}\over3}$.
| 57 | 10 | 27 | 27 | 172 | 11 | 57 | 10 | 0 | The horizontal range $R$ of a projectile is given by $\displaystyle R = \left({{U^2} {\sin{2\alpha}}} \over {g}\right)$, where $U$ is the projection speed, $\alpha$ is the angle of elevation and $g$ is the gravitational acceleration. If $U, \alpha$ are each known to an accuracy of $\pm 0.1\%$ and $g$ is exact, find an expression for the resulting accuracy bounds for $R$ when $\displaystyle \alpha= {{\pi}\over3}$. | 2 |
4e891748-7f57-4df8-9ccb-8e5efb042ee1 | 3 | 0 | 0 | 13 | 4 | 2 | 2 | 0 | 
A weight of $3.2~\mathrm{kN}$ falls $2~\mathrm{m}$ onto a pile weighing $2.4~\mathrm{kN}$. Assuming that the weight and the pile move together after the impact, find:
| The velocity, $v$, with which they begin to move.
\nThe mean resistance of the earth to penetration, $F$, if the pile is driven $20~mm$ into the ground by the blow.
\nThe energy lost in the impact, $\Delta KE$.
| 3 | 0.333333 | 3 | This is a tough one – you’ll have to think back to your school physics lessons! If you happened to miss that particular lesson, check section 3.2 in your notes for the necessary explanation and equations you need. You may want to use the energy method to calculate the velocity of the falling weight right before impact.
\nThe energy method is also useful for solving this part, looking only at the energy directly AFTER impact. Don’t forget to take into account the potential energy added to the system by falling into the ground! Remember, the answer required is a force.
\nThis is why we considered only the energy after impact in part b). You’ll discover a discrepancy between the energy before and after impact – that’s your answer.
| Employing the following SUVAT equation:
$$
v_1^2 = v_0^2 + 2as
$$
***
Rearranging:
$$
v_1 = \sqrt{v_0^2 + 2as}
$$
***
Substituting the values of the parameters in:
$$
v_1 = \sqrt{2g \times 2} = 2\sqrt{g}~~\mathrm{(Equation~4)}
$$
***
Substituting Equation 4 into Equation 3:
$$
v_2 = \cfrac{m_w}{m_w + m_p}\times2\sqrt{g}
$$
***
Noting that $v_2$ is the velocity with which the pile and weight begin to move after the impact:
$$
v = v_2
$$
***
Substituting the values of the parameters gives:
$$
v = \cfrac{(3.2/g)}{(3.2/g) + (2.4/g)} \times 2\sqrt{g} = 3.58~\mathrm{m/s}
$$
, First, an equation needs to be derived which relates the initial velocity of the weight to the final velocity of the weight and pile moving together. This can be done by applying either *impulse-momentum* equation to the system, or *conservation of momentum* equations to the system.
, Employing the following SUVAT equation:
$$
v_1^2 = v_0^2 + 2as
$$
***
Rearranging:
$$
v_1 = \sqrt{v_0^2 + 2as}
$$
***
Substituting the values of the parameters in:
$$
v_1 = \sqrt{2g \times 2} = 2\sqrt{g}~~\mathrm{(Equation~4)}
$$
***
Substituting Equation 4 into Equation 3:
$$
v_2 = \cfrac{m_w}{m_w + m_p}\times2\sqrt{g}
$$
***
Noting that $v_2$ is the velocity with which the pile and weight begin to move after the impact:
$$
v = v_2
$$
Substituting the values of the parameters gives:
$$
v = \cfrac{(3.2/g)}{(3.2/g) + (2.4/g)} \times 2\sqrt{g} = 3.58~\mathrm{m/s}
$$
, $$
\begin{aligned}
\text{Work done} &= \text{Increase in } KE + \text{Increase in } PE \\
0 &= \cfrac{1}{2}~ m_w (v_1^2 - v_0^2) + m_wg (h_1 - h_0) \\
0 &= \cfrac{1}{2}~ m_w v_1^2 + m_wgh_1
\end{aligned}
$$
***
Rearranging:
$$
v_1 = \sqrt{2gh_1}
$$
***
Substituting the values of the parameters gives:
$$
v_1 = \sqrt{2g \times 2} = 2\sqrt{g} ~~\mathrm{(Equation~4)}
$$
***
Substituting Equation 4 into Equation 3:
$$
v_2 = \cfrac{m_w}{m_w + m_p}\times2\sqrt{g}
$$
***
Noting that $v_2$ is the velocity with which the pile and weight begin to move after the impact:
$$
v = v_2
$$
***
Substituting the values of the parameters gives:
$$
v = \cfrac{(3.2/g)}{(3.2/g) + (2.4/g)} \times 2\sqrt{g} = 3.58~\mathrm{m/s}
$$
, Apply conservation of momentum to the system:
***
$$
m_wv_1 + 0 = (m_w+m_p)v_2
$$
Rearranging:
$$
v_2 = \cfrac{m_w}{m_w+m_p}~v_1~~\mathrm{(Equation~1)}
$$
***
Now the velocity of the weight at the point of impact, $v_1$, needs to be calculated. This can be done by either applying the conservation of energy equation, or by using SUVAT equations.
, $$
\begin{aligned}
\text{Work done} &= \text{Increase in } KE + \text{Increase in } PE \\
0 &= \cfrac{1}{2}~ m_w (v_1^2 - v_0^2) + m_wg (h_1 - h_0) \\
0 &= \cfrac{1}{2}~ m_w v_1^2 + m_wgh_1
\end{aligned}
$$
***
Rearranging:
$$
v_1 = \sqrt{2gh_1}
$$
***
Substituting the values of the parameters gives:
$$
v_1 = \sqrt{2g\times 2} = 2\sqrt{g}~~\mathrm{(Equation~2)}
$$
***
Substituting Equation 2 into Equation 1:
$$
v_2 = \cfrac{m_w}{m_w + m_p}\times2\sqrt{g}
$$
***
Noting that $v_2$ is the velocity with which the pile and weight begin to move after the impact:
$$
v = v_2
$$
***
Substituting the values of the parameters gives:
$$
v = \cfrac{(3.2/g)}{(3.2/g) + (2.4/g)} \times 2\sqrt{g} = 3.58~\mathrm{m/s}
$$
, Apply impulse-momentum equations for translational motion to each body.
***
Consider the weight:
$$
\int -F \cdot dt = m_w(v_2 - v_1)~~\mathrm{(Equation~1)}
$$
***
Consider the pile:
$$
\int F \cdot dt = m_p(v_2 - 0)~~\mathrm{(Equation~2)}
$$
***
Sum of Equations 1 and 2:
$$
\begin{aligned}
\int -F \cdot dt + \int F \cdot dt &= m_w(v_2 - v_1) + m_p(v_2 - 0) \\
0 &= v_2(m_w + m_p) - m_wv_1
\end{aligned}
$$
***
Rearranging:
$$
v_2 = \cfrac{m_w}{m_w + m_p}~v_1~~\mathrm{(Equation~3)}
$$
***
Now the velocity of the weight at the point of impact, $v_1$, needs to be calculated. This can be done by either applying the conservation of energy equation, or by using SUVAT equations.
\nApplying conservation of energy:
$$
\text{Work done} = \text{Increase in } KE + \text{Increase in } PE
$$
***
There is no rotational motion, only translational, and using the earth surface as the datum point:
$$
Fd = \cfrac{1}{2}~(m_w + m_p)(0 − v_2^2) + (m_w + m_p)g(d − 0)
$$
***
Substituting the values of the parameters ($d = -0.02$) gives:
$$
F = \cfrac{\cfrac{1}{2}~\left(\cfrac{3.2}{g} + \cfrac{2.4}{g}\right)(0 − 3.58^2)+\left(\cfrac{3.2}{g} + \cfrac{2.4}{g}\right)\times g \times(−0.02 - 0)}{-0.02}
$$
$$
F = 188.6~\mathrm{kN}
$$
\nKinetic energy at the point of impact:
$$
KE_{\text{at impact}} = \cfrac{1}{2}~m_w v_1^2
$$
***
Kinetic energy after impact:
$$
KE_{\text{after impact}} = \cfrac{1}{2}~ (m_w + m_p) v_2^2
$$
***
Therefore the energy loss $\Delta KE$ is given by:
$$
\Delta KE = \left[\cfrac{1}{2}~m_w v_1^2\right] - \left[\cfrac{1}{2}~ (m_w + m_p) v_2^2\right]
$$
***
Substituting the values of the parameters gives:
$$
\Delta KE = \left[\cfrac{1}{2} \times \cfrac{3.2}{g} \times 6.26^2\right] − \left[\cfrac{1}{2}~\left(\cfrac{3.2}{g} + \cfrac{2.4}{g}\right) \times 3.58^2\right]
$$
$$
\Delta KE = 2.74~\mathrm{kJ}
$$
| 
A weight of $3.2~\mathrm{kN}$ falls $2~\mathrm{m}$ onto a pile weighing $2.4~\mathrm{kN}$. Assuming that the weight and the pile move together after the impact, find:
The velocity, $v$, with which they begin to move.
\nThe mean resistance of the earth to penetration, $F$, if the pile is driven $20~mm$ into the ground by the blow.
\nThe energy lost in the impact, $\Delta KE$.
| 66 | 7 | 47 | 47 | 444 | 0 | 40 | 4 | 1 | Assuming that the weight and the pile move together after the impact, find: The velocity, $v$, with which they begin to move. | 1 |
4f074b46-971b-4a1d-976c-cfd514e23116 | 2 | 0 | 3 | 22 | 6 | 1 | 0 | 4 | A damped harmonic oscillator has the equation of motion
$$
\ddot{\psi} + \gamma \dot{\psi} + \omega_0^2 \psi = 0.
$$
| When $\gamma/2 < \omega_0$, the damping is light. Show, by substitution, that for light damping the solution is
$$
\psi(t) = A e^{-\gamma t /2}\cos(\omega_{\rm d} t + \phi),
$$
where $\omega_{\rm d} = \sqrt{\omega_0^2 - \gamma^2/4}$.
\nThe amplitude decay time, $\tau_A$, is the time taken for the amplitude of the oscillations to reduce by a factor of $e$. The energy decay time, $\tau_E$, is the time taken for the energy of the oscillations to reduce by a factor of $e$. Find expressions for $\tau_A$ and $\tau_E$ (for light damping).
\nThe quality factor is $Q = \omega_0/\gamma$. For very light damping ($Q \gg 1$), show the following:
* (i) $\frac{\omega_0 - \omega_{\rm d}}{\omega_0} \approx \frac{1}{8Q^2}$.
* (ii) The number of oscillations for the amplitude to fall by a factor of $e$ is $N_e\approx Q/\pi$.
* (iii) $\frac{|\Delta E|}{E}\approx \frac{2\pi}{Q}$ where $E$ is the energy and $\Delta E$ is the energy lost in one period of oscillation.
\nCritical damping is defined by $\gamma/2 = \omega_0$. Show, by substitution, that in the case of critical damping, the solution is
$$
\psi = (A + B t)e^{-\gamma t /2}.
$$
| 4 | 0.666667 | 3 | Find the derivatives $\dot{\psi}$ and $\ddot{\psi}$.
***
You should get
$\dot{\psi} = A e^{-\gamma t /2}\left[-\omega_{\rm d}\sin(\omega_{\rm d} t + \phi) -\frac{\gamma}{2}\cos(\omega_{\rm d} t + \phi)\right]$
and
$\ddot{\psi} = A e^{-\gamma t/2}\left[-\omega_{\rm d}^2\cos(\omega_{\rm d} t + \phi) + 2 \frac{\gamma}{2}\omega_{\rm d} \sin(\omega_{\rm d} t + \phi) + \frac{\gamma^2}{4}\cos(\omega_{\rm d} t + \phi)\right]$  
***
Using $\psi$, $\dot{\psi}$ and $\ddot{\psi}$, form the left hand side of the equation, and simplify. 
***
You should get
$A e^{-\gamma t/2} \left[\omega_0^2 - \frac{\gamma^2}{4}-\omega_{\rm d}^2\right] \cos(\omega_{\rm d} t + \phi)$
***
How can you make this zero?
\nCalculate the amplitude at the beginning. Then find the value of $t$ needed to make the amplitude a factor of $e$ smaller
***
The energy is proportional to the square of the amplitude
\n(i) Start from the expression for $\omega_{\rm d}$ given in the question. Factor out $\omega_0$. Rewrite in terms of $Q$.
***
At this point, you should have $\omega_{\rm d} = \omega_0\sqrt{1-\frac{1}{4Q^2}}$. Expand this result for large $Q$, remembering that $(1+x)^{1/2} \approx 1+\frac{1}{2}x$ for small $x$. 
***
(ii) From part (a) you should already have $\tau_A = 2/\gamma$. The period is $T=2\pi/\omega_{\rm d}$ and $\omega_{\rm d}$ is very close to $\omega_0$ when $Q$ is large.
***
The quantity needed is the number of oscillations in time $\tau_A$, i.e. $N_e = \tau_A/T$.
***
(iii) The energy drops as $E=E_0 e^{-\gamma t}$.
***
Calculate $\frac{dE}{dt}$, and thus find an expression for $\Delta E$. You should find that you can write this as $\Delta E = - \gamma E T$. You can reach the desired result from here.
\nCalculate $\dot{\psi}$ and $\ddot{\psi}$.
***
Substitute into the equation of motion
***
Is it zero?
| Find the derivatives $\dot{\psi}$ and $\ddot{\psi}$:
***
$\psi = A e^{-\gamma t /2}\cos(\omega_{\rm d} t + \phi)$
$\dot{\psi} = A e^{-\gamma t /2}\left[-\omega_{\rm d}\sin(\omega_{\rm d} t + \phi) -\frac{\gamma}{2}\cos(\omega_{\rm d} t + \phi)\right]$
$\ddot{\psi} = A e^{-\gamma t/2}\left[-\omega_{\rm d}^2\cos(\omega_{\rm d} t + \phi) + 2 \frac{\gamma}{2}\omega_{\rm d} \sin(\omega_{\rm d} t + \phi) + \frac{\gamma^2}{4}\cos(\omega_{\rm d} t + \phi)\right]$
***
Substitute these into the equation of motion:
***
$$
\begin{aligned}
\ddot{\psi} + \gamma \dot{\psi} + \omega_0^2 \psi &= A e^{-\gamma t/2} [-\omega_{\rm d}^2 \cos(\omega_{\rm d} t + \phi) + \gamma \omega_{\rm d} \sin(\omega_{\rm d} t + \phi)+\frac{\gamma^2}{4}\cos(\omega_{\rm d} t + \phi) \nonumber\\ &- \gamma\omega_{\rm d} \sin(\omega_{\rm d} t + \phi) - \frac{\gamma^2}{2}\cos(\omega_{\rm d} t + \phi) + \omega_0^2 \cos(\omega_{\rm d} t + \phi)] \nonumber \\
&= A e^{-\gamma t/2} \left[\omega_0^2 - \frac{\gamma^2}{4}-\omega_{\rm d}^2\right] \cos(\omega_{\rm d} t + \phi).\nonumber
\end{aligned}
$$
***
For the equation of motion to be satisfied, this last expression needs to be equal to zero. 
***
This means that the part in square brackets needs to be zero. 
***
Thus, the equation is satisfied provided $\omega_{\rm d} = \sqrt{\omega_0^2 - \gamma^2/4}$.
\nAt $t=0$ the amplitude of the oscillations is $A$. When $\gamma t = 2$ the amplitude is $A e^{-1}$...
***
...thus $\tau_A = 2/\gamma$.
***
The energy of the oscillations is proportional to the square of the amplitude...
***
... i.e. to $A^2 e^{-\gamma t}$. 
***
This is $A^2$ at $t=0$ and has fallen to $A^2 e^{-1}$ when $\gamma t =1$...
***
 Thus, $\tau_E = 1/\gamma$.
\n(i) $ \omega_{\rm d} = \omega_0\sqrt{1-\frac{\gamma^2}{4\omega_0^2}} $
***
$= \omega_0\sqrt{1-\frac{1}{4Q^2}}$
***
Now use the binomial expansion:
***
$ \omega_d\approx \omega_0\left(1-\frac{1}{8Q^2}\right) $
***
$\therefore \frac{\omega_0 - \omega_{\rm d}}{\omega_0} \approx \frac{1}{8Q^2}$
***
(ii) We already found above that the time for the amplitude to fall by $e$ is $\tau_A = 2/\gamma$. 
The period is 
***
$T=2\pi/\omega_d \approx 2\pi/\omega_0$. 
***
So $N_e$ is
***
 $N_e = \tau_A/T \approx 2\omega_0/(2\pi\gamma) \approx Q/\pi$
***
(iii) As above, energy drops as $E=E_0 e^{-\gamma t}$. So $\frac{d E}{d t}$ is
 $\frac{d E}{d t} = -\gamma E_0 e^{-\gamma t} = -\gamma E$.
***
 $\Delta E$ is the energy change in one period, so we have $\Delta E = - \gamma E T = -\gamma E 2\pi/\omega_{\rm d} \approx -\gamma E 2\pi/\omega_0.$ 
***
This gives
$$
\frac{|\Delta E|}{E} \approx \gamma \frac{2\pi}{\omega_0} = \frac{2\pi}{Q}.
$$
\n$\psi = (A + B t)e^{-\gamma t /2}$
$\dot{\psi}=\left(-A \frac{\gamma}{2}-B t \frac{\gamma}{2}+B\right)e^{-\gamma t/2}$
***
$\ddot{\psi} = \left( A \frac{\gamma^2}{4} + B t \frac{\gamma^2}{4}-B \frac{\gamma}{2} - B \frac{\gamma}{2}\right)e^{-\gamma t/2}$
***
$\ddot{\psi} + \gamma \dot{\psi} + \omega_0^2 \psi = \ddot{\psi} + \gamma \dot{\psi} + \frac{\gamma^2}{4} \psi = \left(A\frac{\gamma^2}{4} + B t \frac{\gamma^2}{4} - B \gamma - A\frac{\gamma^2}{2} - B t \frac{\gamma^2}{2} + B\gamma + A \frac{\gamma^2}{4} + B t \frac{\gamma^2}{4} \right)e^{-\gamma t/2}$.
In the last expression all the terms in the round bracket cancel, leaving zero. So this is indeed the solution.
| A damped harmonic oscillator has the equation of motion
$$
\ddot{\psi} + \gamma \dot{\psi} + \omega_0^2 \psi = 0.
$$
When $\gamma/2 < \omega_0$, the damping is light. Show, by substitution, that for light damping the solution is
$$
\psi(t) = A e^{-\gamma t /2}\cos(\omega_{\rm d} t + \phi),
$$
where $\omega_{\rm d} = \sqrt{\omega_0^2 - \gamma^2/4}$.
\nThe amplitude decay time, $\tau_A$, is the time taken for the amplitude of the oscillations to reduce by a factor of $e$. The energy decay time, $\tau_E$, is the time taken for the energy of the oscillations to reduce by a factor of $e$. Find expressions for $\tau_A$ and $\tau_E$ (for light damping).
\nThe quality factor is $Q = \omega_0/\gamma$. For very light damping ($Q \gg 1$), show the following:
* (i) $\frac{\omega_0 - \omega_{\rm d}}{\omega_0} \approx \frac{1}{8Q^2}$.
* (ii) The number of oscillations for the amplitude to fall by a factor of $e$ is $N_e\approx Q/\pi$.
* (iii) $\frac{|\Delta E|}{E}\approx \frac{2\pi}{Q}$ where $E$ is the energy and $\Delta E$ is the energy lost in one period of oscillation.
\nCritical damping is defined by $\gamma/2 = \omega_0$. Show, by substitution, that in the case of critical damping, the solution is
$$
\psi = (A + B t)e^{-\gamma t /2}.
$$
| 167 | 20 | 37 | 37 | 228 | 30 | 157 | 19 | 0 | Find expressions for $\tau_A$ and $\tau_E$ for light damping. For very light damping $\gamma/2 < \omega_0$1, show the following: i $\gamma/2 < \omega_0$2. | 2 |
500b8b88-75d7-4d91-8703-ec76a9cadce6 | 10 | 0 | 1 | 16 | 6 | 1 | 7 | 3 | \
$f(x, y) = {(x + 2y)} \cos {(2x + y)}$.
| Find all the partial derivatives required for a third-order Taylor-Maclaurin expansion of $f(x, y)$.
\nHence find the expansion of this function to cubic terms about the origin $(0, 0)$.
\nCheck your results using the usual small-angle expansion of the cosine function for a single variable.
| 3 | 0.666667 | 3 | For a third-order Taylor Maclaurin expansion, you will need to find the third-order derivatives, up to terms such as $\displaystyle {\partial ^3 f \over\partial x^3}$.
$$
f(x,y) = (x+2y)\cos(2x+y)
$$
First, find $\displaystyle f_x = {\partial f\over\partial x}$ and $\displaystyle f_y = {\partial f\over\partial y}$. You can use the product rule.
***
Now using $\displaystyle f_{xx} = {\partial\over\partial x}({\partial f\over\partial x} )$ find the second-order derivatives, again using the product rule.
\[Remember that $\displaystyle {\partial^2f\over \partial x\partial y} = {\partial^2 f\over\partial y\partial x}$]
***
Now find the third-order derivatives by finding the derivatives of the above with respect to $x$ and $y$.
\[Remember that $\displaystyle {\partial^3f\over \partial x\partial y\partial y} = {\partial^3f\over \partial y\partial x\partial y} = {\partial^3\over \partial y\partial y\partial x}$ and similarly for $\displaystyle {\partial^3f\over \partial x\partial x\partial y}$]
\nThis is a Maclaurin expansion, since it is around the origin. The form of a Maclaurin expansion is (look at section 5.4 on double Taylor series) :
$$
\begin{aligned}
f(x,y) = u_0 + [x({\partial f\over\partial x})_0 + y({\partial f\over\partial y})_0] + {1\over 2!}[x^2({\partial^2f\over\partial x^2})_0 + 2xy({\partial^2 f \over \partial x \partial y})_0 + y^2({\partial^2f\over\partial y^2})_0 ] +\ldots
\end{aligned}
$$
Can you find this expansion now?
***
As the expansion is around the origin, substitute $x=0, y=0$ into the derivatives found in part (a), and then substitute into the above equation.
\nThe small angle approximation of $\cos \theta$, to third order in $\theta$, is
$$
\displaystyle \cos\theta \approx 1-{\theta^2\over2}
$$
Can you check your result now?
***
Use $\theta = 2x+y$ to find the small angle approximation of $\cos(2x+y)$. Then substitute this to find an approximation for $f(x,y)$. Does it agree with your answer from part (b)?
***
Try expanding this. Does it agree with your previous answer?
| For a third-order Taylor Maclaurin expansion, you will need to find the third-order derivatives, up to terms such as $\displaystyle {\partial ^3 f \over\partial x^3}$.
$$
f(x,y) = (x+2y)\cos(2x+y)
$$
First, find $\displaystyle f_x = {\partial f\over\partial x}$ and $\displaystyle f_y = {\partial f\over\partial y}$. You can use the product rule.
***
$$
f_x = \cos(2x+y) - 2(x+2y)\sin(2x+y) \\
f_y = 2\cos(2x+y)-(x+2y)\sin(2x+y)
$$
Can you find all the derivatives now? 
***
Now using $ \displaystyle f_{xx} = {\partial\over\partial x}\left({\partial f\over\partial x}\right) $ find the second-order derivatives, again using the product rule.
$\bigg($Remember that $\displaystyle {\partial^2f\over \partial x\partial y} = {\partial^2 f\over\partial y\partial x}\bigg)$
***
$$
\begin{aligned}
f_{xx} ={\partial^2f\over\partial x^2}
&= -2\sin(2x+y) - 2\sin(2x+2y) -4(x+2y)\cos(2x+y) \\
&= -4\sin(2x+y)-4(x+2y)\cos(2x+y) \\
f_{xy} = f_{yx} = {\partial^2f\over\partial x\partial y}
&= -4\sin(2x+y) -\sin(2x+y) -2(x+2y)\cos(2x+y) \\
&= -5\sin(2x+y)-2(x+2y)\cos(2x+y) \\
f_{yy} = {\partial^2f\over\partial y^2}
&= -2\sin(2x+y) - 2\sin(2x+y) - (x+2y)\cos(2x+y) \\
&= -4\sin(2x+y)-(x+2y)\cos(2x+y)
\end{aligned}
$$
Can you find all the derivatives now?
***
Now find the third-order derivatives by finding the derivatives of the above with respect to $x$ and $y$.
\[Remember that $\displaystyle {\partial^3f\over \partial x\partial y\partial y} = {\partial^3f\over \partial y\partial x\partial y} = {\partial^3\over \partial y\partial y\partial x}$ and similarly for $\displaystyle {\partial^3f\over \partial x\partial x\partial y}$]
***
$$
\begin{aligned}
f_{xxx} &= -8\cos(2x+y) -4\cos(2x+y) +8(x+2y)\sin(2x+y) \\
&= -12\cos(2x+y)+8(x+2y)\sin(2x+y) \\
f_{yxx} = &= -12\cos(2x+y)+4(x+2y)\sin(2x+y) \\
f_{xyy} &= -9\cos(2x+y)+2(x+2y)\sin(2x+y) \\
f_{yyy} &= -6\cos(2x+y)+(x+2y)\sin(2x+y)
\end{aligned}
$$
\nThis is a Maclaurin expansion since it is around the origin. The form of a Maclaurin expansion is (look at section 5.4 on double Taylor series) :
$$
\begin{aligned}
f(x,y) = u_0 +
\left[x\left({\partial f\over\partial x}\right)_0 + y\left({\partial f\over\partial y}\right)_0\right]
+
{1\over 2!}\left[x^2\left({\partial^2f\over\partial x^2}\right)_0 + 2xy\left({\partial^2 f \over \partial x \partial y}\right)_0 + y^2\left({\partial^2f\over\partial y^2}\right)_0 \right] +\ldots
\end{aligned}
$$
Can you find this expansion now?
***
As the expansion is around the origin, substitute $x=0, y=0$ into the derivatives found in part (a), and then substitute into the above equation.
***
$$
\begin{aligned}
f|_0 &= 0 \\
f_x|_0 &= \cos(2x+y)-2(x+2y)\sin(2x+y) \Big|_0 = 1 \\
f_y|_0 &= 2\cos(2x+y)-(x+2y)\sin(2x+y)\Big|_0 = 2\\
f_{xx}|_0 &= -4\sin(2x+y)-4(x+2y)\cos(2x+y)\Big|_0 = 0\\
f_{xy}|_0 = f_{yx} &= -5\sin(2x+y)-2(x+2y)\cos(2x+y)\Big|_0 =0\\
f_{yy}|_0 &= -4\sin(2x+y)-(x+2y)\cos(2x+y)\Big|_0 =0\\
f_{xxx}|_0 &= -12\cos(2x+y)+8(x+2y)\sin(2x+y)\Big|_0 =-12\\
f_{yxx}|_0 &= -12\cos(2x+y)+4(x+2y)\sin(2x+y)\Big|_0 =-12 \\
f_{xyy}|_0 &= -9\cos(2x+y)+2(x+2y)\sin(2x+y)\Big|_0 =-9\\
f_{yyy}|_0 &= -6\cos(2x+y)+(x+2y)\sin(2x+y)\Big|_0 =-6
\end{aligned}
$$
Now substitute these in,
***
$$
\begin{aligned}
f(x,y) &= 0+ {(1x+2y)\over 1!} + {0x^2+2(0)xy+0y^2\over 2!} + {(-12x^3+3(-12x^2y)+3(-9xy^2)-6y^3)\over 3!} + \ldots
\\
&= (x+2y) + \left(-2x^3-6x^2y-{9\over2}xy^2-y^3\right)+\ldots
\end{aligned}
$$
\nThe small angle approximation of $\cos \theta$ to third order in $\theta$ is 
$$
\displaystyle \cos\theta \approx 1-{\theta^2\over2}
$$
Can you check your result now?
***
Use $\theta = 2x+y$ to find the small angle approximation of $\cos(2x+y)$. Then substitute this to find an approximation for $f(x,y)$. Does it agree with your answer from part (b)?
***
$$
\displaystyle \cos(2x+y) \approx 1-{(2x+y)^2\over2}
$$
Now substituting this,
$$
\displaystyle f(x,y) \approx (x+2y)\left[1-{(2x+y)^2\over2}\right]
$$
Try expanding this. Does it agree with your previous answer?
***
$$
\begin{aligned}
f(x,y) &\approx (x+2y)-{1\over2}(x+2y)(4x^2+4xy+y^2) \\
&\approx (x+2y)-{1\over2}(4x^3+4x^2y+xy^2+8x^2y+8xy^2+2y^3) \\
&\approx (x+2y)-{1\over2}(4x^3+12x^2y+9xy^2+2y^3) \\
&\approx (x+2y)+\left(-2x^3-6x^2y-{9\over2}xy^2-y^3\right)
\end{aligned}
$$
which gives the same result as before to the third order.
| \
$f(x, y) = {(x + 2y)} \cos {(2x + y)}$.
Find all the partial derivatives required for a third-order Taylor-Maclaurin expansion of $f(x, y)$.
\nHence find the expansion of this function to cubic terms about the origin $(0, 0)$.
\nCheck your results using the usual small-angle expansion of the cosine function for a single variable.
| 48 | 3 | 27 | 27 | 244 | 18 | 45 | 2 | 0 | Find all the partial derivatives required for a third-order Taylor-Maclaurin expansion of $f(x, y)$. Hence find the expansion of this function to cubic terms about the origin $(0, 0)$. Check your results using the usual small-angle expansion of the cosine function for a single variable. | 3 |
50141b29-3919-4963-a5ed-1fb35af93870 | 1 | 0 | 0 | 24 | 6 | 1 | 1 | 2 | **\[Boas 5.5.3]:** Find the area of the paraboloid $x^2 + y^2 = z$ inside the cylinder $x^2 + y^2 = 9$. 
***
*Hint:* Don't use the surface of revolution formula here (instead, perform a double integral). Leave your answer in fractional form, or as a number to 1 decimal place.
| **\[Boas 5.5.3]:** Find the area of the paraboloid $x^2 + y^2 = z$ inside the cylinder $x^2 + y^2 = 9$. 
***
*Hint:* Don't use the surface of revolution formula here (instead, perform a double integral). Leave your answer in fractional form, or as a number to 1 decimal place.
| 1 | 0.666667 | 2 | The surface area of the paraboloid within the cylinder is given by $\iint_S{|d\vec{S}|}$, the sum of the magnitude of the surface area elements. Start by trying to understand the geometry of the problem...
***
... First, what is the shape of the paraboloid? ...
***
... If $x^2+y^2=z$, when $z=0$ the paraboloid is at the origin $(0,0)$. For a fixed $z$ (e.g., $z=1,2,...$) what is the equation of the paraboloid? As $z$ increases, the paraboloid expands outwards. 
***
The cylinder defines the boundary of the paraboloid. 
***
Due to the cylindrical symmetry of this problem, the best way to approach the surface integral is using *Cylindrical coordinates*.
***
Find the position vector of a point on the paraboloid in cylindrical coordinates:
$$
\vec{r} = \rho\boldsymbol{\hat{\rho}}(\phi) + z\mathbf{\hat{k}}
$$
Make this a function of two variables only by using the equation of the paraboloid in cylindrical form.
***
Next, find $d\vec{S}$:
$$
\begin{aligned}
d\vec{S} &= \left(\frac{\partial \vec{r}}{\partial \rho} \times \frac{\partial \vec{r}}{\partial \phi}\right)d\rho\,d\phi \\
\end{aligned}
$$
When taking the derivate with respect to $\phi$, remember that the unit vector $\boldsymbol{\hat{\rho}}$ is a function of $\phi$. 
***
Now, you can evaluate:
$$
\iint_S{|d\vec{S}|}
$$
The limits of your integral should not be difficult: this is why we switched to cylindrical coordinates.
| At the origin $(0,0,0)$, the paraboloid is a point. For a fixed $z$: $x^2 + y^2 = z$ $\implies$the paraboloid is a circle of radius $\sqrt{z}$. As $z$ increases, the paraboloid expands outwards in proportion to $\sqrt{z}$.
***
The boundary of the paraboloid is defined by the cylinder. This occurs at $z=9$, where the paraboloid is a circle of radius $3$.
***
The cylindrical symmetry of this problem makes cylindrical coordinates the most favourable. In cylindrical coordinates, the equation of the paraboloid is:
***
$$
z=\rho^2
$$
and a position vector on the paraboloid surface is therefore:
***
$$
\vec{r} = \rho\boldsymbol{\hat{\rho}}(\phi) + z\mathbf{\hat{k}} =\rho\boldsymbol{\hat{\rho}}(\phi) + \rho^2\mathbf{\hat{k}}
$$
To find a $d\vec{S}$ on the paraboloid, we find the cross product:
***
$$
d\vec{S} = \left(\frac{\partial\vec{r}}{\partial\rho}\times\frac{\partial \vec{r}}{\partial \phi}\right)d\rho\, d\phi
$$
**Note:** For the derivative with respect to $\phi$, remember the derivative of the unit vector $\boldsymbol{\hat{\rho}}=\boldsymbol{\hat{\rho}}(\phi)$:
$$
\frac{\partial\boldsymbol{\hat{\rho}}}{\partial \phi}=\boldsymbol{\hat{\phi}}
$$
***
$$
\frac{\partial\vec{r}}{\partial \rho} = \boldsymbol{\hat{\rho}} + 2\rho\mathbf{\hat{k}}
$$
***
$$
\frac{\partial \vec{r}}{\partial \phi} = \rho\boldsymbol{\hat{\phi}}
$$
Finding the cross product:
***
$$
d\vec{S} = \begin{vmatrix}\boldsymbol{\hat{\rho}}&\boldsymbol{\hat{\phi}}&\mathbf{\hat{k}}\\1&0&2\rho\\0&\rho&0\end{vmatrix}d\rho\,d\phi = (-2\rho^2\boldsymbol{\hat{\rho}}+\rho\mathbf{\hat{k}})d\rho\,d\phi
$$
***
The surface area is the double integral of the *magnitude* of $d\vec{S}$:
$$
S = \iint_{S}|d\vec{S}|
$$
The magnitude is given by:
***
$$
|d\vec{S}| = \sqrt{\rho^2 + 4\rho^4}\,d\rho\,d\phi = \rho\sqrt{1+4\rho^2}\,d\rho\, d\phi
$$
Finally, the limits of the paraboloid are:
***
$$
\begin{aligned}
&\rho: 0\to 3\\
&\phi: 0\to 2\pi
\end{aligned}
$$
Evaluating the integral:
***
$$
\begin{aligned}
S = \iint_{S}|d\vec{S}| &= \int_{\phi=0}^{2\pi}\int_{\rho=0}^{3}{\rho\sqrt{1+4\rho^2}\,d\rho\, d\phi}\\
&=2\pi\int_{\rho=0}^{3}{\rho\sqrt{1+4\rho^2}\,d\rho}
\end{aligned}
$$
***
Using a substitution $u=1+4\rho^2$ to evaluate the integral:
* $d\rho = \frac{1}{8\rho}du$ 
* $u: 1\to 37$
***
$$
\begin{aligned}
S &= 2\pi\int_{u=1}^{37}{\rho u^{1/2}\frac{1}{8\rho}du}\\
&=\frac{\pi}{4}\int_{u=1}^{37}u^{1/2}du \\
& = \frac{\pi}{4}\left[\frac{2}{3}u^{3/2}\right]_{u=1}^{37}\\
& = \frac{\pi}{6}(37^{3/2}-1)
\end{aligned}
$$
| **\[Boas 5.5.3]:** Find the area of the paraboloid $x^2 + y^2 = z$ inside the cylinder $x^2 + y^2 = 9$. 
***
*Hint:* Don't use the surface of revolution formula here (instead, perform a double integral). Leave your answer in fractional form, or as a number to 1 decimal place.
| 43 | 2 | 28 | 28 | 196 | 14 | 43 | 2 | 0 | Boas 5.5.3: Find the area of the paraboloid $x^2 + y^2 = z$ inside the cylinder $x^2 + y^2 = 9$. Hint: Don't use the surface of revolution formula here instead, perform a double integral. Leave your answer in fractional form, or as a number to 1 decimal place. | 3 |
50559ad7-6651-4857-94b5-b39e3334d5a6 | 1 | 0 | 0 | 9 | 4 | 2 | 0 | 5 | Calculate the total resistance as measured between points **a** and **b** of the network below. You will have to apply Kirchhoff’s laws and set-up a set of linear equations to solve this. It is also advantageous to apply a virtual voltage source across **ab**. This question is a bonus question for strong students.
|   
| 1 | 1 | 4 | null | Firstly, resolve any branches that are in series or parallel to simplify the network:
***
The highlighted branches are in parallel:
***
This becomes:
  
***
Which is essentially a Wheatstone Bridge.
***
KVL can now be applied to the top and bottom loops respectively:
***
Top loop:
  
$(I-I_1)R_2 + (I_2 - I_1)R_3 = I_1R_1$ (Eq 1)
***
Bottom loop:
  
$(I_2 -I_1)R_3 + I_2R_4 = (I-I_2)R_5$ (Eq 2)
***
These can be rearranged to obtain two equations in terms of $I$:
***
$R_2I = (R_2 + R_3 + R_1)I_1 - R_3I_2$  
   
$50I = 200I_1 - 50I_2$ (Eq 3)
  
$R_5I = -R_3I_1 + (R_3 + R_4 +R_5)I_2$  
  
$100I = -50I_1 + 200I_2$ (Eq 4)
***
***
Solving equations 3 and 4 in terms of $I_1$ and $I_2$:
***
$I_1 = \frac{2}{5}I$ (Eq 5)
  
$I_2 = \frac{3}{5}I$ (Eq 6)
***
There are now two equations with three unknowns. Therefore, a third equation is required.
***
Notice that a closed loop can be drawn through $R_1$ and $R_4$ as highlighted in the figure below:
***
Applying KVL:
  
$ V = R_1I_1 + R_4I_2 $ (Eq 7)
***
Substituting equations 5 and 6 into equation 7:
  
$V = 100(\frac{2}{5}I) + 50(\frac{3}{5}I)$
***
Dividing through by $I$ to obtain $R$:
$R = 70~ \mathrm{k}\Omega$
| Calculate the total resistance as measured between points **a** and **b** of the network below. You will have to apply Kirchhoff’s laws and set-up a set of linear equations to solve this. It is also advantageous to apply a virtual voltage source across **ab**. This question is a bonus question for strong students.
  
| 55 | 0 | 18 | 18 | 176 | 0 | 2 | 0 | 1 | Calculate the total resistance as measured between points a and b of the network below. You will have to apply Kirchhoff’s laws and set-up a set of linear equations to solve this. It is also advantageous to apply a virtual voltage source across ab. | 3 |
50a9bdb3-2c61-4406-aa69-0ebd5649ac30 | 1 | 0 | 0 | 8 | 4 | 2 | 2 | 2 | A helicopter drive shaft is $110\text{ mm}$ diameter, with a wall thickness of $4\text{ mm}$. It is made of titanium alloy ($E=110\text{ GPa}$, failure stress $=650\text{ MPa}$). The engine power output is $297\text{ kW}$ and the rotor speed $190 \text{ RPM}$. The helicopter weighs $86 \text{ kN}$. As it flies along, aerodynamic forces induce shaft bending of $\pm 100\%$ of the mean axial stress. 
| What is the safety factor according to the shear-strain energy criterion, if fatigue considerations are ignored?
| 1 | 0.666667 | 3 | null | First the axial stress needs to be calculated:
   
$$
\begin{align*}
\sigma_z&=\frac{F}{A} \\
&=\frac{W}{2\pi rt} \\
&=\frac{86\times10^3}{2\pi\times0.055\times0.004} \\
&=62.21 \text{ MPa}
\end{align*}
$$
   
***
However, the bending **along the axial axis** induces $\pm100 \%$ of this value. And therefore, the **maximum** bending stress is:
   
$$
\sigma_z'=2\sigma_z=124.4\text{ MPa}
$$
   
***
Following this, the torque through the shaft needs to be calculated:
   
$$
\text{Power}=\text{Torque}\times\text{Angular Velocity} \\
\hspace{10pt} \\
T=\frac{P}{\omega} = \frac{297\times10^3\times60}{190\times2\pi}=14.93\text{ kNm}
$$
   
***
From there, the shear stress can be calculated from the equation below:
   
$$
\tau_{z\theta}=\frac{T}{2\pi r^2t}=\frac{14.93\times10^3}{2\pi\times0.055^2\times0.004}=196.4\text{ MPa}
\hspace{10pt} \\
\hspace{5pt} \\
{altnernatively, using J:}
\hspace{10pt} \\
\hspace{5pt} \\
\tau_{z\theta}=219.1\text{ MPa}
$$
   
***
Consider the element as seen below:
***
Find the principal stresses using the equation below:
   
$$
\begin{align*}
\sigma_{1,2}&=\frac{\sigma_z+\sigma_\theta}{2}\pm \sqrt{\frac{\sigma_z-\sigma_\theta}{2}+\tau_{z\theta}^2} \\
&=\frac{124.2}{2}\pm\sqrt{\left(\frac{124.2}{2}\right)^2+196.4^2}\\
\end{align*}
$$
$$
\sigma_1=268.2\text{ MPa} \hspace{30pt} \sigma_2=-143.8\text{ MPa}
$$
   
***
And using the results from above:
   
$$
\text{SF}=\frac{\sigma_{\text{failure}}}{\sqrt{\sigma_1^2+\sigma_2^2-\sigma_1\sigma_2}}=\frac{650}{\sqrt{268.2^2+(-143.8)^2-(268.2)(-143.8)}}=\boxed{1.795 (or 1.621exact)}
$$
| A helicopter drive shaft is $110\text{ mm}$ diameter, with a wall thickness of $4\text{ mm}$. It is made of titanium alloy ($E=110\text{ GPa}$, failure stress $=650\text{ MPa}$). The engine power output is $297\text{ kW}$ and the rotor speed $190 \text{ RPM}$. The helicopter weighs $86 \text{ kN}$. As it flies along, aerodynamic forces induce shaft bending of $\pm 100\%$ of the mean axial stress. 
What is the safety factor according to the shear-strain energy criterion, if fatigue considerations are ignored?
| 76 | 8 | 8 | 8 | 106 | 0 | 16 | 0 | 0 | What is the safety factor according to the shear-strain energy criterion, if fatigue considerations are ignored? | 1 |
50bcbdb1-5795-41dd-966b-7cf9abd9ad84 | 3 | 0 | 0 | 14 | 4 | 2 | 4 | 5 | Wind with speed $U$ blows perpendicularly to a suspension bridge of length $l$. The bridge deck of thickness $h$ disturbs the flow, leading to the periodic release of vortices, the frequency of which is called the vortex shedding frequency, and is denoted by $f$. The shedding frequency depends on the deck thickness, the bridge length, the wind velocity, the air density $\rho$ and dynamic viscosity $\mu$.
[https://www.youtube.com/watch?v=lXyG68\_caV4](https://www.youtube.com/watch?v=lXyG68_caV4 "https://www.youtube.com/watch?v=lXyG68_caV4")
| Considering the shedding frequency as the main quantity of interested, suggest a list of dimensionless groups to describe the problem.
\nIt is believed that on 07-11-1940, the vortex shedding frequency produced by the Tacoma Narrows bridge (USA) matched one the natural frequency of the bridge and led to its dramatic collapse. Since then, bridge designers test their designs in wind tunnels. If a $1/100$ scale model were to be tested in a wind tunnel and full dynamic similarity was required:
  
What air velocity would be necessary in the wind tunnel compared to the wind velocity experienced in real life (use $\mathrm(U)$ for velocity)? Using you own judgement, would such a wind-tunnel test be realistic? If not, what could you do to make it realistic?
\nIt is believed that on 07-11-1940, the vortex shedding frequency produced by the Tacoma Narrows bridge (USA) matched one the natural frequency of the bridge and led to its dramatic collapse. Since then, bridge designers test their designs in wind tunnels. If a $1/100$ scale model were to be tested in a wind tunnel and full dynamic similarity was required:
What shedding frequency would be observed in the air-filled wind tunnel compared to the shedding frequency generated by the full scale bridge? | 3 | 0.5 | 2 | \n\n | \n\n | Wind with speed $U$ blows perpendicularly to a suspension bridge of length $l$. The bridge deck of thickness $h$ disturbs the flow, leading to the periodic release of vortices, the frequency of which is called the vortex shedding frequency, and is denoted by $f$. The shedding frequency depends on the deck thickness, the bridge length, the wind velocity, the air density $\rho$ and dynamic viscosity $\mu$.
[https://www.youtube.com/watch?v=lXyG68\_caV4](https://www.youtube.com/watch?v=lXyG68_caV4 "https://www.youtube.com/watch?v=lXyG68_caV4")
Considering the shedding frequency as the main quantity of interested, suggest a list of dimensionless groups to describe the problem.
\nIt is believed that on 07-11-1940, the vortex shedding frequency produced by the Tacoma Narrows bridge (USA) matched one the natural frequency of the bridge and led to its dramatic collapse. Since then, bridge designers test their designs in wind tunnels. If a $1/100$ scale model were to be tested in a wind tunnel and full dynamic similarity was required:
  
What air velocity would be necessary in the wind tunnel compared to the wind velocity experienced in real life (use $\mathrm(U)$ for velocity)? Using you own judgement, would such a wind-tunnel test be realistic? If not, what could you do to make it realistic?
\nIt is believed that on 07-11-1940, the vortex shedding frequency produced by the Tacoma Narrows bridge (USA) matched one the natural frequency of the bridge and led to its dramatic collapse. Since then, bridge designers test their designs in wind tunnels. If a $1/100$ scale model were to be tested in a wind tunnel and full dynamic similarity was required:
What shedding frequency would be observed in the air-filled wind tunnel compared to the shedding frequency generated by the full scale bridge? | 278 | 9 | 0 | 0 | 1 | 0 | 207 | 3 | 2 | Wind with speed $U$ blows perpendicularly to a suspension bridge of length $l$. Considering the shedding frequency as the main quantity of interested, suggest a list of dimensionless groups to describe the problem. If a $1/100$ scale model were to be tested in a wind tunnel and full dynamic similarity was required: What air velocity would be necessary in the wind tunnel compared to the wind velocity experienced in real life use $\mathrm(U)$ for velocity? Using you own judgement, would such a wind-tunnel test be realistic? If not, what could you do to make it realistic? If a $1/100$ scale model were to be tested in a wind tunnel and full dynamic similarity was required: What shedding frequency would be observed in the air-filled wind tunnel compared to the shedding frequency generated by the full scale bridge? | 6 |
50c4692a-590e-40c2-a344-c1754d13a64d | 3 | 0 | 0 | 13 | 4 | 2 | 2 | 1 | 
A ship of mass $2.5~\mathrm{Gg}$ is towing another of mass $1.25~\mathrm{Gg}$ using a steel wire rope whose effective elasticity is such that a tension of $5~\mathrm{MN}$ produced a $10~\mathrm{\%}$ increase in length. At the instant the rope becomes taut, the velocities (in the same direction) of the two vessels are $3~\mathrm{m/s}$ and $1.5~\mathrm{m/s}$ respectively.
| Find the speed of the ships, $v_f$, as they move together.
\nFind the strain energy, $SE$, stored in the rope.
\nFind the minimum length of the rope, $L_{min}$, if the tension in it is not to exceed $500~\mathrm{kN}$.
| 3 | 0.666667 | 4 | Think of this as your usual impact but in reverse, where the impacting body is somehow ahead of the receiving body. Same rules apply.
\nExamine the kinetic energies of the ships before and after the rope became taut. Is there a deficit?
***
Be careful with the units in this question. The masses are given in $\mathrm{Gg}$, which is the same as $\mathrm{Kg} \times 10^6$. 
\nYou’ll have to express the elastic modulus of the rope in terms of the rope length in order to work out the percentage length increase the given maximum force ($500~\mathrm{kN}$) would cause, as the information given to you states the rope extension as a percentage and not an absolute value (hint: the proportion of the $5~\mathrm{MN}$ to the $500~\mathrm{kN}$ force might help here). This can then be used with the stored spring energy equation to derive the required rope length. It is tempting to think that the tension in the rope would be unaffected by its length, however as the value of the strain energy stored in the rope is fixed, the instantaneous tension in the rope will indeed depend on its length.
| Applying conservation of momentum:
$$
m_sv_s + m_bv_b = (m_s+m_b)v_f
$$
***
Rearranging and simplifying:
$$
v_f = \cfrac{m_sv_s + m_bv_b}{m_s + m_b}
$$
***
Substituting the values of the parameters gives:
$$
v_f = \cfrac{2.5 \times 3 + 1.25 \times 1.5}{2.5 + 1.25}
$$
$$
v_f = 2.5~\mathrm{m/s}
$$
\nThere is no impact, so conservation of energy can be applied:
$$
\text{Work done} = \text{Increase in } KE + \text{Increase in } SE + \text{Increase in } PE
$$
***
There is no change in vertical position so the change in gravitational potential energy is zero, there is no force or moment applied so the work done is zero, and the initial strain energy is zero:
$$
\begin{aligned}
0 &= KE_{\text{after}} - KE_{\text{before}} + SE_{\text{after}} \\
0 &= \cfrac{1}{2}~(m_s + m_b)v^2_f - \left(\cfrac{1}{2}~m_sv^2_s + \cfrac{1}{2}~m_bv^2_b\right) + SE
\end{aligned}
$$
***
Rearranging and simplifying:
$$
SE = \cfrac{1}{2}~[m_sv^2_s + m_bv^2_b - (m_s + m_b)v^2_f]
$$
***
Substituting the values of the parameters gives:
$$
SE=\cfrac{1}{2}\left[\left(2.5\times3^2\right)+\left(1.25\times{1.5}^2\right)-\left(2.5+1.25\right)\times{2.5}^2\right]
$$
$$
SE = 0.9375~\mathrm{MJ}
$$
\nThe tension $F$ in the rope cannot exceed $500~\mathrm{kN}$, therefore:
$$
F_{\text{max}} = 5 \times 10^5~\mathrm{N}~~\mathrm{(Equation~1)}
$$
***
The question states that a tension $F = 5~\mathrm{MN}$ on the rope produced a $10~\mathrm{\%}$ increase in length. The extension $x$ is given as a percentage, and therefore:
$$
x = 0.1L_{min}
$$
***
The spring constant $k$ of the rope can be calculated using Hooke's Law:
$$
k = \cfrac{F}{x} = \cfrac{5 \times 10^6}{0.1L_{min}} = \cfrac{5 \times 10^7}{L_{min}}~~\mathrm{N/m}
$$
***
$F_{\text{max}}$ is then given by:
$$
F_{\text{max}} = kx_{\text{max}} = \left(\cfrac{5 \times 10^7}{L_{min}}\right)~x_{\text{max}}~~\mathrm{(Equation~2)}
$$
***
Substituting Equation 2 into Equation 1:
$$
\left(\cfrac{5 \times 10^7}{L_{min}}\right)~x_{\text{max}} = 5 \times 10^5
$$
***
Rearranging for $L$:
$$
L_{min} = 100x_{\text{max}}~~\mathrm{(Equation~3)}
$$
***
We now need to find an expression for $x_{max}$ in terms of the strain energy $SE$ stored in the rope.
***
The strain energy, $SE$, in the rope is given by:
$$
SE = \cfrac{1}{2}~kx_{\text{max}}^2
$$
***
Rearranging for $x_{\text{max}}$:
$$
x_{\text{max}} = \sqrt{\cfrac{2 \cdot SE}{k}}~~\mathrm{(Equation~4)}
$$
***
Substituting Equation 4 into Equation 3:
$$
L_{min} = 100~\sqrt{\cfrac{2 \cdot SE}{k}}
$$
***
Substituting in the value for $k$ from earlier and the value for $SE$ from part b:
$$
L_{min} = 100~\sqrt{\cfrac{2 \times 9.375 \times 10^5 \times l_{min}}{5 \times 10^7}}
$$
***
Rearranging for $L_{min}$ leads to:
$$
L_{min} = 375~\mathrm{m}
$$
| 
A ship of mass $2.5~\mathrm{Gg}$ is towing another of mass $1.25~\mathrm{Gg}$ using a steel wire rope whose effective elasticity is such that a tension of $5~\mathrm{MN}$ produced a $10~\mathrm{\%}$ increase in length. At the instant the rope becomes taut, the velocities (in the same direction) of the two vessels are $3~\mathrm{m/s}$ and $1.5~\mathrm{m/s}$ respectively.
Find the speed of the ships, $v_f$, as they move together.
\nFind the strain energy, $SE$, stored in the rope.
\nFind the minimum length of the rope, $L_{min}$, if the tension in it is not to exceed $500~\mathrm{kN}$.
| 97 | 10 | 35 | 35 | 234 | 5 | 42 | 4 | 1 | A ship of mass $2.5~\mathrm{Gg}$ is towing another of mass $1.25~\mathrm{Gg}$ using a steel wire rope whose effective elasticity is such that a tension of $5~\mathrm{MN}$ produced a $10~\mathrm{\%}$ increase in length. Find the speed of the ships, $v_f$, as they move together. Find the strain energy, $SE$, stored in the rope. Find the minimum length of the rope, $L_{min}$, if the tension in it is not to exceed $500~\mathrm{kN}$. | 4 |
50ea3efd-48ab-4e3a-8b84-24e3650c9f37 | 0 | 1 | 0 | 2 | 1 | 2 | 0 | 7 | Which of the following properties are scalars?
| Which of the following properties are scalars?
| 1 | 0.333333 | 0 | null | null | Which of the following properties are scalars?
| 7 | 0 | 0 | 0 | 0 | 0 | 7 | 0 | 0 | Which of the following properties are scalars? | 1 |
514ab4e7-9f2e-41fc-aee7-89caf911312e | 0 | 1 | 0 | 2 | 1 | 2 | 0 | 3 | If the continuum assumption holds for a particular fluid which of the following statements are true?
| If the continuum assumption holds for a particular fluid which of the following statements are true?
| 1 | 0.333333 | 0 | null | null | If the continuum assumption holds for a particular fluid which of the following statements are true?
| 16 | 0 | 0 | 0 | 0 | 0 | 16 | 0 | 0 | If the continuum assumption holds for a particular fluid which of the following statements are true? | 1 |
5189ceb5-42a2-4adc-8722-db1cbd820349 | 6 | 1 | 0 | 11 | 4 | 2 | 5 | 7 | The stator exit flow angle of an axial-flow turbine is $70^{\circ}$. The rotor inlet and exit relative flow angles are symmetrical, i.e. $\beta_3 = -\beta_2$ . The mean radius and axial velocity are constant. If the stator exit gas velocity is $400 ~\mathrm{m/s}$ and the blade speed is $200 ~\mathrm{m/s}$, calculate:  
| The rotor inlet and exit flow angles (relative to the blade).
\nThe rotor exit gas relative velocity.
\nThe rotor exit tangential gas velocity component (magnitude and sign).
\nThe rotor exit absolute gas velocity and flow angle.
| 4 | 0.666667 | 3 | \n\n\n | Draw an initial velocity triangle for state $2$:
*  $ C_{\theta2} $ must be in the same direction as $U$ because $\alpha_2$ is positive;
* Since $ \beta_2 $ is unknown, we choose an arbitrary direction for $ W_{\theta2} $ as our initial guess. 
  
***
The relative velocity angle, $\beta_2$, can be found as follows:
  
$\beta_2 = \mathrm{cos}^{-1}(\frac{C_\mathrm{a2}}{W_2})$  
***
$C_\mathrm{a2} = 400\mathrm{cos}(70^{\circ})$
***
$W_2$ can be found by adding the axial and tangential components of $U$ and $C_2$ nose-to-tail (where $W = C - U$) and taking the magnitude:
***
Axial:
  
$U_\mathrm{a} = 0$
  
$C_{\mathrm{a2}} = 400\mathrm{cos}(70^{\circ})$ (from above)
***
Tangential:
  
$U_\mathrm{\theta2} = 200~\mathrm{m/s}$
  
$C_{\theta} = 400\mathrm{sin}(70^{\circ})$
***
Hence:
  
$W_2 = \sqrt{(400\mathrm{cos}(70^{\circ})-0)^2 +({400\mathrm{sin}(70^{\circ}) -200})^2} = 222.82~\mathrm{m/s}$ 
***
Substituting $W_2$ and $C_\mathrm{2a}$ into the equation for $\beta_2$:
  
$\beta_2 = \mathrm{cos}^{-1}(\frac{400\mathrm{cos}(70^{\circ})}{222.82}) = 52.1^{\circ}$
  
$ \beta_2 $ is positive based the above calculations, that means that our initial guess for the direction of $ W_{\theta2} $ is incorrect.
  
Hence, we re-draw the velocity triangle to ensure that all tangential components and angles are in the correct direction.
  
***
$\beta_2 = -\beta_3$
***
Hence:
  
$\beta_3 = -52.1^{\circ}$ 
\nThe rotor exit is state $3$.
***
Draw a velocity triangle for state $3$, bearing in mind that the axial velocity is constant ($C_\mathrm{a3} = C_\mathrm{a2})$:
  \
***
Hence:
  
$\mathrm{cos}(\beta_3) =\frac{C_\mathrm{a3}}{W_3}$
***
Substituting in numbers and rearranging for $W_3$:
  
$W_3 = \frac{400\mathrm{cos}(70)}{\mathrm{cos}(-52.1)} = 223~\mathrm{m/s}$ 
\nFrom the velocity triangle in part (b):
  
$C_{\theta3} = U + W_{\theta3}$ 
***
Find $W_{\theta3}$:
  
$W_{\theta3} = W_3~\mathrm{sin}(\beta_3)$ 
***
Substituting in numbers:
  
$W_{\theta3} = 223\mathrm{sin}(-52.1) = -175.97~\mathrm{m/s}$
***
Substituting this into the equation for $C_{\theta3}$:
  
$C_{\theta3} = 200-175.97 = +24~\mathrm{m/s}$
\nFrom the velocity triangle in part (b):
***
$ C_3 = \sqrt{C_{\theta3}^2 + C_{\mathrm{a}3}^2} $ (Pythagoras' theorem)
***
From part (b), $C_{\mathrm{a}3} = 400\mathrm{cos}(70)~\mathrm{m/s}$.
***
From part (c), $C_{\theta3} = 24~\mathrm{m/s}$
***
Substituting numbers into the equation for $C_3$:
  
$C_3 = \sqrt{24^2 + (400\mathrm{cos}(70))^2} = 138.9~\mathrm{m/s}$
***
Find the flow angle, $\alpha_3$, using the velocity triangle in (b):
***
$\alpha_3 = \mathrm{tan}^{-1}(\frac{C_{\theta3}}{C_{\mathrm{a}}})$
***
Substituting in numbers:
  
$\alpha_3 = \mathrm{tan}^{-1}(\frac{24}{400\mathrm{cos}(70)}) = 9.95^{\circ}$
| The stator exit flow angle of an axial-flow turbine is $70^{\circ}$. The rotor inlet and exit relative flow angles are symmetrical, i.e. $\beta_3 = -\beta_2$ . The mean radius and axial velocity are constant. If the stator exit gas velocity is $400 ~\mathrm{m/s}$ and the blade speed is $200 ~\mathrm{m/s}$, calculate:  
The rotor inlet and exit flow angles (relative to the blade).
\nThe rotor exit gas relative velocity.
\nThe rotor exit tangential gas velocity component (magnitude and sign).
\nThe rotor exit absolute gas velocity and flow angle.
| 86 | 4 | 46 | 46 | 300 | 0 | 36 | 0 | 0 | The rotor inlet and exit flow angles (relative to the blade).
\nThe rotor exit gas relative velocity.
\nThe rotor exit tangential gas velocity component (magnitude and sign).
\nThe rotor exit absolute gas velocity and flow angle.
| 0 |
52282ac7-5b8c-4d36-9581-c81512c36a62 | 0 | 8 | 0 | 4 | 3 | 3 | 0 | 3 | In Cartesian coordinates ($x$,$y$,z) and for a mean flow with velocity components ($u$,$v$,$w$) select the equations governing pollutant transport under the conditions that follow.
| Diffusion only and in $x$-direction.
\nDiffusion only and in both $x$ and z-directions.
\nAdvection only and in $x$-direction.
\nAdvection only and in both $x$ and $y$-directions.
\nAdvection and diffusion; both in $x$, $y$, and $z$ directions.
\nSteady flow, with advection and diffusion; both in $x$, $y$, and $z$ directions.
\nSteady flow in 3 dimensions, negligible diffusion.
\nWhen diffusion is negligible.
| 8 | 0.666667 | 2 | \n\n\n\n\n\n\n | We assume that diffusion is isotropic and homogeneous.
Here, temporal change is balanced by diffusion in one direction, and so
$$
\underbrace{\dfrac{\partial C}{\partial t}}_{\text{unsteady term}}=
\underbrace{D\dfrac{\partial^2 C}{\partial x^2}}_{\text{diffusion in $x$}}.
$$
\nWe assume that diffusion is isotropic and homogeneous.
Here, temporal change is balanced by diffusion in two directions, and so
$$
\underbrace{\frac{\partial C}{\partial t}}_{\text{unsteady term}}= \underbrace{D\frac{\partial^2 C}{\partial x^2}+D\frac{\partial^2 C}{\partial z^2}}_{\text{diffusion in $x$ and $z$}}.
$$
\nWe assume that diffusion is isotropic and homogeneous.
Here, temporal change is balanced by advection in one direction, and so
$$
\underbrace{\frac{\partial C}{\partial t}}_{\text{unsteady term}}
+\underbrace{u\frac{\partial C}{\partial x}}_{\text{advection in $x$}}=0.
$$
\nWe assume that diffusion is isotropic and homogeneous.
Here, temporal change is balanced by advection in two directions, and so
$$
\underbrace{\frac{\partial C}{\partial t}}_{\text{unsteady term}}
+\underbrace{u\frac{\partial C}{\partial x}+v\frac{\partial C}{\partial y}}_{\text{advection in $x$ and $y$}}=0.
$$
\nWe assume that diffusion is isotropic and homogeneous.
Here, temporal change is balanced by a combination of advection and diffusion in all directions, and so
$$
\underbrace{\frac{\partial C}{\partial t}}_{\text{unsteady term}}
+\underbrace{u\frac{\partial C}{\partial x}
+v\frac{\partial C}{\partial y}
+w\frac{\partial C}{\partial z}}_{\text{advection in $x$,$y$ and $z$}}=
\underbrace{D\frac{\partial^2 C}{\partial x^2}
+D\frac{\partial^2 C}{\partial y^2}
+D\frac{\partial^2 C}{\partial z^2}}_{\text{diffusion in $x$,$y$ and $z$}}.
$$
\nWe assume that diffusion is isotropic and homogeneous.
In this case the flow is steady, which means that $ \bm{u} = \bm{u}(x,y,z) $ (i.e. it does not depend on time). In spite of this condition, in general $\partial C/\partial t\neq 0$, and so
$$
\underbrace{\frac{\partial C}{\partial t}}_{\text{unsteady term}}
+\underbrace{u\frac{\partial C}{\partial x}
+v\frac{\partial C}{\partial y}
+w\frac{\partial C}{\partial z}}_{\text{advection in $x$,$y$ and $z$}}=
\underbrace{D\frac{\partial^2 C}{\partial x^2}
+D\frac{\partial^2 C}{\partial y^2}
+D\frac{\partial^2 C}{\partial z^2}}_{\text{diffusion in $x$,$y$ and $z$}}.
$$
\nWe assume that diffusion is isotropic and homogeneous.
Again, $\bm{u}=\bm{u}(x,y,z)$, and the equations reduce to a balance between temporal change and transport due to advection, and so
$$
\underbrace{\frac{\partial C}{\partial t}}_{\text{unsteady term}}
+\underbrace{u\frac{\partial C}{\partial x}
+v\frac{\partial C}{\partial y}
+w\frac{\partial C}{\partial z}}_{\text{advection in $x$,$y$ and $z$}}=
0.
$$
\nWhen diffusion is negligible the balance is the same as the previous question, thus leading to
$$
\underbrace{\frac{\partial C}{\partial t}}_{\text{unsteady term}}
+\underbrace{u\frac{\partial C}{\partial x}
+v\frac{\partial C}{\partial y}
+w\frac{\partial C}{\partial z}}_{\text{advection in $x$,$y$ and $z$}}=
0.
$$
| In Cartesian coordinates ($x$,$y$,z) and for a mean flow with velocity components ($u$,$v$,$w$) select the equations governing pollutant transport under the conditions that follow.
Diffusion only and in $x$-direction.
\nDiffusion only and in both $x$ and z-directions.
\nAdvection only and in $x$-direction.
\nAdvection only and in both $x$ and $y$-directions.
\nAdvection and diffusion; both in $x$, $y$, and $z$ directions.
\nSteady flow, with advection and diffusion; both in $x$, $y$, and $z$ directions.
\nSteady flow in 3 dimensions, negligible diffusion.
\nWhen diffusion is negligible.
| 101 | 16 | 11 | 11 | 194 | 0 | 67 | 11 | 0 | In Cartesian coordinates $x$,$y$,z and for a mean flow with velocity components $u$,$v$,$w$ select the equations governing pollutant transport under the conditions that follow. When diffusion is negligible. | 2 |
52f515ac-7136-49bb-938e-1af478bac08c | 0 | 0 | 1 | 9 | 4 | 2 | 6 | 7 | Show that a series LR circuit shown below is a Low-Pass filter if the output is the voltage across the resistor. Set up the differential equation relating $v_\mathrm{o}$ to $v_\mathrm{i}$ and hence derive the corresponding transfer function.
| 
| 1 | 0.666667 | 2 | null |
, Apply KVL round the closed loop, recalling that the voltage across an inductor is given by $L\frac{\mathrm{d}i}{\mathrm{d}t}$:
  
$v_\mathrm{i} = L\frac{\mathrm{d}i}{\mathrm{d}t}+v_\mathrm{o}$
***
***
Apply Ohm's law to eliminate $i$:
  
$v_\mathrm{i} = \frac{L}{R}\frac{\mathrm{d}v_\mathrm{o}}{\mathrm{d}t}+v_\mathrm{o}$  
***
Convert to the Laplace Domain using the Laplace Transforms table in the Data and Formulae book. Assuming that all systems are at rest at $t = 0:$
  
$V_\mathrm{i}(s) = \frac{L}{R}sV_\mathrm{o}(s) + V_\mathrm{o}(s)$
***
Divide by $V_\mathrm{i}(s)$ and rearrange to obtain the transfer function:
  
$\frac{V_\mathrm{o}(s)}{V_\mathrm{i}(s)} = \frac{1}{1+s\frac{L}{R}}$
***
Substituting $s = j\omega$:
  
$\frac{V_\mathrm{o}(j\omega)}{V_\mathrm{i}(j\omega)} = \frac{1}{1+j\omega \frac{L}{R}}$  
  
which is the same as for a low-pass filter \[EXPLAIN HOW]
, Find an expression relating $v_\mathrm{i}$ and $v_\mathrm{o}$ using the potential divider equation:
  
$v_\mathrm{o} = \frac{Z_\mathrm{R}}{Z_\mathrm{R}+Z_\mathrm{L}}v_\mathrm{i}$
***
$Z_\mathrm{R} = R$
  
$Z_\mathrm{L} = jL\omega$
***
Substitute the impedance expressions back into the potential divider equation:
  
$v_\mathrm{o} = \frac{R}{R+jL\omega}v_\mathrm{i}$
***
Multiply the fraction by $\frac{R}{R}$:
  
$v_\mathrm{o} = \dfrac{1}{1+j\omega\frac{L}{R}}v_\mathrm{i}$
***
Convert $v_\mathrm{o}$ and $v_\mathrm{i}$ to the Laplace domain, replacing $s$ with $j\omega$:
  
$V_\mathrm{0}(j\omega) = \frac{1}{1+j\omega\frac{L}{R}}V_\mathrm{i}(j\omega)$
***
Divide by $V_\mathrm{i}(j\omega)$ to obtain the transfer function:
  
$|H| = \frac{V_\mathrm{o}(j\omega)}{V_\mathrm{i}(\omega)} = \frac{1}{1+j\omega\frac{L}{R}}$
  
which is the same as for a low-pass filter \[EXPLAIN HOW]
| Show that a series LR circuit shown below is a Low-Pass filter if the output is the voltage across the resistor. Set up the differential equation relating $v_\mathrm{o}$ to $v_\mathrm{i}$ and hence derive the corresponding transfer function.
| 38 | 2 | 25 | 25 | 173 | 0 | 1 | 0 | 1 | Show that a series LR circuit shown below is a Low-Pass filter if the output is the voltage across the resistor. Set up the differential equation relating $v_\mathrm{o}$ to $v_\mathrm{i}$ and hence derive the corresponding transfer function. | 2 |
534d68b9-b8cb-4db1-9a55-0583dbe97fc5 | 6 | 0 | 1 | 8 | 4 | 2 | 5 | 2 | A thick-walled cylinder with an outer to inner diameter ratio $K$ and **closed ends** is subjected to increasing internal pressure until yielding first occurs. For a steel cylinder, $\sigma_Y = 310\text{ MPa}$ and $K = 2.4$.
| Show that, if the Von Mises criterion applies, yielding will occur at a pressure:
   
$$
P=\sigma_Y\frac{K^2-1}{\sqrt{3K^2}}
$$
   
Where $\sigma_Y$ is the tensile yield stress. 
\n Find the values of $P$ for both the Von Mises and Tresca yield criteria when the cylinder has **closed ends.**
\nFind the values of $P$ for both the Von Mises and Tresca yield criteria when the cylinder has **closed zero end load.**
\n
 Find the values of $P$ for both the Von Mises and Tresca yield criteria when the cylinder has **its ends fixed to rigid abutments**.
| 4 | 1 | 4 | \n\n\n | The cylinder will be subjected to the following stresses at the bore (which is where the cylinder is most likely to yield first):
   
The hoop stress can be written as below (from Q6.1b):
   
$$
\begin{align}
\sigma_\theta=P\frac{K^2+1}{K^2-1}
\end{align}
$$
   
***
The radial stress at the bore (from Q6.1b):
   
$$
\begin{align}
\sigma_r=-P
\end{align}
$$
   
***
The axial stress for the uniform cross-section can be written as:
   
$$
\begin{align}
\sigma_z=\frac{F}{A}&=\frac{P\pi r_\text{i}^2}{\pi(r_\text{o}^2-r_i^2)}
=\frac{P}{K^2-1}
\end{align}
$$
   
***
The Von Mises criterion for stresses in 3D:
   
$$
\begin{align}
2\sigma_Y^2=(\sigma_\theta-\sigma_r)^2+(\sigma_r-\sigma_z)^2+(\sigma_z-\sigma_\theta)^2
\end{align}
$$
   
***
Substitute $(1), (2)$ and $(3)$ into the $(4)$ and then rearranging $(5)$:
   
$$
\begin{align}
2\sigma_Y^2=\left[ \frac{P}{K^2-1}(1+K^2)-(-P)\right]^2+\left[-P-\frac{P}{K^2-1}\right]^2
+\left[\frac{P}{K^2-1}-\frac{P}{K^2-1}(1+K^2)\right]^2
\end{align}
$$
   
***
   
$$
\begin{align}
\frac{2\sigma_Y^2}{P^2}(K^2-1)&=[(K^2)^2+(K^2)^2+(-2K^2)^2] \\
&=6\left(\frac{PK^2}{K^2-1}\right)^2
\end{align}
$$
   
***
Continue to rearrange and solve for $P$:
   
$$
\begin{align*}
\sigma_Y=\sqrt{3}\frac{PK^2}{K^2-1} \\
\boxed{P=\frac{K^2-1}{\sqrt{3}K^2}}
\end{align*}
$$
   
\nLooking at the **Von Mises** criterion, using the solution from (a):
   
$$
P_\text{Von Mises}=\sigma_Y\frac{K^2-1}{\sqrt{3}K^2}=(310)\frac{2.4^2-1}{\sqrt{3}(2.4)^2}=\boxed{147.9\text{ MPa}}
$$
   
***
Looking at the **Tresca** criterion:
   
$$
\sigma_Y=\text{max}[|\sigma_\theta-\sigma_z|, |\sigma_\theta-\sigma_r|, |\sigma_z-\sigma_r|]
$$
   
***
Using the $\sigma_\theta, \sigma_r, \sigma_z$ from part (a), the maximum for $\sigma_Y$ is:
   
$$
\begin{align*}
\sigma_Y=|\sigma_\theta-\sigma_r|&=\frac{P_\text{Tresca}}{K^2-1}(1+K^2)-(-P_\text{Tresca}) \\
\end{align*}
$$
   
***
Rearrange for $P$ and substitute values from the question:
   
$$
\begin{align*}
P_\text{Tresca}=\frac{\sigma_Y}{\frac{1+K^2}{K^2-1}+1}=\frac{310}{\frac{1+2.4^2}{2.4^2-1}+1}=\boxed{128.1\text{ MPa}}
\end{align*}
$$
\nIn this scenario, there is zero end load in the cylinder, meaning $\sigma_z=0$.
   
***
Looking at the **Von Mises** criterion for stress in 2D, the following equation is derived. Then substituting in the expressions for $\sigma_r$ and $\sigma_\theta$ from (a).
   
$$
\begin{align*}
\sigma_Y^2&=\sigma_\theta^2+\sigma_r^2-\sigma_\theta\sigma_r \\
&=\left[ P\frac{1+K ^2}{K^2-1}\right]^2+\left[-P\right]^2-\left[ P\frac{1+K ^2}{K^2-1}\right]\left[-P\right] \\
&=P^2\left[ \left(\frac{1+K^2}{K^2-1} \right)^2 +1+ \left(\frac{1+K^2}{K^2-1} \right)\right]
\end{align*}
$$
   
***
Rearrange and solve for $P_\text{Von Mises}$:
   
$$
P_\text{Von Mises}=\sqrt\frac{\sigma_Y^2}{\left(\frac{1+K^2}{K^2-1} \right)^2 +1+ \left(\frac{1+K^2}{K^2-1} \right)}=\boxed{147.2\text{ MPa}}
$$
 
***
Looking at the **Tresca** criterion, the pressure at failure would be the same as in (b) because $\sigma_Y=|\sigma_\theta-\sigma_r|$ would still be true.
   
$$
P_\text{Tresca}=\boxed{128.1\text{ MPa}}
$$
\nIf the cylinder is fixed at the ends, then it will experience plane strain in the $z$ direction:
   
$$
e_z=0
$$
   
***
Hooke's Law can therefore be used to solve for the stresses and therefore find the pressure. 
   
$$
\begin{align*}
e_z=\frac{1}{E}[\sigma_z-\nu(\sigma_r+\sigma_\theta)]&=0 \\
\sigma_z-\nu(\sigma_r+\sigma_\theta)&=0\\
\sigma_z&=\nu(\sigma_r+\sigma_\theta)
\end{align*}
$$
   
***
Substituting $\sigma_r$ and $\sigma_\theta$ from (a) produces an expression for $\sigma_z$ in this particular scenario.
   
$$
\sigma_z=\nu\left(-P+\frac{P(1+K^2)}{K^2-1}\right)=\frac{2P\nu}{K^2-1}
$$
   
***
Looking at the **Von Mises** failure criteria for stresses in 3D, using $\sigma_r$ and $\sigma_\theta$ from (a):
   
$$
\begin{align*}
2\sigma_Y^2&=(\sigma_\theta-\sigma_z)^2+(\sigma_z-\sigma_r)^2+(\sigma_r-\sigma_\theta)^2\\
&=\left( P\frac{1+K^2}{K^2-1}-\frac{2P\nu}{K^2-1}\right)^2+\left(\frac{2P\nu}{K^2-1}-(-P)\right)^2+\left((-P)-P\frac{1+K^2}{K^2-1}\right)^2
\end{align*}
$$
   
***
Rearrange and simplify, then substitute in values and solve for $P$.
   
$$
\sigma_Y=\frac{P^2}{2(K^2-1)^2}[(1+K^2-2\nu)^2+(2\nu-1+K^2)^2+(-2K^2)^2]
$$
   
***
For steel, $\nu=0.3$.
   
$$
P_\text{Von Mises}=\boxed{147.8\text{ MPa}}
$$
   
***
Looking at the **Tresca** criteria, failure would occur when $\sigma_Y=|\sigma_\theta-\sigma_r|$, substituting the appropriate expression, $P_\text{Tresca}$ can be calculated.
   
$$
\sigma_Y=|\sigma_\theta-\sigma_r|=P\frac{1+K^2}{K^2-1}-(-P)
$$
   
$$
P_\text{Tresca}=\boxed{128.1\text{ MPa}}
$$
| A thick-walled cylinder with an outer to inner diameter ratio $K$ and **closed ends** is subjected to increasing internal pressure until yielding first occurs. For a steel cylinder, $\sigma_Y = 310\text{ MPa}$ and $K = 2.4$.
Show that, if the Von Mises criterion applies, yielding will occur at a pressure:
   
$$
P=\sigma_Y\frac{K^2-1}{\sqrt{3K^2}}
$$
   
Where $\sigma_Y$ is the tensile yield stress. 
\n Find the values of $P$ for both the Von Mises and Tresca yield criteria when the cylinder has **closed ends.**
\nFind the values of $P$ for both the Von Mises and Tresca yield criteria when the cylinder has **closed zero end load.**
\n
 Find the values of $P$ for both the Von Mises and Tresca yield criteria when the cylinder has **its ends fixed to rigid abutments**.
| 125 | 8 | 45 | 45 | 408 | 0 | 93 | 5 | 0 | Show that, if the Von Mises criterion applies, yielding will occur at a pressure: $ P=\sigma_Y\frac{K^2-1}{\sqrt{3K^2}} $ Where $\sigma_Y$ is the tensile yield stress. Find the values of $P$ for both the Von Mises and Tresca yield criteria when the cylinder has closed ends. Find the values of $P$ for both the Von Mises and Tresca yield criteria when the cylinder has closed zero end load. Find the values of $P$ for both the Von Mises and Tresca yield criteria when the cylinder has its ends fixed to rigid abutments. | 4 |
536be15b-e639-4855-b0e9-1e840852c65e | 4 | 0 | 0 | 0 | 0 | 2 | 2 | 0 | A symmetrical aerofoil with a rounded leading edge is flying at $M=3$ at zero incidence and at sea level conditions ($T_{1}=288 \ \mathrm{K}$,
 $p_1 = 1.013 \cdot 10^{5} \ \mathrm{N/m^2}$). 
  
Hint: a curved shock forms in front of the aerofoil, however just in front of the nose it is a normal shock. 
| Calculate the temperature and pressure at the nose of the aerofoil. 
  
Hint: the nose of the aerofoil is a stagnation point. 
  
\[Note: to enter numbers in scientific notation, use format 1.5e6, for instance]
\nDetermine the pressure coefficient at the nose of the aerofoil. 
\nWhat is the pressure coefficient at a point on the aerofoil where the local Mach number is equal to 1?
| 3 | 0.333333 | 2 | The stagnation temperature does not change across the shock. So we can use the isentropic relation to find $ T_{01} $. 
***
The stagnation pressure is not conserved across the shock, so we also need to determine the change across the normal shock wave. 
\nRemember that the pressure coefficient can be expressed as a function of $\gamma$ and the Mach number:
 
$C_{p} = \frac{p-p_{\infty}}{p_{\infty}} \frac{1}{\frac{1}{2} \gamma M_{\infty}^{2}}=(\frac{p}{p_{\infty}}-1)\frac{1}{\frac{1}{2} \gamma M_{\infty}^{2}}$
***
What do we know about the value of $ p $ at the nose (stagnation point)? How can this be used to find $ \frac{p}{p_{\infty}} $ from tables at $ M=3 $?
\nThe pressure coefficient can again obtained with 
   
$ C_{p} = \frac{p-p_{\infty}}{p_{\infty}} \frac{1}{\frac{1}{2} \gamma M_{\infty}^{2}}=(\frac{p}{p_{\infty}}-1)\frac{1}{\frac{1}{2} \gamma M_{\infty}^{2}} $
  
What is not known in this expression and how can we obtain it?
| The nose of the airfoil is a stagnation point, therefore we need to find the stagnation pressure and temperature behind a shock at $ M=3 $. Since it is air, we can use the tables (though it can also be done with the normal shock equations).
   
***
The stagnation temperature does not change across the shock. So we can use the isentropic relation to find $ T_{01} $. 
   
$T_{02} = T_{01} = \frac{T_{01}}{T_{1}} \cdot T_{1}$
   
***
At $ M=3 $, from the tables we get:
  \
$ \frac{T_{01}}{T_{1}}=2.8 $
   
  
 so $ T_{02} =(2.8)(2.88) = 806.4 $ K.
   
***
The stagnation pressure is not conserved across the shock, so we also need to determine the change across the normal shock wave. 
   
$ p_{02} =\frac{p_{02}}{p_{01}} \cdot \frac{p_{01}}{p_{1}} \cdot p_{1} = (0.3283)(36.733)(1.013 \cdot 10^{5})=1.22 \cdot 10^{6} \ \mathrm{N/m^2}. $
   
Alternatively: 
   
$p_{02} = \frac{p_{02}}{p_{1}} \cdot p_{1} = (12.06)(1.013 \cdot 10^{5}) = 1.22 \cdot 10^{6} \ \mathrm{N/m^2}$. 
\nThe pressure coefficient at the nose can be calculated with this expression, as a function of $\gamma$ and $M$ instead of the velocity $V_{\infty}$:
  
$ C_{p} = \frac{p-p_{\infty}}{p_{\infty}} \frac{1}{\frac{1}{2} \gamma M_{\infty}^{2}}=(\frac{p}{p_{\infty}}-1)\frac{1}{\frac{1}{2} \gamma M_{\infty}^{2}} $. 
   
The nose is a stagnation point, therefore the pressure at the nose, $ p=p_{02} $.
   
From the tables we can get $p_{02}$:
   
 $\frac{p_{02}}{p_{1}} =12.06$
   
***
As $p_1$ is the freestream condition, we can now use the $C_p$ expression directly:
   
$ C_{p} = (12.06 -1)\frac{1}{\frac{1}{2} (1.4)(3^{2})}=1.76. $
\nRecall the expression for $C_p$:
  
$C_{p} = \frac{p-p_{\infty}}{p_{\infty}} \frac{1}{\frac{1}{2} \gamma M_{\infty}^{2}}=(\frac{p}{p_{\infty}}-1)\frac{1}{\frac{1}{2} \gamma M_{\infty}^{2}}$
  
***
   
We need to know the pressure at the point on the airfoil where $ M=1 $ in order to calculate the pressure coefficient. 
  
Behind the shock wave, the flow is isentropic. We can use the isentropic relations or the tables:
   
 $\frac{p_{01}}{p_{A}} =(1+\frac{\gamma -1}{2})^{\frac{\gamma}{\gamma-1}}$
   
or from tables $ \frac{p_{02}}{p_{A}} =1.8929 $.
  
***
  
So we can obtain the pressure at point A with respect to the free stream with:
  
 $ \frac{p_{A}}{p_{\infty}} =\frac{p_{A}}{p_{02}} \cdot \frac{P_{02}}{P_{\infty}}=\frac{1}{1.893} \cdot 12.06=6.37. $
   
***
And we can with this obtain the pressure coefficient at A:  
  
$C_{p}=(6.37-1)\frac{1}{\frac{1}{2}(1.4)(3^{2})} =0.852$.
| A symmetrical aerofoil with a rounded leading edge is flying at $M=3$ at zero incidence and at sea level conditions ($T_{1}=288 \ \mathrm{K}$,
 $p_1 = 1.013 \cdot 10^{5} \ \mathrm{N/m^2}$). 
  
Hint: a curved shock forms in front of the aerofoil, however just in front of the nose it is a normal shock. 
Calculate the temperature and pressure at the nose of the aerofoil. 
  
Hint: the nose of the aerofoil is a stagnation point. 
  
\[Note: to enter numbers in scientific notation, use format 1.5e6, for instance]
\nDetermine the pressure coefficient at the nose of the aerofoil. 
\nWhat is the pressure coefficient at a point on the aerofoil where the local Mach number is equal to 1?
| 114 | 3 | 25 | 25 | 309 | 7 | 65 | 0 | 0 | Calculate the temperature and pressure at the nose of the aerofoil. Note: to enter numbers in scientific notation, use format 1.5e6, for instance Determine the pressure coefficient at the nose of the aerofoil. What is the pressure coefficient at a point on the aerofoil where the local Mach number is equal to 1? | 3 |
53e5ea9f-f53c-486d-87b9-6c0930b58f67 | 2 | 0 | 0 | 18 | 6 | 1 | 1 | 6 | This question is about space-like and time-like separations. Either answer the following questions by drawing out a space-time diagram, or solve algebraically using the Lorentz transformations. | Two events are separated by $(c\Delta t, \Delta x, \Delta y, \Delta z) = (3,4,0,0)$ in frame $S$. Is there a reference frame $S^\prime$ in which the events are simultaneous, or at the same position? What is the relative velocity between $S$ and the relevant frame $S^\prime$ (give your answer in terms of $c$)?\nTwo events are separated by $(c\Delta t, \Delta x, \Delta y, \Delta z) = (-8,2,0,0)$ in frame $S$. Again, find the relative velocity between $S$ and $S^\prime$ where $S^\prime$ is the frame in which the two events are either simultaneous or at the same position (give your answer in terms of $c$). | 2 | 0.333333 | 2 | First, calculate the value of the invariant interval in this case.
***
   
You should find that this is negative number - is this a space-like or time-like interval?
***
   
This is a *space-like* interval and so there is a frame where the two events are simultaneous...
***
   
In the frame where the two events are simultaneous, $(\Delta t)' = 0$. How can you use the Lorentz transform to find the boost you need to apply to move to this frame?\nFirst, calculate the value of the invariant interval in this case.
***
   
You should find that this is positive number - is this a space-like or time-like interval?
***
   
This is a *time-like* interval and so there is a frame where the two events are in the same position...
***
   
In the frame where the two events occur at the same place, $(\Delta x)' = 0$. How can you use the Lorentz transform to find the boost you need to apply to move to this frame? | The invariant interval is $s^2=9-16=-7 < 0$.
***
Thus, the events are space-like separated.  
***
This means a reference frame can be found in which the events are simultaneous. 
***
For this frame: $\Delta t^\prime = \gamma(\Delta t-v\Delta x /c^2)=0$. 
***
Therefore we find $v\Delta x/c^2=\Delta t$ and $v=c^2 \Delta t/\Delta x=3c/4$.
There is no frame in which the events are at the same position as $\Delta x^\prime$ would be zero and the invariant interval could not then be negative\nThe invariant interval is $s^2=60 > 0$. 
***
Thus, the events are time-like separated.  
This means a reference frame can be found in which the events occur at the same position. 
***
For this frame: $\Delta x^\prime = \gamma(\Delta x-v\Delta t)=0$. 
***
Therefore we find $v\Delta t=\Delta x$ and $v=\Delta x/\Delta t=-2c/8=-c/4$.
***
There is no frame in which the events are simultaneous as $\Delta t^\prime$ would be zero and the invariant interval could not then be positive | This question is about space-like and time-like separations. Either answer the following questions by drawing out a space-time diagram, or solve algebraically using the Lorentz transformations.Two events are separated by $(c\Delta t, \Delta x, \Delta y, \Delta z) = (3,4,0,0)$ in frame $S$. Is there a reference frame $S^\prime$ in which the events are simultaneous, or at the same position? What is the relative velocity between $S$ and the relevant frame $S^\prime$ (give your answer in terms of $c$)?\nTwo events are separated by $(c\Delta t, \Delta x, \Delta y, \Delta z) = (-8,2,0,0)$ in frame $S$. Again, find the relative velocity between $S$ and $S^\prime$ where $S^\prime$ is the frame in which the two events are either simultaneous or at the same position (give your answer in terms of $c$). | 116 | 12 | 4 | 4 | 72 | 2 | 91 | 12 | 0 | Either answer the following questions by drawing out a space-time diagram, or solve algebraically using the Lorentz transformations.Two events are separated by $(c\Delta t, \Delta x, \Delta y, \Delta z) = (3,4,0,0)$ in frame $S$. Is there a reference frame $S^\prime$ in which the events are simultaneous, or at the same position? What is the relative velocity between $S$ and the relevant frame $S^\prime$ give your answer in terms of $c$? Again, find the relative velocity between $S$ and $S^\prime$ where $S^\prime$ is the frame in which the two events are either simultaneous or at the same position give your answer in terms of $c$. | 4 |
544ecba6-40bd-43ba-96c7-d9d4725fd5fd | 2 | 0 | 0 | 20 | 6 | 1 | 0 | 2 | Define $z=(5+7i)(5+bi)$:
| If $b$ and $z$ are both real, find $b$. 
\nIf $\rm{Im}(b)=4/5$ and $z$ is pure imaginary, find $\rm{Re}(b)$. 
| 2 | 0.666667 | 1 | Expand the brackets...
***
... collect the imaginary terms
***
If $b$ and $z$ are both purely real, what is the imaginary part equal to?
\nStart by expanding the brackets...
***
It is useful to let $b=x+\frac{4}{5}i$, 
***
our goal is to solve for $x$.
***
If $z$ is pure imaginary, what is the real part of $z$?
***
Collect the real terms and make use of the previous hint.
| If $b$ and $z$ are both real then the imaginary part is 0,
***
expanding the brackets,
***
$(5+7i)(5+bi)= 25 + 35i + 5bi - 7b$
***
Collecting the imaginary terms,
***
and setting equal to 0,
***
$35+5b=0$, so:
***
$b=-7$
\nExpanding the brackets,
***
$$
z=(5+7i)(5+bi)=25+35i+5bi+7b i^2 = 25-7b+35i+5bi
$$
***
Next, we can let $b=x+\frac{4}{5}i$, where $\rm{Re}(b)=$ $x$.
***
$$
z = 25-7(x+\frac{4}{5}i)+35i+5i(x+\frac{4}{5}i),
$$
***
because $z$ is pure imaginary, $\rm{Re}(z)=0$. 
***
So, we will collect the real terms only and set them equal to 0,
***
$$
\begin{equation}
{\rm Re}(z)=25-7x+4i^2 = 21-7x = 0
\end{equation}
$$
***
Solving for $x$, we have:
***
$$
x=\rm{Re}(b)=3
$$
| Define $z=(5+7i)(5+bi)$:
If $b$ and $z$ are both real, find $b$. 
\nIf $\rm{Im}(b)=4/5$ and $z$ is pure imaginary, find $\rm{Re}(b)$. 
| 23 | 7 | 15 | 15 | 88 | 6 | 20 | 6 | 0 | Define $z=(5+7i)(5+bi)$: If $b$ and $z$ are both real, find $b$. If $\rm{Im}(b)=4/5$ and $z$ is pure imaginary, find $\rm{Re}(b)$. | 2 |
545a7370-9d91-4c92-a99c-92b64099ab2f | 5 | 0 | 0 | 4 | 3 | 3 | 2 | 4 | Reconsider the sea-river system below.
For this question, the river is discharging $\dot{M}=1\ \mathrm{kg/s}$ of phosphorus into the sea as a continuous release. The sea has an average depth of $6\ \mathrm{m}$ and it may be assumed that the mixing at the river mouth is sufficient to distribute it uniformly across the depth. Assume the tidal current $U=0.5\ \mathrm{m/s}$, and that the flow has an eddy diffusivity $D_T=1.3\ \mathrm{m^2/s}$.
| Give the solution to this problem in symbolic form. You may use the far-field solution.
\nCalculate the phosphorous concentration at $x = 1\ \mathrm{km}$ at $y=0,\ 50,\ 100,\ \mathrm{and\ }200\ \mathrm{m}$. Finally, plot the concentration as a function of $y$.
| 2 | 0.666667 | 2 | \n | We use the far-field solution for a continuous release in 2D as
$$
C = \dfrac{\dot{M}}{\sqrt{2 \pi} h U \sigma} \exp(-y^2/(2 \sigma^2)),
$$
where $\sigma^2 = 2 D x / U$.
***
Applying the boundary condition at $y=0$ results in
$$
C = \dfrac{2\dot{M}}{\sqrt{2 \pi} h U \sigma} \exp(-y^2/(2 \sigma^2)).
$$
\nWe use the far-field solution for a continuous release in 2D as
$$
C = \dfrac{\dot{M}}{\sqrt{2 \pi} h U \sigma} \exp(-y^2/(2 \sigma^2)),
$$
where $\sigma^2 = 2 D x / U$.
***
Applying the boundary condition at $y=0$ results in
$$
C = \dfrac{2\dot{M}}{\sqrt{2 \pi} h U \sigma} \exp(-y^2/(2 \sigma^2)).
$$
Using the parameter values $\dot{M}=1000\ \mathrm{g/m^3}$, $h=6\ \mathrm{m}$, $U=0.5\ \mathrm{m/s}$ and $D_T=1.3\ \mathrm{m^2/s}$, we have that $\sigma^2 = 5200\ \mathrm{m^2}$ at the required $x$-location. The resulting table and figure is shown below. 
| $y\ [\mathrm{m}]$ | $C\ [\mathrm{g/m^3}]$ |
| :---------------- | :-------------------- |
| 0 | 3.688 |
| 50 | 2.900 |
| 100 | 1.410 |
| 200 | 0.079 |
| Reconsider the sea-river system below.
For this question, the river is discharging $\dot{M}=1\ \mathrm{kg/s}$ of phosphorus into the sea as a continuous release. The sea has an average depth of $6\ \mathrm{m}$ and it may be assumed that the mixing at the river mouth is sufficient to distribute it uniformly across the depth. Assume the tidal current $U=0.5\ \mathrm{m/s}$, and that the flow has an eddy diffusivity $D_T=1.3\ \mathrm{m^2/s}$.
Give the solution to this problem in symbolic form. You may use the far-field solution.
\nCalculate the phosphorous concentration at $x = 1\ \mathrm{km}$ at $y=0,\ 50,\ 100,\ \mathrm{and\ }200\ \mathrm{m}$. Finally, plot the concentration as a function of $y$.
| 102 | 7 | 14 | 14 | 84 | 0 | 34 | 3 | 1 | Reconsider the sea-river system below. The sea has an average depth of $6\ \mathrm{m}$ and it may be assumed that the mixing at the river mouth is sufficient to distribute it uniformly across the depth. Assume the tidal current $U=0.5\ \mathrm{m/s}$, and that the flow has an eddy diffusivity $D_T=1.3\ \mathrm{m^2/s}$. Give the solution to this problem in symbolic form. You may use the far-field solution. Calculate the phosphorous concentration at $x = 1\ \mathrm{km}$ at $y=0,\ 50,\ 100,\ \mathrm{and\ }200\ \mathrm{m}$. Finally, plot the concentration as a function of $y$. | 7 |
5500f226-1737-4c9c-938a-a0eadd134831 | 2 | 0 | 2 | 16 | 6 | 1 | 6 | 5 |  If $u = u(x, y)$ and $x$ and $y$ are related to two new independent variables $s$ and $t$ by
$$
\displaystyle x = st,\quad y = {{s + t} \over {s - t}}
$$
so that $u = \bar u(s, t)$, 
| Use the chain rule to find $\displaystyle{\partial\bar u\over\partial s}$ and $\displaystyle{\partial \bar u\over \partial t}$ in terms of $\displaystyle{\partial u\over\partial x}$ and $\displaystyle{\partial u\over\partial y}$. 
\nHence show that 
$$
\displaystyle 2x{{\partial u}\over {\partial x}} = s{{\partial\bar u}\over {\partial s}} + t{{\partial \bar u}\over {\partial t}}
$$
\nHence, show that
$$
\displaystyle 4y{{\partial u}\over{\partial y}} = {(s^2 - t^2)} \biggl({1\over s}{{\partial \bar u}\over{\partial t}} - {1\over t}{{\partial\bar u}\over{\partial s}}\biggr).
$$
| 3 | 0.666667 | 2 | When working on these equivalences, we need to have pairs of variables confined on each side. That is to say $\mathrm{function\ of\ 'old'} \equiv \mathrm{function\ of\ 'new'}$ so aim to find equations of the form  
$$
\displaystyle
\\[5pt]
{\partial\bar{u}\over\partial s} = ... \enspace {\partial\bar{u}\over\partial t}=...
\\[10pt] \Rightarrow s{\partial\bar{u}\over\partial s} + t{\partial\bar{u}\over\partial t} \equiv x,y\ \mathrm{variables\ only}
$$
***
Using the chain rule,
$$
\displaystyle
{\partial\bar{u}\over\partial s} = {\partial u\over\partial x}{\partial x\over\partial s} + {\partial u\over\partial y}{\partial y\over\partial s}
$$
Try expressing $\displaystyle {\partial\bar{u}\over\partial t}$ in a similar form using the chain rule.
***
$$
$$
Now can you find $\displaystyle {\partial\bar{u}\over\partial s}$ and $\displaystyle {\partial\bar{u}\over\partial t}$?
***
Remember you can leave them in terms of $\displaystyle {\partial u\over\partial x}$ and $\displaystyle {\partial u\over\partial y}$. You can find $\displaystyle {\partial x\over\partial s}$ and $\displaystyle {\partial y\over\partial s}$ from the definitions of $x$ and $y$.
\n$$
\begin{aligned}
{\partial\bar{u}\over\partial s} = t{\partial u\over\partial x} - {2t\over (s-t)^2}{\partial u\over\partial y}
\\
{\partial\bar{u}\over\partial t} = s{\partial u\over\partial x}+{2s\over(s-t)^2}{\partial u\over\partial y}
\end{aligned}
$$
Try substituting these into the RHS of the equation.
\n$$
\begin{aligned}
{\partial\bar{u}\over\partial s} = t{\partial u\over\partial x} - {2t\over (s-t)^2}{\partial u\over\partial y}
\\
{\partial\bar{u}\over\partial t} = s{\partial u\over\partial x}+{2s\over(s-t)^2}{\partial u\over\partial y}
\end{aligned}
$$
Try substituting these into the RHS of the equation.
***
Remember that $(s^2-t^2) = (s+t)(s-t)$. Can you finish the proof now?
| Using the chain rule,
$$
\displaystyle
{\partial\bar{u}\over\partial s} = {\partial u\over\partial x}{\partial x\over\partial s} + {\partial u\over\partial y}{\partial y\over\partial s}
$$
Try expressing $\displaystyle {\partial\bar{u}\over\partial t}$ in a similar form using the chain rule.
***
$$
\displaystyle {\partial\bar{u}\over\partial t} ={\partial u\over\partial x}{\partial x\over\partial t} + {\partial u\over\partial y}{\partial y\over\partial t}
$$
Now can you find $\displaystyle {\partial\bar{u}\over\partial s}$ and $\displaystyle {\partial\bar{u}\over\partial t}$?
***
Remember you can leave them in terms of $\displaystyle {\partial u\over\partial x}$ and $\displaystyle {\partial u\over\partial y}$. You can find $\displaystyle {\partial x\over\partial s}$ and $\displaystyle {\partial y\over\partial s}$ from the definitions of $x$ and $y$.
***
$$
\begin{aligned}
{\partial\bar{u}\over\partial s} = {\partial u\over\partial x}{\partial x\over\partial s} + {\partial u\over\partial y}{\partial y\over\partial s} = t{\partial u\over\partial x} - {2t\over (s-t)^2}{\partial u\over\partial y}
\end{aligned}
$$
$$
\begin{aligned}
{\partial\bar{u}\over\partial t}
={\partial u\over\partial x}{\partial x\over\partial t} + {\partial u\over\partial y}{\partial y\over\partial t}
=s{\partial u\over\partial x}+{2s\over(s-t)^2}{\partial u\over\partial y}
\end{aligned}
$$
\n$$
\begin{aligned}
{\partial\bar{u}\over\partial s} = t{\partial u\over\partial x} - {2t\over (s-t)^2}{\partial u\over\partial y}
\\
{\partial\bar{u}\over\partial t} = s{\partial u\over\partial x}+{2s\over(s-t)^2}{\partial u\over\partial y}
\end{aligned}
$$
Try substituting these into the RHS of the equation.
***
$$
\displaystyle
s{{\partial\bar u}\over {\partial s}} + t{{\partial \bar u}\over {\partial t}}
= st{\partial u\over\partial x} - {2st\over (s-t)^2}{\partial u\over\partial y} + st{\partial u\over\partial x}+{2st\over(s-t)^2}{\partial u\over\partial y}
$$
Can you finish the proof now?
***
$$
\begin{aligned}
s{{\partial\bar u}\over {\partial s}} + t{{\partial \bar u}\over {\partial t}}
&= 2st{\partial u\over\partial x}
\\
&= 2x{\partial u\over\partial x}
\end{aligned}
$$
as required.
\n$$
\begin{aligned}
{\partial\bar{u}\over\partial s} = t{\partial u\over\partial x} - {2t\over (s-t)^2}{\partial u\over\partial y}
\\
{\partial\bar{u}\over\partial t} = s{\partial u\over\partial x}+{2s\over(s-t)^2}{\partial u\over\partial y}
\end{aligned}
$$
Try substituting these into the RHS of the equation.
***
$$
\begin{aligned}
{1\over s}{\partial\bar{u}\over\partial t} &= {\partial u\over\partial x} + {2\over(s-t)^2}{\partial u\over\partial y}
\\
{1\over t}{\partial\bar{u}\over\partial s} &= {\partial u\over\partial x} - {2\over(s-t)^2}{\partial u\over\partial y}
\\
{1\over s}{\partial\bar{u}\over\partial t} - {1\over t}{\partial\bar{u}\over\partial s} &= {4\over(s-t)^2}{\partial u\over\partial y}
\end{aligned}
$$
Can you finish the proof now?
***
Remember that $(s^2-t^2) = (s+t)(s-t)$. Can you finish the proof now?
***
$$
\begin{aligned}
(s^2-t^2)\left({1\over s}{\partial\bar{u}\over\partial t} - {1\over t}{\partial\bar{u}\over\partial s}\right) &= 4{s^2-t^2\over(s-t)^2}{\partial u\over\partial y}
\\
&= 4{s+t\over s-t}{\partial u\over\partial y}
\\
&= 4y{\partial u\over\partial y}
\end{aligned}
$$
as required.
|  If $u = u(x, y)$ and $x$ and $y$ are related to two new independent variables $s$ and $t$ by
$$
\displaystyle x = st,\quad y = {{s + t} \over {s - t}}
$$
so that $u = \bar u(s, t)$, 
Use the chain rule to find $\displaystyle{\partial\bar u\over\partial s}$ and $\displaystyle{\partial \bar u\over \partial t}$ in terms of $\displaystyle{\partial u\over\partial x}$ and $\displaystyle{\partial u\over\partial y}$. 
\nHence show that 
$$
\displaystyle 2x{{\partial u}\over {\partial x}} = s{{\partial\bar u}\over {\partial s}} + t{{\partial \bar u}\over {\partial t}}
$$
\nHence, show that
$$
\displaystyle 4y{{\partial u}\over{\partial y}} = {(s^2 - t^2)} \biggl({1\over s}{{\partial \bar u}\over{\partial t}} - {1\over t}{{\partial\bar u}\over{\partial s}}\biggr).
$$
| 46 | 13 | 20 | 20 | 113 | 16 | 24 | 6 | 0 | If $u = u(x, y)$ and $x$ and $y$ are related to two new independent variables $s$ and $t$ by $ \displaystyle x = st,\quad y = {{s + t} \over {s - t}} $ so that $u = \bar u(s, t)$, Use the chain rule to find $\displaystyle{\partial\bar u\over\partial s}$ and $\displaystyle{\partial \bar u\over \partial t}$ in terms of $\displaystyle{\partial u\over\partial x}$ and $x$0. | 1 |
5594e094-50b1-4072-8bbf-e6c394fb67dc | 1 | 0 | 0 | 14 | 4 | 2 | 8 | 0 | If we denote $\vec{\mathcal{T}}$ as the `traction force' per unit area on each face of a particle; $\underline{\underline{{\sigma}}}$ as the stress tensor; and $\hat{n}$ as a unit normal; what is the order and physical dimensions ([$M$],[$L$],[$T$]) of each of the following terms? | As an example, the answer for $\vec{u}$ is Order 1, $[M]^0$, $[L]^1$, $[T]^{ -1}$.
| 1 | 0.333333 | 1 | null | null | If we denote $\vec{\mathcal{T}}$ as the `traction force' per unit area on each face of a particle; $\underline{\underline{{\sigma}}}$ as the stress tensor; and $\hat{n}$ as a unit normal; what is the order and physical dimensions ([$M$],[$L$],[$T$]) of each of the following terms?As an example, the answer for $\vec{u}$ is Order 1, $[M]^0$, $[L]^1$, $[T]^{ -1}$.
| 63 | 10 | 0 | 0 | 0 | 0 | 16 | 4 | 0 | If we denote $\vec{\mathcal{T}}$ as the `traction force' per unit area on each face of a particle; $\underline{\underline{{\sigma}}}$ as the stress tensor; and $\hat{n}$ as a unit normal; what is the order and physical dimensions $M$,$L$,$T$ of each of the following terms?As an example, the answer for $\vec{u}$ is Order 1, $[M]^0$, $[L]^1$, $[T]^{ -1}$. | 1 |
55a12f5a-681a-4375-a9ac-5546802168e2 | 0 | 0 | 3 | 24 | 6 | 1 | 4 | 4 | Verify the following vector operator identities, e.g., by:
* Expanding out each side in cartesian coordinates:
* It is possible to avoid expanding in cartesians by using the vector and differential properties of $\nabla$. One "trick,'' e.g., to evaluate $\nabla\times(\Omega\vec{V})$ is to write $\nabla \equiv \nabla_\Omega + \nabla_{\vec{V}}$ where $\nabla_\Omega$ differentiates $\Omega$ but gives zero when operating on $\vec{V}$ - i.e., it treats $\vec{V}$ as constant - and vice versa for $\nabla_{\vec{V}}$. This allows you to split the problem into pieces in which you can move some parts past the derivative aspects of $\nabla$.
| $$
\nabla\times(\Omega\vec{V})=\Omega(\nabla\times\vec{V})-\vec{V}\times\nabla\Omega
$$
\n$$
\nabla\cdot(\vec{U}\times\vec{V}) = \vec{V}\cdot(\nabla\times\vec{U})-\vec{U}\cdot(\nabla\times\vec{V})
$$
\n\[BONUS]:
$$
\nabla\times(\nabla\times\vec{E})=\nabla(\nabla\cdot \vec{E})-\nabla^2\vec{E}
$$
| 3 | 1 | 4 | There are two methods to approach this problem:
1. Separating $\nabla$ into two parts (see hint in question). This is a non-brute force approach and does not require you to evaluate in Cartesian coordinates.
2. Evaluate the curl $\nabla\times\nabla\vec{V}$ by brute force (i.e., by expressing the curl in Cartesian coordinates). 
See further hints below\...
***
**(Method 1)** Let $\nabla = (\nabla_\Omega + \nabla_V)$. 
***
**(Method 1)** Expand out the cross product using the distributive nature of the cross product. 
***
**(Method 1)** By considering which term is held constant (e.g. $\vec{V}$ is constant with respect to $\nabla_\Omega$), you should be able to remove the dependence of the grad on a certain variable, e.g., $\nabla_\Omega \to \nabla$. 
***
**(Method 2)** Evaluate the curl of the field $\Omega \vec{V}$ (remember that $\Omega$ is a scalar field), in Cartesian coordinates, using the cross product...
***
**(Method 2)** ...Apply the differentiation product rule when evaluating each partial derivative. 
***
**(Method 2)** Compare your curl expression to $\Omega(\nabla\times\vec{V})-\vec{V}\times\nabla\Omega$... are they equivalent? You may have to write out each of these terms in determinant form to see the equivalence.
\nHere, the best approach is to apply the hint in the question. The $\color{green}\textsf{Structured Tutorial}$ and $\color{red}\textsf{Worked Solutions}$ will cover this method only, but you can also attempt to prove the relation by brute force (you will need to consider the formulae for divergence and curl in Cartesian coordinates).
***
Let $\nabla = \nabla_U + \nabla_V$. 
***
Expand out the dot product using its distributive property. 
***
Make use of the scalar triple product identity:
$$
\vec{A}\cdot(\vec{B}\times\vec{C})=\vec{B}\cdot(\vec{C}\times\vec{A}) = \vec{C}\cdot(\vec{A}\times\vec{B})
$$
Think of this in terms of volumes of parallelepiped's (does the ordering of the vectors matter)?
***
$\vec{V}$ is treated as constant with respect to $\nabla_U$... 
***
... This allows you to factor $\vec{V}$ out from the cross product, and let $\nabla_U \to \nabla$. Likewise for $\vec{U}$ and $\nabla_V$.
***
Hence, re-arrange the dot/cross product into the required form.
\nUnlike parts (a) and (b), this identity consists of a single vector field, rather than two. Therefore, you cannot use the method suggested in the question hint. Instead, take the Cartesian coordinates 'brute force' approach.
***
Start on the LHS. First, write down $\nabla\times\vec{E}$ in Cartesian coordinates.
***
Take the curl of this result, $\nabla\times(\nabla\times\vec{E})$ (i.e., use the same formula, but replace $E_x,E_y,E_z$ with the components of the curl).
***
Now, consider the RHS. $\nabla(\nabla\cdot\vec{E})$ is the gradient of the divergence. Start by writing the divergence, then take the gradient of this scalar field.
***
$\nabla^2\vec{E}$ is the Laplacian (see Lectures). You should notice term cancellation when taking $\nabla(\nabla\cdot\vec{E})-\nabla^2\vec{E}$. Are the LHS and RHS of the equation now equivalent?
|
, Letting $\nabla = \nabla_\Omega + \nabla_V$:
***
$$
\nabla\times(\Omega\vec{V}) = (\nabla_\Omega+\nabla_V)\times(\Omega\vec{V})
$$
***
Applying the distributive property of the cross product, 
***
$$
= (\nabla_\Omega \Omega \times \vec{V}) + (\Omega\nabla_V \times\vec{V})
$$
In the first term, $\nabla_\Omega$ treats $\vec{V}$ as a constant, so $\nabla_\Omega \Omega = \nabla\Omega$. Likewise, in the second term $\nabla_V \times \vec{V} = \nabla\times\vec{V}$, because $\Omega$ is treated as constant.  
***
Therefore, switching the order of the first term around (which introduces a negative sign), we have:
$$
(\nabla_\Omega \Omega \times \vec{V}) + (\Omega\nabla_V \times\vec{V}) = \Omega(\nabla\times\vec{V})-\vec{V}\times\nabla\Omega
$$
QED.
, Evaluating the curl of the field $\Omega \vec{V}$ in Cartesian coordinates:
***
$$
\nabla\times(\Omega\vec{V})=\begin{vmatrix}\mathbf{\hat{i}}&\mathbf{\hat{j}}&\mathbf{\hat{k}}\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\\Omega V_x&\Omega V_y&\Omega V_z\end{vmatrix}
$$
***
$$
=\mathbf{\hat{i}}\left(\frac{\partial\Omega V_z}{\partial y}-\frac{\partial\Omega V_y}{\partial z}\right)+\mathbf{\hat{j}}\left(\frac{\partial\Omega V_x}{\partial z}-\frac{\partial\Omega V_z}{\partial x}\right)+\mathbf{\hat{k}}\left(\frac{\partial\Omega V_y}{\partial x}-\frac{\partial\Omega V_x}{\partial y}\right)
$$
Evaluating each partial derivative using the product rule, the number of terms doubles:
***
$$
\begin{aligned}
&=\Omega\left(\mathbf{\hat{i}}\left(\frac{\partial V_z}{\partial y}-\frac{\partial V_y}{\partial z}\right)+\mathbf{\hat{j}}\left(\frac{\partial V_x}{\partial z}-\frac{\partial V_z}{\partial x}\right)+\mathbf{\hat{k}}\left(\frac{\partial V_y}{\partial x}-\frac{\partial V_x}{\partial y}\right)\right)\\
&-\mathbf{\hat{i}}\left(V_y\frac{\partial \Omega}{\partial z}-V_z\frac{\partial \Omega}{\partial y}\right)-\mathbf{\hat{j}}\left(V_z\frac{\partial \Omega}{\partial x}-V_x\frac{\partial \Omega}{\partial z}\right)-\mathbf{\hat{k}}\left(V_x\frac{\partial \Omega}{\partial y}-V_y\frac{\partial \Omega}{\partial x}\right)
\end{aligned}
$$
***
The first term is $\Omega \nabla\times\vec{V}$. The second term is $\vec{V}\times\nabla\Omega$ since:
$$
\vec{V}\times\nabla\Omega = \begin{vmatrix}\mathbf{\hat{i}}&\mathbf{\hat{j}}&\mathbf{\hat{k}}\\V_x&V_y&V_z\\\frac{\partial \Omega}{\partial x}&\frac{\partial \Omega}{\partial y}&\frac{\partial \Omega}{\partial z}\end{vmatrix}
$$
Therefore, we have proven that:
$$
\nabla\times(\Omega\vec{V})=\Omega(\nabla\times\vec{V})-\vec{V}\times\nabla\Omega
$$
\nLet:
$$
\nabla = \nabla_U + \nabla_V
$$
Thus:
$$
\nabla\cdot(\vec{U}\times\vec{V})=(\nabla_U+\nabla_V)\cdot(\vec{U}\times\vec{V})
$$
***
Expanding this out using the distributive nature of the dot product:
***
$$
= \nabla_U\cdot(\vec{U}\times\vec{V})-\nabla_V\cdot(\vec{V}\times\vec{U})
$$
***
Applying the triple scalar product identity:
$$
\vec{A}\cdot(\vec{B}\times\vec{C})=\vec{B}\cdot(\vec{C}\times\vec{A}) = \vec{C}\cdot(\vec{A}\times\vec{B})
$$
***
$$
\nabla_U\cdot(\vec{U}\times\vec{V})-\nabla_V\cdot(\vec{V}\times\vec{U})= \vec{V}\cdot(\nabla_U\times\vec{U})-\vec{U}\cdot(\nabla_V\times\vec{V})
$$
***
* $\vec{V}$ is treated as a constant with respect to $\nabla_U$.
* $\vec{U}$ is treated as a constant with respect to $\nabla_V$.
So, $\nabla_U\times\vec{U}=\nabla\times\vec{U}$ since $\nabla_U$ is already with respect to $U$ only. Likewise $\nabla_V\times\vec{V}=\nabla\times\vec{V}$. This gives us the required form:
***
$$
=\vec{V}\cdot(\nabla\times\vec{U})-\vec{U}(\nabla\times\vec{V})
$$
Therefore,
$$
\nabla\cdot(\vec{U}\times\vec{V}) = \vec{V}\cdot(\nabla\times\vec{U})-\vec{U}\cdot(\nabla\times\vec{V})
$$
\nStarting with the LHS and $\nabla\times\vec{E}$:
***
$$
\nabla\times\vec{E}=\mathbf{\hat{i}}\left(\frac{\partial E_z}{\partial y}-\frac{\partial E_y}{\partial z}\right)+\mathbf{\hat{j}}\left(\frac{\partial E_x}{\partial z}-\frac{\partial E_z}{\partial x}\right)+\mathbf{\hat{k}}\left(\frac{\partial E_y}{\partial x}-\frac{\partial E_x}{\partial y}\right)
$$
***
Taking the curl of this result (i.e., the same formula, but the $x$, $y$ and $z$ components replaced with the $x$, $y$ and $z$ components of the curl):
***
$$
\begin{aligned}
\nabla\times(\nabla\times\vec{E})=&\,\mathbf{\hat{i}}\left(\frac{\partial}{\partial y}\left(\frac{\partial E_y}{\partial x}-\frac{\partial E_x}{\partial y}\right)-\frac{\partial}{\partial z}\left(\frac{\partial E_x}{\partial z}-\frac{\partial E_z}{\partial x}\right)\right)\\
&\mathbf{\hat{j}}\left(\frac{\partial}{\partial z}\left(\frac{\partial E_z}{\partial y}-\frac{\partial E_y}{\partial z}\right)-\frac{\partial}{\partial x}\left(\frac{\partial E_y}{\partial x}-\frac{\partial E_x}{\partial y}\right)\right)\\
&\mathbf{\hat{k}}\left(\frac{\partial}{\partial x}\left(\frac{\partial E_x}{\partial z}-\frac{\partial E_z}{\partial x}\right)-\frac{\partial}{\partial y}\left(\frac{\partial E_z}{\partial y}-\frac{\partial E_y}{\partial z}\right)\right)
\end{aligned}
$$
***
Now compare this to the RHS. Start with $\nabla(\nabla\cdot\vec{E})$:
***
$$
\begin{aligned}
\nabla(\nabla\cdot\vec{E})=&\frac{\partial}{\partial x}\left(\frac{\partial E_x}{\partial x}+\frac{\partial E_y}{\partial y}+\frac{\partial E_z}{\partial z}\right)\mathbf{\hat{i}}+\frac{\partial}{\partial y}\left(\frac{\partial E_x}{\partial x}+\frac{\partial E_y}{\partial y}+\frac{\partial E_z}{\partial z}\right)\mathbf{\hat{j}}\\
& \frac{\partial}{\partial z}\left(\frac{\partial E_x}{\partial x}+\frac{\partial E_y}{\partial y}+\frac{\partial E_z}{\partial z}\right)\mathbf{\hat{k}}
\end{aligned}
$$
and then the Laplacian:
***
$$
\begin{aligned}
\nabla^2\vec{E} = &\left(\frac{\partial^2 E_x}{\partial x^2}+\frac{\partial^2 E_x}{\partial y^2}+\frac{\partial^2 E_x}{\partial z^2}\right)\mathbf{\hat{i}}+\left(\frac{\partial^2 E_y}{\partial x^2}+\frac{\partial^2 E_y}{\partial y^2}+\frac{\partial^2 E_y}{\partial z^2}\right)\mathbf{\hat{j}}\\
&\left(\frac{\partial^2 E_z}{\partial x^2}+\frac{\partial^2 E_z}{\partial y^2}+\frac{\partial^2 E_z}{\partial z^2}\right)\mathbf{\hat{k}}
\end{aligned}
$$
***
So, when taking $\nabla(\nabla\cdot \vec{E})-\nabla^2\vec{E}$, notice that the $\partial^2 E_x/\partial x^2$, $\partial^2 E_y/\partial y^2$, $\partial^2 E_z/\partial z^2$ terms cancel. This leaves us with $\nabla\times(\nabla\times\vec{E})$.
| Verify the following vector operator identities, e.g., by:
* Expanding out each side in cartesian coordinates:
* It is possible to avoid expanding in cartesians by using the vector and differential properties of $\nabla$. One "trick,'' e.g., to evaluate $\nabla\times(\Omega\vec{V})$ is to write $\nabla \equiv \nabla_\Omega + \nabla_{\vec{V}}$ where $\nabla_\Omega$ differentiates $\Omega$ but gives zero when operating on $\vec{V}$ - i.e., it treats $\vec{V}$ as constant - and vice versa for $\nabla_{\vec{V}}$. This allows you to split the problem into pieces in which you can move some parts past the derivative aspects of $\nabla$.
$$
\nabla\times(\Omega\vec{V})=\Omega(\nabla\times\vec{V})-\vec{V}\times\nabla\Omega
$$
\n$$
\nabla\cdot(\vec{U}\times\vec{V}) = \vec{V}\cdot(\nabla\times\vec{U})-\vec{U}\cdot(\nabla\times\vec{V})
$$
\n\[BONUS]:
$$
\nabla\times(\nabla\times\vec{E})=\nabla(\nabla\cdot \vec{E})-\nabla^2\vec{E}
$$
| 98 | 12 | 49 | 49 | 273 | 25 | 5 | 3 | 0 | Verify the following vector operator identities, e.g., by: Expanding out each side in cartesian coordinates: It is possible to avoid expanding in cartesians by using the vector and differential properties of $ abla$. One "trick,'' e.g., to evaluate $ abla\times(\Omega\vec{V})$ is to write $ abla \equiv abla_\Omega + abla_{\vec{V}}$ where $ abla_\Omega$ differentiates $\Omega$ but gives zero when operating on $\vec{V}$ - i.e., it treats $\vec{V}$ as constant - and vice versa for $ abla_{\vec{V}}$. | 2 |
55a2d487-3921-4465-85fa-8aac1bf0f672 | 0 | 0 | 2 | 16 | 6 | 1 | 2 | 6 | Consider the integral
$$
\displaystyle I = \int_0^1 {{{{ {{x}^4} {{(1 - x)}^4 }}} }\over{ {(1 + {x}^2)}}}\ \mathrm{d}x
$$
| Use partial fractions to expand the integrand in order to show that:
$$
\displaystyle I = {22\over7} - \pi.
$$
\n By noting that $1\leq {1 + {x}^2}\leq 2$ in the denominator of $I$, show that:
$$
\displaystyle {{22}\over7} - {1\over{630}} < \pi < {{22}\over7} - {1\over{1260}},
$$
that is to say:
$$
3.14126... < \pi < 3.14206...
$$
| 2 | 1 | 2 | First, use the binomial expansion on the numerator.
***
Now divide using whichever method you prefer (e.g. long division or inspection)
***
Remember that $\displaystyle \int {1 \over 1+x^2} \,\mathrm{d}x = \tan^{-1} x + c$
***
You need to divide out the integrand, which is currently in the form $\displaystyle \frac{\mathrm{8th \,degree}}{\mathrm{quadratic}}$, so that it is in the form $\displaystyle \mathrm{6th\,degree\,polynomial}+\frac{\mathrm{linear}}{\mathrm{quadratic}}$ which you can then integrate as normal.
\nThe bounds on $1+x^2$ in the integral limit $[0,1]$ allow you to bound $I$ suitably, giving a good approximation to $\pi$.
Write an inequality for $ I={22\over7} - \pi $ using that $1\leq 1+x^2 \leq 2$ in the denominator
***
Use the binomial expansion, and then evaluate the two integrals.
| You need to divide out the integrand, which is currently in the form $\displaystyle \frac{\mathrm{8th \,degree}}{\mathrm{quadratic}}$, so that it is in the form $\displaystyle \mathrm{6th\,degree\,polynomial}+\frac{\mathrm{linear}}{\mathrm{quadratic}}$ which you can then integrate as normal.
***
First, use the binomial expansion on the numerator.
***
$$
\begin{aligned}
\frac{x^4(1-x)^4}{1+x^2}
&= \frac{x^4(1-4x+6x^2-4x^3+x^4)}{1+x^2} \\
&= \frac{x^8-4x^7+6x^6-4x^5+x^4}{1+x^2} \\
\end{aligned}
$$
***
Now you can divide using whichever method you prefer (e.g. long division or inspection)
***
Using inspection,
$$
\begin{aligned}
\frac{x^4(1-x)^4}{1+x^2}
&= \frac{x^6(1+x^2)-4x^5(1+x^2)+5x^4(1+x^2)-4x^2(1+x^2)+4(1+x^2)-4}{1+x^2} \\
&= x^6-4x^5+5x^4-4x^2+4-{4\over 1+x^2}
\end{aligned}
$$
Now can you find the integral?
***
Remember that $\displaystyle \int {1 \over 1+x^2} \,\mathrm{d}x= \tan^{-1} x + c$
***
$$
\begin{aligned}
I &= \left[ {1\over7}x^7 - {2\over3}x^6 + x^5 - {4\over3}x^3 + 4x - 4\tan^{-1}x \right]^1_0
\\
&= {1\over7}-{2\over3}+1-{4\over3} + 4 - \pi
\\
&= {22\over 7} - \pi
\end{aligned}
$$
\nThe bounds on $1+x^2$ in the integral limit $[0,1]$ allow you to bound $I$ suitably, giving a good approximation to $\pi$.
Write an inequality for $ I={22\over7} - \pi $ using that $1\leq 1+x^2 \leq 2$ in the denominator
***
$$
\displaystyle \int_0^1 \frac{x^4(1-x)^4}{2} \mathrm{d}x < I={22\over7} - \pi < \int_0^1 \frac{x^4(1-x)^4}{1}\mathrm{d}x
$$
***
Use the binomial expansion, and then evaluate the two integrals.
***
$$
\begin{aligned}
{1\over2}\int_0^1 x^8-4x^7+6x^6-4x^5+x^4 \,\mathrm{d}x < {22\over7} - \pi < \int_0^1 x^8-4x^7+6x^6-4x^5+x^4 \,\mathrm{d}x
\end{aligned}
$$
$$
\begin{aligned}
{1\over2}\left[{1\over5}-{2\over3}+{6\over7}-{1\over2}+{1\over9}\right] &< {22\over7}-\pi<\left[{1\over5}-{2\over3}+{6\over7}-{1\over2}+{1\over9}\right]
\\
{1\over1260} &< {22\over7}-\pi < {1\over630}
\\
{22\over 7}-{1\over1260} &< \pi < {22\over 7}-{1\over630}
\\
3.14126\ldots &< \pi < 3.14206\ldots
\end{aligned}
$$
***
Historical interest:
Archimedes famously obtained $3\frac{1}{7} > \pi > 3\frac{10}{71}$ using a 96 sided polygon (see below for the $n = 5$ case)! This is equivalent to 6 d.p. accuracy. 
| Consider the integral
$$
\displaystyle I = \int_0^1 {{{{ {{x}^4} {{(1 - x)}^4 }}} }\over{ {(1 + {x}^2)}}}\ \mathrm{d}x
$$
Use partial fractions to expand the integrand in order to show that:
$$
\displaystyle I = {22\over7} - \pi.
$$
\n By noting that $1\leq {1 + {x}^2}\leq 2$ in the denominator of $I$, show that:
$$
\displaystyle {{22}\over7} - {1\over{630}} < \pi < {{22}\over7} - {1\over{1260}},
$$
that is to say:
$$
3.14126... < \pi < 3.14206...
$$
| 35 | 6 | 5 | 5 | 76 | 9 | 31 | 5 | 0 | Consider the integral $ \displaystyle I = \int_0^1 {{{{ {{x}^4} {{(1 - x)}^4 }}} }\over{ {(1 + {x}^2)}}}\ \mathrm{d}x $ Use partial fractions to expand the integrand in order to show that: $ \displaystyle I = {22\over7} - \pi. | 1 |
561094ab-75df-481c-a5f2-7a7bf711a46d | 0 | 0 | 3 | 21 | 6 | 1 | 1 | 3 | In this question we are looking at the power spectrum of a Fourier series and why that is a useful quantity.
Let $P_n$ be given by: $P_n = a_n^2+b_n^2$. The distribution of the values of, $\lbrace{P_n\rbrace}$, is called the **power spectrum** of the Fourier series.
| Show that writing down the Fourier series in the form:
$$
f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty}{\alpha_n \cos(nx - \theta_n)} \
$$
$$
$$
is equivalent to the trigonometric Fourier series if:
$$
\begin{aligned}
a_n &= \alpha_n\cos\theta_n, ~~~~~ b_n &= \alpha_n\sin\theta_n \\
\alpha_n^2 &= a_n^2 + b_n^2, ~~~~~~~\tan\theta_n &= b_n/a_n \ .
\end{aligned}
$$
\nShow that the form of ${P_n}$ is unchanged by a phase shift in $f$. That is, the power spectrum is the same if $f(x)$ is replaced by $f(x+\phi)$, where $\phi$ is an arbitrary phase.
\nShow that:
$$
\int_{-\pi}^{\pi}{|f(x)|^2dx}
$$
can be written in terms of the power spectrum $\alpha_n$ as:
$$
\int_{-\pi}^{\pi}{|f(x)|^2dx}=2\pi\left(\frac{a_0^2}{4}
+ \frac{1}{2}\sum_{n=1}^{\infty}{\alpha_n^2}\right)
$$
| 3 | 0.666667 | 3 | This question can be separated into two parts. 
1. Start from the given form and prove equivalence to the trigonometric Fourier series by using the definitions of $a_n$ and $b_n$. 
2. Start from the trigonometric Fourier series and prove equivalence to the given equation using the definitions of $\alpha_n$ and $\tan\theta_n$. 
***
(**Part 1**) Use the compound angle formula to expand $\cos(nx-\theta_n)$. 
***
(**Part 1**) You should now be able to use the definition of $a_n$ and $b_n$ to show that the given form is equivalent to the standard trigonometric Fourier Series equation.
***
(**Part 2**) Start with the trigonometric Fourier series equation. 
***
(**Part 2**) It may help you to factor $\sqrt{a_n^2 + b_n^2}$ into the equation (i.e., divide top and bottom by this)
***
(**Part 2**) Can you see how this manipulation allows you to introduce $\sin\theta_n$ and $\cos\theta_n$ into the expression
\nEvaluate $f(x+\phi)$ for the power spectrum formula. 
***
Expand out the brackets as in part (a). 
***
Is the form of the power spectrum changed?
\nStart by inputting the expression from part (a) into $\int_{-\pi}^{\pi}{|f(x)|^2dx}$.
***
Can you expand the brackets $|f(x)|^2$?
***
Now evaluate each term of the integral independently... 
***
... Do not be alarmed by the occurrence of a summation within an integral (each term in the summation can be treated independently). 
***
One of the terms in your integral should be the *square* of a summation. You can use the summation rules covered in the lectures to combine these summations into one...
***
... Make use of the orthogonality of trigonometric functions (see *PS1 Question 2/6*) to evaluate the integral. 
| To begin, we expand the $\cos(nx-\theta_n)$ term using the compound angle formula:
***
$$
\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B \; .
$$
***
Inputting this expression into the given formula, we see that:
***
$$
\begin{aligned}
\alpha_n\cos(nx-\theta_n)
&= \alpha(\cos nx \cos \theta_n + \sin nx \sin \theta_n) \\
&= \alpha\cos\theta_n \cos nx + \alpha\sin\theta_n \sin nx \\
\end{aligned}
$$
***
$$
~~~~~~~~~= a_n \cos nx + b_n \sin nx
$$
where we defined $a_n$ and $b_n$ as in the question. This shows that the two forms are equivalent.
***
Now, we move in the opposite direction to make use of the other two Power spectrum terms. A good place to start is to factor $\alpha_n$ in terms of $a_n,b_n$ into the expression:
***
$$
\begin{aligned}
a_n \cos nx + b_n \sin nx
= \sqrt{a_n^2 + b_n^2}
\left(
\frac{a_n}{\sqrt{a_n^2 + b_n^2}} \cos nx
+ \frac{b_n}{\sqrt{a_n^2 + b_n^2}} \sin nx
\right) \; ,\\
\end{aligned}
$$
***
The advantage of this approach is that we can now introduce $\theta_n$. Consider that $\tan\theta_n = b_n/a_n$. 
From the above triangle, we see that:
$$
\begin{aligned}
&\sin\theta_n = \frac{b_n}{\sqrt{a_n^2+b_n^2}}\\
&\cos\theta_n = \frac{a_n}{\sqrt{a_n^2+b_n^2}}
\end{aligned}
$$
***
As the square of the terms inside the parenthesis are now normalised, we can represent them as $\sin$ and $\cos$ of an angle:
***
$$
a_n \cos nx + b_n \sin nx= \alpha_n(\cos\theta_n \cos nx + \sin\theta_n\sin nx) = \alpha_n \cos(nx - \theta_n) \; ,
$$
with $\alpha_n$ and $\theta_n$ defined as in the question. Hence we have used the definitions to transition from the trigonometric Fourier series to the given form, QED.
\nReplacing $x$ with $x+\phi$:
***
$$
\begin{aligned}
\alpha_n\cos(n(x+\phi)-\theta_n)
&= \alpha(\cos (n(x+\phi)) \cos \theta_n + \sin (n(x+\phi)) \sin \theta_n) \\
&= \alpha\cos\theta_n \cos (n(x+\phi)) + \alpha\sin\theta_n \sin (n(x+\phi)) \\
\end{aligned}
$$
The magnitude and form of the power Spectrum is unchanged; the $x$ values are simply shifted by $\phi$. 
\nWe have:
$$
\begin{aligned}
\int_{-\pi}^{\pi}{|f(x)|^2dx}
&= \int_{-\pi}^{\pi}{
\left|
\frac{a_0}{2} + \sum_{n=1}^{\infty}{\alpha_n \cos(nx - \theta_n)}
\right|^2 dx} \\
\end{aligned}
$$
Expanding the brackets (treat the moduli as brackets):
***
$$
\begin{align}
\int_{-\pi}^{\pi}{
\left(\frac{a_0^2}{4} + a_0\sum_{n=1}^{\infty}{\alpha_n \cos(nx - \theta_n)} + \left(\sum_{n=1}^{\infty}{\alpha_n \cos(nx - \theta_n)}\right)^2
\right) \text{d}x} \\
\end{align}
$$
Let us evaluate the final term in the integral independently: 
***
$$
\begin{aligned}
&\int^{\pi}_{-\pi}{\left(\sum_{n=1}^{\infty}{\alpha_n \cos(nx - \theta_n)}\right)^2~\text{d}x}\\
&= \int^{\pi}_{-\pi}{\sum_{n=1}^{\infty}{\alpha_n \cos(nx - \theta_n)}\sum_{m=1}^{\infty}{\alpha_m \cos (mx - \theta_n)}~\text{d}x}\\
\end{aligned}
$$
We have expressed one of the summations in terms of $m$, so that we can distinguish between the two summations.
***
As we have seen in the Fourier lectures, the summations can be combined as follows:
$$
= \sum_{n,m=1}^{\infty}\alpha_n \alpha_m\int^{\pi}_{-\pi}{{ \cos(nx-\theta_n)\cos(mx-\theta_m)}~\text{d}x}
$$
The summation is brought to the outside of the expression, and the constants factored out of the integral. Manipulations such as the one shown above are covered in the lectures.
***
Next, we invoke the orthogonality of the $\cos$ function (see *PS1 Question 2 & 6*). When $n \ne m$, the cosine functions are orthogonal, and so the integral evaluates to $0$. When $n=m$, the integral evaluates to $1$. Can you think why the orthogonality rules still apply despite the $\theta_n$ and $\theta_m$ being different? We are left with:
***
$$
\int^{\pi}_{-\pi}{\left(\sum_{n=1}^{\infty}{\alpha_n \cos(nx - \theta_n)}\right)^2~\text{d}x}=\pi{\sum_{n=1}^{\infty}{\alpha_n^2}}
$$
We recombined $n,m$ into a single term, and $\alpha_n\alpha_m=\alpha_n^2$, because the integral only exists for $n=m$. 
***
Returning to eq.(1) to evaluate the other two parts of the integral:
***
The first term evaluates simply to $2\pi(\frac{a_0^2}{4})$. 
***
The second term evaluates to $0$. This is because the area beneath the $\cos(nx)$ curve for any $n$ is 0 (the positive and negative areas cancel). **Note**: do not be thrown off by the summation in the integral: because each individual term in the summation would evaluate independently to 0. 
***
Hence we have for the final integral:
***
$$
\int_{-\pi}^{\pi}{|f(x)|^2dx} = 2\pi\left(\frac{a_0^2}{4}
+ \frac{1}{2}\sum_{n=1}^{\infty}{\alpha_n^2}\right) \; .
$$
| In this question we are looking at the power spectrum of a Fourier series and why that is a useful quantity.
Let $P_n$ be given by: $P_n = a_n^2+b_n^2$. The distribution of the values of, $\lbrace{P_n\rbrace}$, is called the **power spectrum** of the Fourier series.
Show that writing down the Fourier series in the form:
$$
f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty}{\alpha_n \cos(nx - \theta_n)} \
$$
$$
$$
is equivalent to the trigonometric Fourier series if:
$$
\begin{aligned}
a_n &= \alpha_n\cos\theta_n, ~~~~~ b_n &= \alpha_n\sin\theta_n \\
\alpha_n^2 &= a_n^2 + b_n^2, ~~~~~~~\tan\theta_n &= b_n/a_n \ .
\end{aligned}
$$
\nShow that the form of ${P_n}$ is unchanged by a phase shift in $f$. That is, the power spectrum is the same if $f(x)$ is replaced by $f(x+\phi)$, where $\phi$ is an arbitrary phase.
\nShow that:
$$
\int_{-\pi}^{\pi}{|f(x)|^2dx}
$$
can be written in terms of the power spectrum $\alpha_n$ as:
$$
\int_{-\pi}^{\pi}{|f(x)|^2dx}=2\pi\left(\frac{a_0^2}{4}
+ \frac{1}{2}\sum_{n=1}^{\infty}{\alpha_n^2}\right)
$$
| 117 | 14 | 43 | 43 | 446 | 13 | 72 | 11 | 0 | Let $P_n$ be given by: $P_n = a_n^2+b_n^2$. Show that writing down the Fourier series in the form: $ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty}{\alpha_n \cos(nx - \theta_n)} \ $ $ $ is equivalent to the trigonometric Fourier series if: $ \begin{aligned} a_n &= \alpha_n\cos\theta_n, ~~~~~ b_n &= \alpha_n\sin\theta_n \\ \alpha_n^2 &= a_n^2 + b_n^2, ~~~~~~~\tan\theta_n &= b_n/a_n \ . \end{aligned} $ Show that the form of ${P_n}$ is unchanged by a phase shift in $f$. Show that: $P_n = a_n^2+b_n^2$1 | 4 |
563d3f4f-c3cb-45c2-8814-261bb1869b10 | 5 | 1 | 0 | 6 | 4 | 1 | 0 | 0 | 
  
You are building a toy car to race in a miniature version of Formula 1. In this racing series, car weight is set at 8kg. You are required to use a small motor with a speed vs torque curve shown below.
  
<https://www.engineeringtoolbox.com/electrical-motors-torques-d_651.html>
| The gearbox you have been supplied only came with the gear part numbers, which are as follows…
  
* G1 - 12
* G1 - 30
* G2 - 8
* G2 - 28
  
What is the step-down ratio of the gearbox?
\nGiven that the regulations for Formula Miniature stipulate that your car is allowed to travel no faster than 3 m/s, you are going to have to limit the speed of the motor. The wheel diameter is 0.04m. Using your answer to part a, determine this maximum speed you will need to limit the motor to. Additionally, find the torque that will be produced at this maximum speed. You can assume the full load operating speed for this motor is 1000rpm and that the maximum torque it can provide is 0.045Nm.
\nAssuming that your car can only operate at its limited motor speed, calculated in part b, determine its instanteous acceleration at the beginning of a race (from a standing start).
\nGiven that the Red Bull car can achieve an acceleration of 1.3 $\mathrm{~m} / \mathrm{s}^2$, but only has a max speed of 2.8 m/s, will you beat them off the line into turn 1, a distance of 8m?
Assume acceleration is uniform up until the point of reaching maximum speed.
| 4 | 0.666667 | 4 | Use your knowledge of gear selection and part numbers from DTP to determine which gears mesh with each other and how many teeth they each have.
\nAs with previous parts of this question set, use the wheel's circumference as a measure of the distance travelled in 1 revolution. This can then be used to find the wheel rotation speed.
\nThis question follows a similar process to previous questions in this set. Use question 2 for help if you are struggling. The key aspect is to determine how to get a force from the torque value.
Remember that the torque is affected by the transmission ratio.
\nHint - break down the total time it will take to cover the 8m into time during acceleration and time at top speed.
| First, you need to determine that these gear part numbers represent a 12-tooth gear meshing with a 30-tooth gear, and an 8-tooth gear meshing with a 28-tooth gear.
Then, you can calculate the step-down ratio produced by the first stage:
step down $=\frac{30}{12}=2.5$
Next, find the step down for the second stage:
step down $=\frac{28}{8}=3.5$
Finally, calculate the total step down ratio achieved:
total step down $=2.5 \times 3.5=8.75$
\nWe first need to find the rotational speed of the wheel which corresponds to the maximum 3m/s speed.
To do this, we first need the circumference of the wheel:
circumference $=\pi D=0.04 \pi$ metres
Using this, we can then find the rotational speed of the wheels:
Rotational speed $=\frac{\text { linear speed }}{\text { circumference }}=\frac{3}{0.04 \pi}=23.87 \ldots \mathrm{s}^{-1}$
Converting this into rpm gives:
$23.87 \ldots \times 60=1432.39 \ldots r p m$
***
We now need to use the transmission ratio from the gears to determine the engine speed.
engine speed $=1432.39 \ldots \times 8.75=12,533.45 \ldots r p m$
***
The next step is to determine the corresponding torque delivered by the motor at this operating point.
To do this we need to use the graph which gives its axis in terms of percentages of the maximum speed and torque, so this % of max speed is what must be found first.
$\%$ of max speed $=\frac{12,533.45 \ldots}{18,000}=69.63 \ldots \%$
From the graph, we then need to identify this point:
***
As can be identified from the graph above, the % of maximum torque that can be expected is 45%.
As the maximum torque is 45mNm, the torque at this operating point can be calculated as follows:
Torque $=0.45 \times 45=20.25 \mathrm{mNm}=0.02025 \mathrm{Nm}$
\nFirst, the torque after the step-down needs to be calculated.
Torque $=0.02025 \times 8.75=0.1771875 \mathrm{Nm}$
***
This torque can then be used to find the force at the wheel.
Force $=\frac{\text { Torque }}{\text { radius }}=\frac{0.177 \ldots}{0.02}=8.859375 \mathrm{~N}$
***
Then, use the force and the weight of the car, to find its acceleration.
Acceleration $=\frac{\text { Force }}{\text { mass }}=\frac{8.859375}{8}=1.107 \ldots \mathrm{m} / \mathrm{s}^2$
\nFirst, we will consider your design:
To do this, we first need to find the time it will take to reach maximum speed.
time to reach max speed $=\frac{\text { max speed }}{\text { acceleration }}=\frac{3}{1.107 \ldots}=2.708 \ldots \mathrm{s}$
We then need to find the distance travelled during this acceleration period.
To find the equation for this, you can either integrate an equation for speed from constant acceleration or use the SUVAT equations.
distance $=\frac{1}{2} a t^2=\frac{1}{2} \times 1.107 \ldots \times 2.708 \ldots^2=4.063 \ldots m$
We can then find the time required to travel the remaining distance (at the top speed).
time $=\frac{\text { remaining distance }}{\text { speed }}=\frac{8-4.063 \ldots}{3}=1.312 \ldots \mathrm{s}$
Finally, we can find the total time it will take your car to reach turn 1.
total time $=2.708 \ldots+1.312 \ldots=4.021 \ldots s$
***
Next, we need to consider Red Bull's design. We have to follow the same process for this.
time to reach max speed $=\frac{\text { max speed }}{\text { acceleration }}=\frac{2.8}{1.3}=2.153 \ldots \mathrm{s}$
Next, find the distance travelled in this time...
distance $=\frac{1}{2} a t^2=\frac{1}{2} \times 1.3 \times 2.153 \ldots^2=3.015 \ldots m$
Next, the remaining time spent can be found.
time $=\frac{\text { remaining distance }}{\text { speed }}=\frac{8-3.015 \ldots}{2.8}=1.780 \ldots \mathrm{s}$
Finally, the total time can be evaluated.
total time $=2.153 \ldots+1.780 \ldots=3.934 \ldots s$
***
From the times for your's and Red Bull's cars, we can find the difference in time between you.
It is apparent that Red Bull are faster, the magnitude of that gap is...
time gap into turn $1=4.063 \ldots-3.934 \ldots=0.0869 \ldots s$
| 
  
You are building a toy car to race in a miniature version of Formula 1. In this racing series, car weight is set at 8kg. You are required to use a small motor with a speed vs torque curve shown below.
  
<https://www.engineeringtoolbox.com/electrical-motors-torques-d_651.html>
The gearbox you have been supplied only came with the gear part numbers, which are as follows…
  
* G1 - 12
* G1 - 30
* G2 - 8
* G2 - 28
  
What is the step-down ratio of the gearbox?
\nGiven that the regulations for Formula Miniature stipulate that your car is allowed to travel no faster than 3 m/s, you are going to have to limit the speed of the motor. The wheel diameter is 0.04m. Using your answer to part a, determine this maximum speed you will need to limit the motor to. Additionally, find the torque that will be produced at this maximum speed. You can assume the full load operating speed for this motor is 1000rpm and that the maximum torque it can provide is 0.045Nm.
\nAssuming that your car can only operate at its limited motor speed, calculated in part b, determine its instanteous acceleration at the beginning of a race (from a standing start).
\nGiven that the Red Bull car can achieve an acceleration of 1.3 $\mathrm{~m} / \mathrm{s}^2$, but only has a max speed of 2.8 m/s, will you beat them off the line into turn 1, a distance of 8m?
Assume acceleration is uniform up until the point of reaching maximum speed.
| 257 | 1 | 22 | 22 | 487 | 0 | 212 | 1 | 1 | You are required to use a small motor with a speed vs torque curve shown below. embedded The gearbox you have been supplied only came with the gear part numbers, which are as follows… G1 - 12 G1 - 30 G2 - 8 G2 - 28 What is the step-down ratio of the gearbox? Given that the regulations for Formula Miniature stipulate that your car is allowed to travel no faster than 3 m/s, you are going to have to limit the speed of the motor. Using your answer to part a, determine this maximum speed you will need to limit the motor to. Additionally, find the torque that will be produced at this maximum speed. You can assume the full load operating speed for this motor is 1000rpm and that the maximum torque it can provide is 0.045Nm. Assuming that your car can only operate at its limited motor speed, calculated in part b, determine its instanteous acceleration at the beginning of a race from a standing start. Given that the Red Bull car can achieve an acceleration of 1.3 $\mathrm{~m} / \mathrm{s}^2$, but only has a max speed of 2.8 m/s, will you beat them off the line into turn 1, a distance of 8m? | 8 |
565ee457-579a-4492-bff9-38cefa3ecdff | 0 | 2 | 1 | 4 | 3 | 3 | 0 | 1 | Consider the mass flux of a contaminant.
| Choose Fick's Law, stating the meaning of the variables on paper.
\nChoose the correct dimensions below of $\boldsymbol{q}$, $D$, and $\nabla C$ in Fick's Law.
\nState the significance of the negative sign in Fick's Law, using a diagram (student's are expected to draw the diagram on paper, and compare with final answer).
| 3 | 0.333333 | 2 | \n\nRefer to the theory of Fick's law, and think visually about plotting concentration with respect to spacial distance from the source, and how it changes with repsect to time.
| The mass flux of a contaminant arising from molecular transport can be described by Fick's Law
$$
\bm{q}=-D\nabla C.
$$
Here,
$$
\begin{array}{ll}
\bm{q} & \text{is the mass flux per unit area} \\
D & \text{is the diffusion coefficient} \\
\nabla C & \text{is the concentration gradient}
\end{array}
$$
\n$$
[\bm{q}] = ML^{-2}T^{-1}\newline
[D]=L^{2}T^{-1}\newline
[\nabla C]=ML^{-4}
$$
* $[\boldsymbol{q}]$ is constructed as the mass flux, which represents the mass per second ($MT^{-1}$), passing through a unit area ($L^{-2}$)
* $[\nabla C]$ is constructed as the concentration gradient, which represents mass per unit volume (concentration), ($ML^{-3}$), per unit length ($L^{-1}$)
* $[D]$ is constructed as the rate at which a contaminant spreads, which can be found by dividing the dimensions of $\boldsymbol{q}$ by those of $\nabla C$
To check, the dimensions of the left and and right hand sides of $ \bm{q}=-D\nabla C $ are consistent, since
$$
[D\nabla C]=L^{2}T^{-1}ML^{-4}=ML^{-2}T^{-1}.
$$
\n | Consider the mass flux of a contaminant.
Choose Fick's Law, stating the meaning of the variables on paper.
\nChoose the correct dimensions below of $\boldsymbol{q}$, $D$, and $\nabla C$ in Fick's Law.
\nState the significance of the negative sign in Fick's Law, using a diagram (student's are expected to draw the diagram on paper, and compare with final answer).
| 60 | 3 | 14 | 14 | 115 | 0 | 53 | 3 | 0 | Consider the mass flux of a contaminant. Choose Fick's Law, stating the meaning of the variables on paper. Choose the correct dimensions below of $\boldsymbol{q}$, $D$, and $ abla C$ in Fick's Law. State the significance of the negative sign in Fick's Law, using a diagram student's are expected to draw the diagram on paper, and compare with final answer. | 4 |
57140949-6e46-4482-9225-df71ca7ee0e7 | 2 | 0 | 0 | 23 | 6 | 1 | 1 | 0 | Consider a point 1 m away from a point charge of +1 $\mu$C.
How are far along the radial direction is the potential due to the point charge ...
| 100 V higher?
| 1 | 0 | 1 | Firstly draw a diagram of what is being asked:
  
***
Equation for Electrostatic Potential:
$$
V = \frac{1}{4\pi\epsilon_0}\frac{q}{r}
$$
 
***
Calculate the potential for 1m using the equation above and then $\pm100$V from this and solve for r
| Firstly draw a diagram of what is being asked:
  
***
Next, solve for the potential $V$ at a point 1m from the charge using the equation for electric potential:
$$
V = \frac{1}{4\pi\epsilon_0}\frac{q}{r}
$$
  
***
If $q=1\mu C$, $r=1m$, and $\epsilon_0 = 8.85 \times 10^{-12} Fm^{-1}$, then $V = 8991.8 \times 10^9 \times 10^{-6} = 8991.8 V$.
***
Next, rearrange the equation for electric potential to make r the subject:
$$
r = \frac{1}{4\pi\epsilon_0}\frac{q}{V}
$$
***
Finally, solve for $r$ to find that when $V=9091.8V$, $r=0.989$m and when $V=8891.8V$, $r=1.011$m.
  
We find at this distance, the change in voltage is small, and so the change in distance is similarly small (consider link to binomial expansion and linearity on small scales)
| Consider a point 1 m away from a point charge of +1 $\mu$C.
How are far along the radial direction is the potential due to the point charge ...
100 V higher?
| 33 | 1 | 12 | 12 | 109 | 2 | 3 | 0 | 0 | Consider a point 1 m away from a point charge of +1 $\mu$C. How are far along the radial direction is the potential due to the point charge ... 100 V higher? | 2 |
57425408-8cdf-43ee-9d17-04abee93bb6a | 6 | 0 | 1 | 18 | 6 | 1 | 0 | 0 | It is common practice to scale velocities by the speed of light, so we define $\beta=v/c$ and $\gamma = 1/ \sqrt{1-\beta^2}$. This means $\beta$ is a dimensionless number between $-1$ and $1$. | For what positive values of $\beta$ is i) $\gamma=1.1$? ii) $\gamma = 2$? iii) $\gamma = 20$?\nWhat values of $\gamma$ are given by i) $\beta=0.09$? ii) $\beta=0.90$? iii) $\beta=0.99$?\nFor small values of $\beta$, derive an approximate expression for $\gamma$ in terms of $\beta$, keeping terms up to $\beta^2$. (Hint: use the binomial expansion.) How close is this approximation to the true value of $\gamma$ for $\beta=0.09$? | 3 | 0.333333 | 2 | Can you rearrange the expression for $\gamma$ to make $\beta$ the subject?
***
   
You should find that $\beta = \pm \sqrt{1-1/\gamma^2}$\n\nThe binomial expansion formula is:
$$
(1+x)^n \approx 1 + nx + \frac{n(n-1)}{2!}x^2 + ...
$$
***
   
For $n = -\frac{1}{2}$, this formula gives (substituting $x = - \beta ^2$ and keeping terms up to $\beta^2$)
$$
(1-\beta^2)^{-1/2} \approx 1 + \frac{1}{2}\beta^2
$$
***
   
Substitute in $\beta = 0.09$ and compare your result to the exact value of $\gamma$ (see part (a) for the formula for $\gamma$). | $\gamma = 1.1$ implies $1/\sqrt{1-\beta^2}=1.1$ , or $1-\beta^2 = 1/1.21$ . 
***
Thus $\beta = 0.4166$.
 Similarly, for $\gamma = 2$, $\beta = 0.8660$...
***
...and for $\gamma = 20$, $\beta = 0.9987$.\n$\gamma = 1/\sqrt{1-\beta^2}$ 
***
So, for $\beta=0.09$ then $\gamma=1/\sqrt{0.9919} = 1.004$. 
***
 Similarly for $\beta=0.90$ then $\gamma = 2.294$...
***
 and for $\beta=0.99$ then $\gamma = 7.089$ \n$\gamma = (1-\beta^2)^{-1/2}$. Using the binomial expansion $(1+x)^n \approx 1+ nx$ gives...
***
...$\gamma \approx 1+ (-1/2)(-\beta^2) \approx 1 + \beta^2/2$
***
Using this approximation, then for $\beta = 0.09$, $\gamma \approx 1 + 0.0081/2 \approx 1.00405$, while the exact value is $1.00407$ so they differ only at the sixth significant figure. | It is common practice to scale velocities by the speed of light, so we define $\beta=v/c$ and $\gamma = 1/ \sqrt{1-\beta^2}$. This means $\beta$ is a dimensionless number between $-1$ and $1$.For what positive values of $\beta$ is i) $\gamma=1.1$? ii) $\gamma = 2$? iii) $\gamma = 20$?\nWhat values of $\gamma$ are given by i) $\beta=0.09$? ii) $\beta=0.90$? iii) $\beta=0.99$?\nFor small values of $\beta$, derive an approximate expression for $\gamma$ in terms of $\beta$, keeping terms up to $\beta^2$. (Hint: use the binomial expansion.) How close is this approximation to the true value of $\gamma$ for $\beta=0.09$? | 102 | 19 | 21 | 21 | 83 | 11 | 72 | 14 | 0 | This means $\beta$ is a dimensionless number between $-1$ and $1$.For what positive values of $\beta$ is i $\gamma=1.1$? ii $\gamma = 2$? iii $\gamma = 20$? What values of $\gamma$ are given by i $\gamma = 1/ \sqrt{1-\beta^2}$0? ii $\gamma = 1/ \sqrt{1-\beta^2}$1? iii $\gamma = 1/ \sqrt{1-\beta^2}$2? For small values of $\beta$, derive an approximate expression for $\gamma$ in terms of $\beta$, keeping terms up to $\gamma = 1/ \sqrt{1-\beta^2}$6. Hint: use the binomial expansion. How close is this approximation to the true value of $\gamma$ for $\gamma = 1/ \sqrt{1-\beta^2}$0? | 9 |
577b2436-099e-43e0-8081-574f038c8b0e | 0 | 0 | 1 | 18 | 6 | 1 | 1 | 0 | It seems obvious that we can find the inverse Lorentz transformations by swapping the primes and reversing the sign of $v$ (or equivalently $\beta$),
that is
$$
ct^\prime = \gamma (ct-\beta x),\qquad
x^\prime = \gamma(x-\beta ct)
$$
becomes
$$
ct = \gamma (ct^\prime+ \beta x^\prime),\qquad
x = \gamma(x^\prime+ \beta ct^\prime)
$$
   
Show this is true by explicitly solving the first set of equations for $ct$ and $x$. | It seems obvious that we can find the inverse Lorentz transformations by swapping the primes and reversing the sign of $v$ (or equivalently $\beta$),
that is
$$
ct^\prime = \gamma (ct-\beta x),\qquad
x^\prime = \gamma(x-\beta ct)
$$
becomes
$$
ct = \gamma (ct^\prime+ \beta x^\prime),\qquad
x = \gamma(x^\prime+ \beta ct^\prime)
$$
   
Show this is true by explicitly solving the first set of equations for $ct$ and $x$. | 1 | 0.333333 | 2 | This question requires no knowledge of relativity.
***
   
There are two equations and two 'unknown' variables, $x$ and $t$ - how do you solve such a system?
***
   
What happens if you multiply the equation for $t'$ by $\beta$ and add it to the equation for $x'$?
***
   
You should find that this eliminates $t$ from the equations and means you can rearrange to find $x(x',t')$. Can you do something similar to eliminate $x$ and thus find $t(x',t')$ | Dividing by $\gamma$ gives
$$
\frac{ct^\prime}{\gamma} = ct-\beta x,\qquad
\frac{x^\prime}{\gamma} = x-\beta ct
$$
***
The second of these gives $\beta x^\prime/\gamma = \beta x-\beta^2 ct$. 
***
Adding this to the first equation leads to...
***
   
$$
\frac{ct^\prime}{\gamma} + \frac{\beta x^\prime}{\gamma} = ct - \beta^2 ct
\qquad\textrm{so}\qquad
\frac{ct^\prime + \beta x^\prime}{\gamma}
= (1-\beta^2) ct = \frac{ct}{\gamma^2}
$$
   
which gives $ct= \gamma (ct^\prime+ \beta x^\prime)$ as required. The calculation for $x^\prime$ is similar. | It seems obvious that we can find the inverse Lorentz transformations by swapping the primes and reversing the sign of $v$ (or equivalently $\beta$),
that is
$$
ct^\prime = \gamma (ct-\beta x),\qquad
x^\prime = \gamma(x-\beta ct)
$$
becomes
$$
ct = \gamma (ct^\prime+ \beta x^\prime),\qquad
x = \gamma(x^\prime+ \beta ct^\prime)
$$
   
Show this is true by explicitly solving the first set of equations for $ct$ and $x$. | 49 | 6 | 6 | 6 | 39 | 9 | 49 | 6 | 0 | It seems obvious that we can find the inverse Lorentz transformations by swapping the primes and reversing the sign of $v$ or equivalently $\beta$, that is $ ct^\prime = \gamma (ct-\beta x),\qquad x^\prime = \gamma(x-\beta ct) $ becomes $ ct = \gamma (ct^\prime+ \beta x^\prime),\qquad x = \gamma(x^\prime+ \beta ct^\prime) $ Show this is true by explicitly solving the first set of equations for $ct$ and $x$. | 1 |
577cc9c4-acd9-4e75-a759-7500c1713a43 | 1 | 0 | 0 | 22 | 6 | 1 | 2 | 2 | Calculate the speed of a longitudinal wave travelling along an aluminium rod. You will need to look up the density and Young’s modulus of the material.
| Calculate the speed of a longitudinal wave travelling along an aluminium rod. You will need to look up the density and Young’s modulus of the material.
| 1 | 0.333333 | 1 | Look up the density and Young's modulus of aluminium. The main purpose here is to look things up correctly and work out how to convert the units if they are not in the units you want. This is an important skill - don't jump to the next hint until you've done this!!
***
The values I found are $Y=69\times 10^9$ Pa and $\rho = 2710$ kg/m$^3$.
***
Look up the expression for the speed of a sound wave in a solid
***
You should have found $v=\sqrt{Y/\rho}$. Put in the numbers.
| Approximate values are $Y=69\times 10^9$ Pa and $\rho = 2710$ kg m$^{-3}$. 
***
The speed is $\sqrt{Y/\rho} = 5046$ m s$^{-1}$.
| Calculate the speed of a longitudinal wave travelling along an aluminium rod. You will need to look up the density and Young’s modulus of the material.
| 26 | 0 | 5 | 5 | 18 | 4 | 26 | 0 | 0 | Calculate the speed of a longitudinal wave travelling along an aluminium rod. | 1 |
5788abb2-f4a6-4bc9-abcb-c2794afe447f | 1 | 0 | 1 | 16 | 6 | 1 | 0 | 4 | If $r(1 + \cos\theta) = 2$, where $r$ and $\theta$ are plane polar coordinates,
| express this equation in terms of Cartesian coordinates $(x, y)$, with $y$ as the subject, and show it is a parabola.
\nSketch the parabola.
| 2 | 0.333333 | 0 | Transformations between Cartesian and Polar coordinates:
\
$r=\sqrt{x^2+y^2}$ $\displaystyle \theta = \arctan{y\over x}$ $x=r\cos\theta$ $y=r\sin\theta$
Now can you express it in Cartesian coordinates?
***
Rearrange to get it into the form of a parabola.
***
Remember that the parabola could be in any direction, so can be in terms of $y^2$ rather than $x^2$.
\nMake sure to include where the parabola crosses the axes, any minimum/maximum points, and consider its orientation.
| $$
r(1+\cos\theta)=2
$$
Transformations between Cartesian and Polar coordinates:
\
$r=\sqrt{x^2+y^2}$ $\displaystyle \theta = \arctan{y\over x}$ $x=r\cos\theta$ $y=r\sin\theta$
Now can you express it in Cartesian coordinates? Rearrange to get it into the form of a parabola.
***
Remember that the parabola could be in any direction, so can be in terms of $y^2$ rather than $x^2$.
***
$$
\begin{aligned}
r(1+{x\over r})&=2 \\
r+x &= 2 \\
\sqrt{x^2+y^2} &= 2-x \\
y^2 &= 4(1-x)
\end{aligned}
$$
This is a parabola. 
\nMake sure to include where the parabola crosses the axes, any minimum/maximum points, and consider its orientation.
***
$$
y^2 = 4(1-x)
$$
By setting $x=0$, it crosses at $(0,2), (0-2)$.
By setting $y=0$, it crosses at $(1,0)$.
By writing in terms of $x$, $ \displaystyle x= 1-{y^2\over 4} $, so $(1,0)$ is a maximum point for $x(y)$.
Now can you sketch the curve?
***
| If $r(1 + \cos\theta) = 2$, where $r$ and $\theta$ are plane polar coordinates,
express this equation in terms of Cartesian coordinates $(x, y)$, with $y$ as the subject, and show it is a parabola.
\nSketch the parabola.
| 35 | 5 | 17 | 17 | 120 | 6 | 24 | 2 | 0 | Sketch the parabola. | 1 |
58763411-9d04-476f-8e8e-7f96f3dc99df | 1 | 0 | 0 | 2 | 1 | 2 | 1 | 4 | The velocity distribution in a pipe with diameter $10\,$cm is given to be $u_z(r)=30(1-\frac{r^2}{r_0^2})\,$m/s, where $r_0\,$is the radius of the pipe. What is the shear stress at the wall if the fluid has a viscosity, $\mu$?
| The velocity distribution in a pipe with diameter $10\,$cm is given to be $u_z(r)=30(1-\frac{r^2}{r_0^2})\,$m/s, where $r_0\,$is the radius of the pipe. What is the shear stress at the wall if the fluid has a viscosity, $\mu$?
| 1 | 0.333333 | 1 | Find the *expression* for the shear stress, which is the viscosity multiplied by the normal gradient of the tangential velocity (i.e. the derivative of $u(r)$ with respect to $r$)
***
Take the derivative to find a general expression for the shear stress.
***
Substitute in $r=r_o$ to find the shear stress at the wall.
***
Substitute in $r_0 = 0.05$ to get your final numerical value.
| At $r=r_0$, the normal gradient of the tangential velocity is
$\frac{\delta u}{\delta r}=\frac{\delta (30(1-r^2/{r_0}^2))}{\delta r}=-60r/{r_0}^2=-60/r_0=-60/0.05=-1200$
so the shear stress there equals $-1200\mu~kg/m/s^2$ (in the $-z$ direction).
| The velocity distribution in a pipe with diameter $10\,$cm is given to be $u_z(r)=30(1-\frac{r^2}{r_0^2})\,$m/s, where $r_0\,$is the radius of the pipe. What is the shear stress at the wall if the fluid has a viscosity, $\mu$?
| 40 | 4 | 4 | 4 | 23 | 4 | 40 | 4 | 0 | The velocity distribution in a pipe with diameter $10\,$cm is given to be $u_z(r)=30(1-\frac{r^2}{r_0^2})\,$m/s, where $r_0\,$is the radius of the pipe. What is the shear stress at the wall if the fluid has a viscosity, $\mu$? | 2 |
58a0c02b-33e9-4db6-9dea-28a2cf85cb34 | 0 | 0 | 2 | 18 | 6 | 1 | 1 | 7 | The four-velocity is given by $(\gamma_u,\gamma_u\vec{\beta}_u)$, where $\vec{u}$ is the usual three-velocity of a particle, $\vec{\beta}_u = \vec{u}/c$ and $\gamma_u = 1/\sqrt{1-|\vec{\beta}_u|^2}$. | What is the squared-length of the four-velocity?\nApply a Lorentz transformation to the four-velocity formed from the three-velocity $\vec{u}=(u,0,0)$. From this, retrieve the velocity addition formula: $u^\prime = \frac{u-v}{1-uv/c^2}.$ | 2 | 0.333333 | 2 | What is the formula for the length squared of a four vector? How is it different to finding the length of a normal three-vector?
***
   
The squared length of the four vector $(\gamma_u,\gamma_u\vec{\beta}_u)$ is given by $\gamma_u^2 - \gamma_u^2|\vec{\beta}_u|^2$\nHow do you apply the Lorentz transform to a four-vector? You can refer back to Lecture 7, if you have forgotten.
***
   
After applying the Lorentz transformations, use the fact that $\frac{\gamma_{u^\prime}u^\prime}{\gamma_{u^\prime} c}=\frac{\gamma_{u^\prime}\beta_{u^\prime}}{\gamma_{u^\prime} }=\beta_u^\prime$
***
   
Remember that $\beta = u/c$ | The squared length is given by $\gamma_u^2 - \gamma_u^2 |\vec{\beta}_u|^2 = \gamma_u^2 (1-|\vec{\beta}_u|^2) = 1$.\nApplying the Lorentz transformation to the given four-velocity gives
***
$$
\gamma_u^\prime = \gamma (\gamma_u - \beta \gamma_u \beta_u),\qquad
\gamma_u^\prime \beta_u^\prime = \gamma (\gamma_u\beta_u - \beta \gamma_u)
$$
***
Hence
$$
\beta_u^\prime = \frac{\gamma_u^\prime \beta_u^\prime}{\gamma_u^\prime}
=\frac{\gamma (\gamma_u\beta_u - \beta \gamma_u)}{\gamma (\gamma_u - \beta \gamma_u \beta_u)}
=\frac{\beta_u - \beta}{1 - \beta \beta_u}
$$
which is the velocity transformation in terms of $\beta_u$. 
***
In terms of $u$ 
$$
u^\prime=\beta_u^\prime c = \frac{u/c - v/c}{1 - (v/c)(u/c)}\,c
=\frac{u - v}{1 - vu/c^2}
$$
which is the desired velocity addition formula. | The four-velocity is given by $(\gamma_u,\gamma_u\vec{\beta}_u)$, where $\vec{u}$ is the usual three-velocity of a particle, $\vec{\beta}_u = \vec{u}/c$ and $\gamma_u = 1/\sqrt{1-|\vec{\beta}_u|^2}$.What is the squared-length of the four-velocity?\nApply a Lorentz transformation to the four-velocity formed from the three-velocity $\vec{u}=(u,0,0)$. From this, retrieve the velocity addition formula: $u^\prime = \frac{u-v}{1-uv/c^2}.$ | 46 | 6 | 6 | 6 | 45 | 4 | 27 | 2 | 0 | The four-velocity is given by $(\gamma_u,\gamma_u\vec{\beta}_u)$, where $\vec{u}$ is the usual three-velocity of a particle, $\vec{\beta}_u = \vec{u}/c$ and $\gamma_u = 1/\sqrt{1-[table]^2}$.What is the squared-length of the four-velocity? Apply a Lorentz transformation to the four-velocity formed from the three-velocity $\vec{u}=(u,0,0)$. | 2 |
58e2a175-e436-40fb-9613-87a62c94fcc9 | 1 | 1 | 0 | 4 | 3 | 3 | 6 | 2 | A company releases a sand mining disposal in the water surface of a wide river of water depth $h =6.91\ \mathrm{m}$ and average river slope of $S = 1.26 \times 10^{-4}$ . The sand disposal is characterised by a medium sediment size $d_{50} = 0.25\ \mathrm{mm}$.
| Due to the potential contamination by the sediment disposal, the company is interested in knowing whether the material will be transported in suspension with the river or whether it will settle on the bed.
\nCompute the maximum river specific flow rate that would lead to sediment deposition.
| 2 | 0.666667 | 2 | \n | The bed shear stress is computed from
$$
\tau_0 = \rho g S h = 1000\cdot 9.81 \cdot 6.91 \cdot 1.26 \times 10^{-4} = 8.541\ \mathrm{N/m^2}.
$$
The shear velocity is given as
$$
u_* = \sqrt{\dfrac{\tau_0}{\rho}} = \sqrt{8.541 \cdot 1000} = 0.09242\ \mathrm{m/s}.
$$
***
We then compute the settling velocity for the given sediment size using
$$
w_s = \sqrt{\dfrac{4g d_{50}}{3 C_D} \left( \dfrac{\rho_s}{\rho_w} -1 \right)},
$$
where $C_D = 1.4 + \dfrac{36}{\text{Re}_p}$, and $\mathrm{Re}_p = \dfrac{w_s d_{50}}{\nu}$.
***
We then proceed iteratively, starting with a initial guess value of $w_s = 0.1\ \mathrm{m/s}$. We have (1) $w_s = 0.04359\ \mathrm{m/s}$, (2) $w_s = 0.03386\ \mathrm{m/s}$ ... and a final value of $w_s = 0.02919\ \mathrm{m/s}$ after 9 iterations.
Therefore $u_* > w_s$ and the sand mining disposal will be transported with the flow in suspension. Usually, this leads to turbidity problems (muddy waters).
\nThe maximum specific flow rate to obtain deposition would be given as
$$
u_{*,\max} = \sqrt{C_f}U_{\max} = w_s \Longrightarrow U_{\max} = \dfrac{w_s}{\sqrt{C_f}}.
$$
***
We could obtain the friction coefficient assuming that a change in the shear velocity will not change the friction coefficient. However, it is more accurate to re-compute the friction coefficient knowing that a reduction in the flow velocity would modify the river water depth.
From $\tau_{\max} = \rho u_{*,\max}^2 = \rho g h_{\max} S$ we get
$$
h_{\max} = \dfrac{u_{*,\max}^2}{gS} = \dfrac{0.02919^2}{9.81 \cdot 1.26 \times 10^{-4}}= 0.6894\ \mathrm{m}.
$$
The roughness Reynolds number at $u_{*, max}$ is 
$$
Re_k = \frac{u_{*, max} \cdot k_s}{\nu} = 18.24 > 10,
$$
where, $k_s=2.5d_{50}$. So, the flow is rough turbulent. 
***
With this, the maximum depth averaged velocity is
$$
U_{\max} = \dfrac{u_{*,\max}}{\kappa} \left[ \ln \left( \dfrac{h_{\max}}{z_0} \right) -1 \right] = \dfrac{0.02919}{0.41} \left[ \ln \left(\dfrac{0.6894}{2.083 \times 10^{-5}} \right) -1 \right] = 0.6698\ \mathrm{m/s},
$$
where the roughness scale that has been used is
$$
z_0 = \dfrac{2.5 \cdot d_{50}}{30} = 2.083 \times 10^{-5}\ \mathrm{m}.
$$
***
The maximum specific flow rate is
$$
q_{\max} = U_{\max} \cdot h_{\max} = 0.6698 \cdot 0.6894= 0.4618\ \mathrm{m^2 /s}.
$$
| A company releases a sand mining disposal in the water surface of a wide river of water depth $h =6.91\ \mathrm{m}$ and average river slope of $S = 1.26 \times 10^{-4}$ . The sand disposal is characterised by a medium sediment size $d_{50} = 0.25\ \mathrm{mm}$.
Due to the potential contamination by the sediment disposal, the company is interested in knowing whether the material will be transported in suspension with the river or whether it will settle on the bed.
\nCompute the maximum river specific flow rate that would lead to sediment deposition.
| 85 | 3 | 19 | 19 | 197 | 0 | 47 | 0 | 0 | Compute the maximum river specific flow rate that would lead to sediment deposition. | 1 |
58eaf562-be8a-41a5-8583-6195c9ac6abf | 2 | 0 | 0 | 9 | 4 | 2 | 6 | 10 | A permanent magnet D.C. motor has the following characteristics: 
  
Moment of inertia: $\hspace{10 mm} J = 2 \times 10^{-4}~\mathrm{kgm^2}$ 
Armature resistance: $\hspace{6 mm} R_\mathrm{a} = 300 ~\Omega$ 
Back e.m.f. const: $\hspace{10 mm} K_\mathrm{e} = 0.4 ~\mathrm{Vs/rad}$ 
Motor constant: $\hspace{13 mm} K_\mathrm{t} = 0.4 ~\mathrm{Nm/A}$ 
Viscous friction:$\hspace{13 mm} K_\mathrm{f} = 0$ 
| Calculate the motor time constant and the final speed with $200 ~\mathrm{V}$ D.C. applied.
| 1 | 0.666667 | 2 | null | From the lecture notes Section 7.3.5, the gain and time constant of a D.C. motor can be written as follows:
  
$K = \frac{K_\mathrm{t}}{K_\mathrm{f}R_\mathrm{a}+K_\mathrm{e}K_\mathrm{t}}$
  
$\tau = \frac{JR_\mathrm{a}}{K_\mathrm{f}R_\mathrm{a}+K_\mathrm{e}K_\mathrm{t}}$
  
The full derivation can be found in the aforementioned section of the lecture notes.
***
Substituting in numbers:
  
$K = \frac{0.4}{0.4\times 0.4} = 2.5$
  
$\tau = \frac{2\times 10^{-4}\times 300}{0.4\times 0.4} = 0.375~\mathrm{s}$
***
The steady state speed can be calculated using the final value theorem:
  
$y_\mathrm{ss} = Kc$
***
The system is excited with a step response of $\frac{c}{s}$. Hence, in this case, $c$ is the applied voltage.
***
Therefore:
  
$y_\mathrm{ss} = 2.5\times 200 = 500~\mathrm{rad/s}$
| A permanent magnet D.C. motor has the following characteristics: 
  
Moment of inertia: $\hspace{10 mm} J = 2 \times 10^{-4}~\mathrm{kgm^2}$ 
Armature resistance: $\hspace{6 mm} R_\mathrm{a} = 300 ~\Omega$ 
Back e.m.f. const: $\hspace{10 mm} K_\mathrm{e} = 0.4 ~\mathrm{Vs/rad}$ 
Motor constant: $\hspace{13 mm} K_\mathrm{t} = 0.4 ~\mathrm{Nm/A}$ 
Viscous friction:$\hspace{13 mm} K_\mathrm{f} = 0$ 
Calculate the motor time constant and the final speed with $200 ~\mathrm{V}$ D.C. applied.
| 45 | 6 | 8 | 8 | 87 | 0 | 13 | 1 | 0 | const: $\hspace{10 mm} K_\mathrm{e} = 0.4 ~\mathrm{Vs/rad}$ Motor constant: $\hspace{13 mm} K_\mathrm{t} = 0.4 ~\mathrm{Nm/A}$ Viscous friction:$\hspace{13 mm} K_\mathrm{f} = 0$ Calculate the motor time constant and the final speed with $200 ~\mathrm{V}$ D.C. applied. | 1 |
5949ab93-b491-4ccb-8e93-b7fb4578d384 | 0 | 0 | 1 | 11 | 4 | 2 | 7 | 5 | How is the enthalpy of formation of $\mathrm{CO}_2$ defined? How is it related to the calorific value of carbon (graphite)?
| How is the enthalpy of formation of $\mathrm{CO}_2$ defined? How is it related to the calorific value of carbon (graphite)?
| 1 | 0.333333 | 0 | null | null | How is the enthalpy of formation of $\mathrm{CO}_2$ defined? How is it related to the calorific value of carbon (graphite)?
| 20 | 1 | 0 | 0 | 0 | 0 | 20 | 1 | 0 | How is the enthalpy of formation of $\mathrm{CO}_2$ defined? How is it related to the calorific value of carbon graphite? | 2 |
598da216-0699-4875-8c08-04fb6362dc7b | 0 | 0 | 1 | 4 | 3 | 3 | 1 | 12 | For the two diffusivities mentioned in the previous question ($D = 0.202\ \mathrm{cm^{2}/s}$ and $D = 10^4\ \mathrm{cm^{2}/s}$), plot the distance in the $x$ direction where the concentration exceeds $C = 4\ \mathrm{mg/m^3}$ ($y$-axis) as a function of time ($x$-axis).
Hint: You may need to use the $\tt erfcinv()$ function in Matlab.
| For the two diffusivities mentioned in the previous question ($D = 0.202\ \mathrm{cm^{2}/s}$ and $D = 10^4\ \mathrm{cm^{2}/s}$), plot the distance in the $x$ direction where the concentration exceeds $C = 4\ \mathrm{mg/m^3}$ ($y$-axis) as a function of time ($x$-axis).
Hint: You may need to use the $\tt erfcinv()$ function in Matlab.
| 1 | 1 | 2 | null | Treating the garage as infinitely long $(L \rightarrow \infty)$, we determine the values of $x = x^*$ for which $C = 4\ \mathrm{mg/m^3}$ as a function of time $t$. Therefore, $C > 4\ \mathrm{mg/m^3}$ for $x < x^{*}$. From $C(x, t) = \frac{C_0}{2}\mathrm{erfc}\left(\frac{x}{\sqrt{2} \sigma}\right)$ we get
$$
C^* - \dfrac{C_0}{2}\mathrm{erfc}\left(\dfrac{x^*}{\sqrt{2} \sigma}\right) = 0,
$$
where $C^* = C(x = x^*, t) = 4\ \mathrm{mg/m^3}$.
***
The above relation can be inverted using $\tt erfcinv(C)$ in MATLAB, but the equation becomes
$$
x^{*}=\sqrt{4Dt}\,\underbrace{\mathrm{erfc}^{-1}\left(\frac{2C^{*}}{C_{0}}\right)}_{\mathrm{constant}},
$$
which reveals the expected power law dependence that $x^{*}$ has on $t$. 
***
The figure below plots $x^*(t)$ for $D = 0.202\ \mathrm{cm^2/s}$ and $D = 10^4\ \mathrm{cm^2/s}$. In the region below the curve $x^*(t)$, the concentration exceeds $C^{*}$, i.e. $C > 4\ \mathrm{mg/m^3}$. A similar plot could also have been obtained by using $\tt contour(t,x,C,[4\ 4])$ in MATLAB.
*Distance $x^*\ \mathrm{[m]}$ at which the concentration $C = 4\ \mathrm{mg/m^3}$ as a function of time $t$ for $D = 0.202\ \mathrm{cm^2/s}$ (left) and $D = 10^{4}\ \mathrm{cm^2/s}$ (right).*
| For the two diffusivities mentioned in the previous question ($D = 0.202\ \mathrm{cm^{2}/s}$ and $D = 10^4\ \mathrm{cm^{2}/s}$), plot the distance in the $x$ direction where the concentration exceeds $C = 4\ \mathrm{mg/m^3}$ ($y$-axis) as a function of time ($x$-axis).
Hint: You may need to use the $\tt erfcinv()$ function in Matlab.
| 48 | 7 | 5 | 5 | 82 | 0 | 48 | 7 | 0 | For the two diffusivities mentioned in the previous question $D = 0.202\ \mathrm{cm^{2}/s}$ and $D = 10^4\ \mathrm{cm^{2}/s}$, plot the distance in the $x$ direction where the concentration exceeds $C = 4\ \mathrm{mg/m^3}$ $y$-axis as a function of time $x$-axis. Hint: You may need to use the $\tt erfcinv()$ function in Matlab. | 2 |
59ab6d91-f676-4b4f-8b80-ffc9c0302de0 | 6 | 0 | 1 | 22 | 6 | 1 | 1 | 6 | Two equal masses (each of mass $m$) are connected to each other and to two adjacent walls by identical springs (spring constant $\kappa$). The walls are a distance $L$ apart. The equilibrium configuration is shown in the figure. The masses can only move perpendicular to the walls.
| Assuming the masses have small displacements ($x_1$ and $x_2$) from equilibrium, show that their equations of motion are
	 $m \ddot{x}_1 = -2\kappa x_1 + \kappa x_2$,
	 $m \ddot{x}_2 = \kappa x_1 - 2\kappa x_2$.	
\nWe now look for normal modes of the system. These are solutions where both masses oscillate at the same constant frequency. Show that the trial solutions $x_1(t) = A_1\cos(\omega t)$, $x_2(t) = A_2\cos(\omega t)$ satisfy the equations of motion and find expressions for the angular frequencies of the two normal modes in terms of $\omega_0 = \sqrt{\kappa/m}$. What are the values of $A_2/A_1$ corresponding to each of these normal modes?
\nCalculate the normal mode frequencies (in Hz), assuming $\kappa = 100$ N m$^{-1}$ and $m=0.5$\~kg
| 3 | 0.666667 | 4 | First figure out the the extensions of each of the three springs in terms of $x_1$ and $x_2$, taking care about the signs.
***
Work out the total force on mass 1, bearing in mind that there is a force coming from the left spring and from the middle spring.
Do the same for mass 2.
Apply Newton's second law.
\nWork out $\ddot{x}_1$ and $\ddot{x}_2$ and substitute into the equations of motion. Use the expression for $\omega_0$ to replace $\kappa$, so that you have a pair of equations involving only the amplitudes $A_1$,$A_2$ and the frequencies $\omega_1$, $\omega_2$.
***
You should now have the following two equations:
$A_1 (2 \omega_0^2 - \omega^2) = \omega_0^2 A_2$,\
$A_2 (2 \omega_0^2 - \omega^2) = \omega_0^2 A_1$.
Use them to form a single equation that doesn't contain $A_1$ or $A_2$.
***
A good way to do the last step is to multiply the two equations together, and then cancel the common factor of $A_1 A_2$. Now you have an equation involving only the $\omega$'s. Solve it to obtain the normal mode frequencies.
***
From the last step, you should now have $\omega_1=\omega_0$ and $\omega_2 = \sqrt{3}\omega_0$.
Next, return to your equations containing $A_1$ and $A_2$. Either one will do. Plug in the normal mode frequencies to find the ratios $A_2/A_1$ in each case.
\nPut in the numbers! Remember that $f=\omega/(2\pi)$
| Let the displacements from equilibrium be $x_1$ and $x_2$. That means that the extensions of the three springs, from left to right are $x_1$, $x_2-x_1$ and $-x_2$. 
***
The total force on mass 1 is 
	 $F_1 = -\kappa x_1 + \kappa(x_2-x_1) = -2\kappa x_1 + \kappa x_2 = m \ddot{x}_1$.
The total force on mass 2 is
	 $F_2 = -\kappa(x_2-x_1)-\kappa x_2 = \kappa x_1 - 2\kappa x_2 = m \ddot{x}_2$.
\n$x_1 = A_1\cos(\omega t), \quad x_2 = A_2\cos(\omega t)$.
$\therefore \dot{x}_1 = -\omega^2 A_1\cos(\omega t), \quad \ddot{x}_2 = -\omega^2 A_2\cos(\omega t)$.
***
Substitute into the equations of motion:
	 $-m \omega^2 A_1 = -2\kappa A_1 + \kappa A_2$, 
	 $-m \omega^2 A_2 = \kappa A_1 - 2\kappa A_2$. 	
***
Rearrange, and use $\kappa = m \omega_0^2$:
	 $A_1 (2 m \omega_0^2 - m \omega^2) = m \omega_0^2 A_2$,
	 $A_2 (2 m \omega_0^2 - m \omega^2) = m \omega_0^2 A_1$.
***
Multiply these together and cancel common factors:
	 $(2 \omega_0^2 - \omega^2)^2 = \omega_0^4$.
***
Take the square root:
	 $(2 \omega_0^2 - \omega^2) = \pm \omega_0^2$.
	 $\therefore \omega^2 = \omega_0^2 \quad {\rm or} \quad \omega^2 = 3\omega_0^2$.
The frequencies should be positive. So we arrive at the result that the normal mode frequencies are $\omega_1 = \omega_0$ and $\omega_2 = \sqrt{3}\omega_0$. 
***
From the equations above, we see that
	 $\frac{A_2}{A_1} = \frac{2 \omega_0^2 - \omega^2}{\omega_0^2} = 2 - \frac{\omega^2}{\omega_0^2}$.
Thus, for normal mode 1 ($\omega^2=\omega_0^2$) we have $A_2/A_1=1$. 
For normal mode 2 ($\omega^2=3\omega_0^2$) we have $A_2/A_1=-1$.
\n$\omega_0 = \sqrt{\kappa/m} = \sqrt{100/0.5}=14.14$ rad s$^{-1}$ . 
The normal mode frequencies are $\omega_0/(2\pi)$ and $\sqrt{3}\omega_0/(2\pi)$. 
The values are 2.25 Hz and 3.90Hz.
		
| Two equal masses (each of mass $m$) are connected to each other and to two adjacent walls by identical springs (spring constant $\kappa$). The walls are a distance $L$ apart. The equilibrium configuration is shown in the figure. The masses can only move perpendicular to the walls.
Assuming the masses have small displacements ($x_1$ and $x_2$) from equilibrium, show that their equations of motion are
	 $m \ddot{x}_1 = -2\kappa x_1 + \kappa x_2$,
	 $m \ddot{x}_2 = \kappa x_1 - 2\kappa x_2$.	
\nWe now look for normal modes of the system. These are solutions where both masses oscillate at the same constant frequency. Show that the trial solutions $x_1(t) = A_1\cos(\omega t)$, $x_2(t) = A_2\cos(\omega t)$ satisfy the equations of motion and find expressions for the angular frequencies of the two normal modes in terms of $\omega_0 = \sqrt{\kappa/m}$. What are the values of $A_2/A_1$ corresponding to each of these normal modes?
\nCalculate the normal mode frequencies (in Hz), assuming $\kappa = 100$ N m$^{-1}$ and $m=0.5$\~kg
| 155 | 14 | 28 | 28 | 181 | 22 | 105 | 11 | 1 | Assuming the masses have small displacements $x_1$ and $x_2$ from equilibrium, show that their equations of motion are 	 $m \ddot{x}_1 = -2\kappa x_1 + \kappa x_2$, 	 $m \ddot{x}_2 = \kappa x_1 - 2\kappa x_2$.	 We now look for normal modes of the system. Show that the trial solutions $x_1(t) = A_1\cos(\omega t)$, $x_2(t) = A_2\cos(\omega t)$ satisfy the equations of motion and find expressions for the angular frequencies of the two normal modes in terms of $\omega_0 = \sqrt{\kappa/m}$. What are the values of $\kappa$0 corresponding to each of these normal modes? Calculate the normal mode frequencies in Hz, assuming $\kappa$1 N m$\kappa$2 and $\kappa$3~kg | 4 |
5a2a2425-d81a-4117-86e5-314af49e923f | 0 | 1 | 0 | 7 | 4 | 2 | 0 | 1 | The heat flux that is applied to the left face of a plane wall is $q^{\prime \prime}=25 \mathrm{~W} / \mathrm{m}^2$. The wall is of thickness $L=12 \mathrm{~mm}$ and of thermal conductivity $k=12 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$. If the surface temperatures of the wall are measured to be $50^{\circ} \mathrm{C}$ on the left side and $30^{\circ} \mathrm{C}$ on the right side, do steady-state conditions exist?
| The heat flux that is applied to the left face of a plane wall is $q^{\prime \prime}=25 \mathrm{~W} / \mathrm{m}^2$. The wall is of thickness $L=12 \mathrm{~mm}$ and of thermal conductivity $k=12 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$. If the surface temperatures of the wall are measured to be $50^{\circ} \mathrm{C}$ on the left side and $30^{\circ} \mathrm{C}$ on the right side, do steady-state conditions exist?
| 1 | 0.333333 | 0 | null | If steady-state conditions exist, Fourier's Law can be applied:
  
$q^{\prime\prime} = k\frac{\Delta T}{L}$
***
Calculate $q^{\prime\prime}$ assuming steady-state conditions:
  
$q^{\prime\prime} = 12\times\frac{50-30}{0.012} = 20000~\mathrm{W/m^2}$
  
which is much higher than the applied heat flux. Therefore, steady-state conditions do not exist.
***
If steady-state conditions did exist, there would be a much smaller temperature difference across the wall:
  
$\Delta T = \frac{q^{\prime\prime}L}{K} = \frac{25\times 0.012}{12} = 0.025~\mathrm{K}$ 
  
| The heat flux that is applied to the left face of a plane wall is $q^{\prime \prime}=25 \mathrm{~W} / \mathrm{m}^2$. The wall is of thickness $L=12 \mathrm{~mm}$ and of thermal conductivity $k=12 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$. If the surface temperatures of the wall are measured to be $50^{\circ} \mathrm{C}$ on the left side and $30^{\circ} \mathrm{C}$ on the right side, do steady-state conditions exist?
| 55 | 5 | 4 | 4 | 56 | 0 | 55 | 5 | 0 | The heat flux that is applied to the left face of a plane wall is $q^{\prime \prime}=25 \mathrm{~W} / \mathrm{m}^2$. If the surface temperatures of the wall are measured to be $50^{\circ} \mathrm{C}$ on the left side and $30^{\circ} \mathrm{C}$ on the right side, do steady-state conditions exist? | 2 |
5a42f703-3644-4f9b-b2f7-12f8d5361b29 | 1 | 0 | 0 | 11 | 4 | 2 | 2 | 2 | A sample of a wet steam is passed through a well-insulated throttling calorimeter in order to measure its dryness fraction. The inlet pressure $P_1$ is measured as $8.99$ bar gauge and the sample is throttled to atmospheric pressure $P_2$, measured as $1010$ mbar. The calorimeter exit temperature $T_2$ is measured as $120^{\circ}\mathrm{C}$. 
| Sketch the process on an $h - s$ diagram and evaluate the dryness fraction $x_1$ of the wet steam sample.
| 1 | 0.666667 | 2 | null | In an $h-s$ diagram, the bell is slightly skewed to the left. The question states that state $1$ is in the wet steam region. Moreover, a throttling process is isenthalpic (see below) but entropy increases. Therefore, the diagram looks something like this:
  
***
***
Why is throttling an isenthalpic process? To answer this question, apply the SFEE between any two states, assuming that kinetic and potential energy changes can be neglected:
  
$\dot{Q} - \dot{W} = \dot{m}(h_2-h_1)$
***
In a throttling process, there is no heat or work transfer. Therefore:
  
$h_2 = h_1$
***
$h$ can be found by looking at the steam tables at state $2$, $1.01$ bar. In Table E20, the value of temperature at $1.01$ bar is below $120^{\circ}\mathrm{C}$. Therefore, the steam at state $2$ must be superheated. The value for $h$ at $120^{\circ}\mathrm{C}$, $1.01$ bar can be found in Table E24:
  
$h = 2716.5~\mathrm{kJ/kg}$
***
To find the dryness fraction at state $1$, find $h_\mathrm{g}$ and $h_{\mathrm{f}}$ at $P_1$ and apply the following equation:
  
$x_1 = \frac{h-h_\mathrm{f}}{h_\mathrm{fg}}$
***
The pressure given for $P_1$ is the gauge pressure. The absolute value is:
  
$8.99 + 1.01 = 10$ bar.
***
Now the values for $h_\mathrm{g}$ and $h_\mathrm{f}$ can be found by reading off the values in Table E21 at $10$ bar. Substituting into the equation for $x_1$:
  
$x_1 = \frac{2716.5-762.7}{2014.4} = 0.970$
  
| A sample of a wet steam is passed through a well-insulated throttling calorimeter in order to measure its dryness fraction. The inlet pressure $P_1$ is measured as $8.99$ bar gauge and the sample is throttled to atmospheric pressure $P_2$, measured as $1010$ mbar. The calorimeter exit temperature $T_2$ is measured as $120^{\circ}\mathrm{C}$. 
Sketch the process on an $h - s$ diagram and evaluate the dryness fraction $x_1$ of the wet steam sample.
| 72 | 8 | 26 | 26 | 219 | 0 | 18 | 2 | 0 | Sketch the process on an $h - s$ diagram and evaluate the dryness fraction $x_1$ of the wet steam sample. | 1 |
5b049ef3-038a-4caa-bff8-a50c9b3964bb | 2 | 0 | 0 | 13 | 4 | 2 | 4 | 1 | A uniform spherical ball rolls without slipping along a horizontal plane towards an incline. The ball neither slips nor rebounds, as pictured below:
The ball's radius $r = 0.2~\mathrm{m}$ and mass $m = 2~\mathrm{kg}$.
| Determine the velocity $U$ of the centre of the ball as it starts to roll up the incline in terms of its initial velocity $v = 10~\mathrm{m/s}$, and the angle of the incline $\theta = 30^o$.
*(Take rotation in the clockwise direction to be positive)*
\nFind the percentage of energy lost in the collision.
| 2 | 0.666667 | 4 | The first step to solving this problem is the same as the previous question –write down the angular momentum equations about a suitable point before and after impact, again in terms of the usual variables.
***
Try choosing the point at which the ball will first come in contact with the incline as it rolls towards it (this point of course depends on the incline angle $\theta$). Applying the conservation of momentum about this point will give you an expression for the tangential velocity up the ramp after impact. Don’tforget to take into account the moment of the linear momentum and angular momentum about the ball’s centre of gravity when constructing your momentum equations.
***
Finally, you should be able to simplify the equation you obtained earlier by substituting in a suitable expression for the mass moment of inertia of the ball (note that the questions states in the beginning that you’re dealing with a uniform spherical ball here, not a flat circular disk).
\nAs for the energy lost, you’ll need to examine the kinetic energy of the ball before and after impact and write an expression for the percentage energy lost (you don’t need to worry too much about the exact expressions for the kinetic energy as they’re essentially the same before and after impact and will get normalised later as you’re dealing with a percentage).
***
Use the expression for the velocity after impact obtained in the first part of the question to simplify your percentage expression to obtain your final numerical answer.
| Free body diagram and kinematic diagram:
***
Starting with the equation for angular momentum about a general point:
$$
\begin{aligned}
H_O=I_G\dot{\theta}+m{\dot{x}}_G\left(y_G-y_O\right)-m{\dot{y}}_G\left(x_G-x_O\right)
\end{aligned}
$$
***
Angular momentum about point $O$ before impact:
$$
\begin{aligned}
H_{O1}=I_G{\dot{\theta}}_1+m{\dot{x}}_1r\cos{\theta}
\end{aligned}
$$
***
Angular momentum about point $O$ after impact:
$$
\begin{aligned}
H_{O2}=I_G{\dot{\theta}}_2+m{\dot{x}}_2r
\end{aligned}
$$
***
No external moments about $O$ so angular momentum is conserved:
$$
\begin{aligned}
H_{O1}=H_{O2}
\end{aligned}
$$
***
Hence from Equations 1 and 2:
$$
\begin{aligned}
I_G{\dot{\theta}}_1+m{\dot{x}}_1r\cos{\theta}=I_G{\dot{\theta}}_2+m{\dot{x}}_2r
\end{aligned}
$$
***
Recall that for a solid sphere:
$$
\begin{aligned}
I_G=\frac{2}{5}mr^2
\end{aligned}
$$
***
And that for a body rolling without slip:
$$
\begin{aligned}
\dot{x}=r\dot{\theta}
\end{aligned}
$$
***
Substituting into Equation 3:
$$
\begin{aligned}
\left(\frac{2}{5}mr^2\right)\frac{{\dot{x}}_1}{r}+m{\dot{x}}_1r\cos{\theta}=\left(\frac{2}{5}mr^2\right)\frac{{\dot{x}}_2}{r}+m{\dot{x}}_2r
\end{aligned}
$$
***
Which simplifies to:
$$
\begin{aligned}
\frac{2}{5}{\dot{x}}_1+{\dot{x}}_1\cos{\theta}=\frac{2}{5}{\dot{x}}_2+{\dot{x}}_2
\end{aligned}
$$
***
Resulting in an expression for the velocity after impact:
$$
\begin{aligned}
{\dot{x}}_2=\frac{{\dot{x}}_1}{7}\left(2+5\cos{\theta}\right)
\end{aligned}
$$
***
Substituting the values of the parameters given:
$$
\begin{aligned}
{\dot{x}}_2=\cfrac{10}{7}~(2+5\cos 30^o)
\end{aligned}
$$
$$
U = {\dot{x}}_2 = 9.04~\mathrm{m/s}
$$
\nThe kinetic energy at any time is:
$$
\begin{aligned}
KE=\frac{1}{2}m{\dot{x}}^2+\frac{1}{2}I_G{\dot{\theta}}^2
\end{aligned}
$$
$$
\begin{aligned}
KE=\frac{1}{2}m{\dot{x}}^2+\frac{1}{2}\left(\frac{2}{5}mr^2\right)\left(\frac{\dot{x}}{r}\right)^2
\end{aligned}
$$
$$
\begin{aligned}
KE=\frac{7}{10}m{\dot{x}}^2
\end{aligned}
$$
Hence the percentage kinetic energy loss is simply:
$$
\begin{aligned}
\text{Energy loss} &= \cfrac{{KE}_1 - KE_2}{KE_1} \\
\\
&= \frac{{{\dot{x}}_1}^2-{{\dot{x}}_2}^2}{{{\dot{x}}_1}^2}
\end{aligned}
$$
Substituting the values of the two speeds: 
$$
\text{Energy loss} = \cfrac{10^2 - 9.04^2}{10^2}
$$
$$
\text{Energy loss} = 18.3~\%
$$
| A uniform spherical ball rolls without slipping along a horizontal plane towards an incline. The ball neither slips nor rebounds, as pictured below:
The ball's radius $r = 0.2~\mathrm{m}$ and mass $m = 2~\mathrm{kg}$.
Determine the velocity $U$ of the centre of the ball as it starts to roll up the incline in terms of its initial velocity $v = 10~\mathrm{m/s}$, and the angle of the incline $\theta = 30^o$.
*(Take rotation in the clockwise direction to be positive)*
\nFind the percentage of energy lost in the collision.
| 82 | 5 | 21 | 21 | 137 | 1 | 52 | 3 | 1 | Determine the velocity $U$ of the centre of the ball as it starts to roll up the incline in terms of its initial velocity $v = 10~\mathrm{m/s}$, and the angle of the incline $\theta = 30^o$. Take rotation in the clockwise direction to be positive Find the percentage of energy lost in the collision. | 2 |
5b97641a-6952-44f2-b98a-388f8b0f6d82 | 0 | 0 | 4 | 14 | 4 | 2 | 9 | 1 | Consider a generic velocity vector $\vec{u}=\left[u ~~ v ~~ w\right]^{T}$ in a Cartesian frame of reference. | Write the velocity gradient tensor in Cartesian coordinates\nWrite the deformation tensor in Cartesian coordinates\nWrite the spin tensor in Cartesian coordinates\nWrite the vorticity vector in Cartesian coordinates | 4 | 0.333333 | 2 | \n\n\n | \n\n\n | Consider a generic velocity vector $\vec{u}=\left[u ~~ v ~~ w\right]^{T}$ in a Cartesian frame of reference.Write the velocity gradient tensor in Cartesian coordinates\nWrite the deformation tensor in Cartesian coordinates\nWrite the spin tensor in Cartesian coordinates\nWrite the vorticity vector in Cartesian coordinates | 37 | 1 | 0 | 0 | 1 | 0 | 26 | 0 | 0 | Consider a generic velocity vector $\vec{u}=\left[u ~~ v ~~ w\right]^{T}$ in a Cartesian frame of reference.Write the velocity gradient tensor in Cartesian coordinates Write the deformation tensor in Cartesian coordinates Write the spin tensor in Cartesian coordinates Write the vorticity vector in Cartesian coordinates | 1 |
5c9b67cf-710b-407c-96f6-c44f08c658fc | 3 | 3 | 0 | 2 | 1 | 2 | 6 | 4 | Let us revisit the problem on the sudden expansion in a previous problem set.
| Write down an expression for the total head at surface 1, $p_{T1}$, if gravitational effects can be neglected.
\nWrite down an expression for the total head at surface 2, $p_{T2}$, if gravitational effects can be neglected.
\nRecall from the previous problem that we have $u_2=\frac{A_1}{A_2}u_1$ and $p_2=p_1+\rho u_1^2\frac{A_1}{A_2}(1-\frac{A_1}{A_2})$. Use this to give an expression for the total head on surface 2 in terms of the geometrical variables $A_1$ and $A_2$, and the upstream variables $p_1$, $\rho$ and $u_1$. (i.e. eliminate $p_2$ and $u_2$ from your expression for $p_{T2}$)
\nCompare the values of the two heads. When will $p_{T1}$= $p_{T2}$?
\nCalculate the power lost at the expansion, and express it as a fraction of both the initial kinetic power and the final kinetic power.
Power lost as a fraction of the initial kinetic power:
| 5 | 0.333333 | 2 | Your gravitational term can be left out.
\n\nSubstitute the values given in the question into the equation for $p_{T2}$ and simplify.
\n\nThe volumetric flow is $u_1A_1$, and the energy lost per unit volume is $p_{T1}-p_{T2}$.
So if we multiply the volumetric flow by the energy lost per unit volume, we will get the total power lost. Substitute the values in for this and calculate the final expression.
***
The initial power due to the kinetic energy can be calculated by multiplying the volumetric flow by the kinetic energy term calculated using the initial velocity, $u_1$. Similarly the final power due to kinetic energy can be calculated using the kinetic energy term calculated using the final velocity.
***
To calculate the power lost as a fraction of the final and initial kinetic power, simply divide the total power by the terms obtained for the final and initial powers, respectively.
| \n\n$p_{T2}=p_2+\frac12\rho u_2^2$
$=p_1+\rho u_1^2\frac{A_1}{A_2}(1-\frac{A_1}{A_2})+\frac12\rho \frac{A_1^2}{A_2^2}u_1^2$
$=p_1+\frac12\rho u_1^2\frac{A_1}{A_2}(2-\frac{A_1}{A_2}).$
\nA sketch of the quadratic function appearing at the end of the expression for $p_{T2}$ is shown below:
Thus we have $p_{T2}=p_{T1}$ when the areas $A_1$ and $A_2$ are equal, otherwise $p_{T1}>p_{T2}$.
\nThe volumetric flow is $u_1A_1$, and the energy lost per unit volume is $p_{T1}-p_{T2}$. Hence the energy lost is
$(p_{T1}-p_{T2})u_1A_1=\frac{\rho u_1^3A_1}{2}(1-\frac{A_1}{A_2})^2.$
The initial power due to the kinetic energy is
$\frac{\rho u_1^2}{2}u_1A_1=\frac{\rho u_1^3A_1}{2},$
and hence the power lost is a fraction
$(1-\frac{A_1}{A_2})^2$
of the initial kinetic power. Likewise the final kinetic power is $\rho u_1^3A_1^3/(2A_2^2)$, giving the power loss as a fraction
$(\frac{A_2}{A_1}-1)^2$
of the final kinetic power.
| Let us revisit the problem on the sudden expansion in a previous problem set.
Write down an expression for the total head at surface 1, $p_{T1}$, if gravitational effects can be neglected.
\nWrite down an expression for the total head at surface 2, $p_{T2}$, if gravitational effects can be neglected.
\nRecall from the previous problem that we have $u_2=\frac{A_1}{A_2}u_1$ and $p_2=p_1+\rho u_1^2\frac{A_1}{A_2}(1-\frac{A_1}{A_2})$. Use this to give an expression for the total head on surface 2 in terms of the geometrical variables $A_1$ and $A_2$, and the upstream variables $p_1$, $\rho$ and $u_1$. (i.e. eliminate $p_2$ and $u_2$ from your expression for $p_{T2}$)
\nCompare the values of the two heads. When will $p_{T1}$= $p_{T2}$?
\nCalculate the power lost at the expansion, and express it as a fraction of both the initial kinetic power and the final kinetic power.
Power lost as a fraction of the initial kinetic power:
| 156 | 14 | 15 | 15 | 105 | 4 | 141 | 14 | 1 | Let us revisit the problem on the sudden expansion in a previous problem set. Write down an expression for the total head at surface 1, $p_{T1}$, if gravitational effects can be neglected. Write down an expression for the total head at surface 2, $p_{T2}$, if gravitational effects can be neglected. Use this to give an expression for the total head on surface 2 in terms of the geometrical variables $A_1$ and $A_2$, and the upstream variables $p_1$, $\rho$ and $u_1$. eliminate $p_2$ and $p_{T2}$0 from your expression for $p_{T2}$ Compare the values of the two heads. When will $p_{T1}$= $p_{T2}$? Calculate the power lost at the expansion, and express it as a fraction of both the initial kinetic power and the final kinetic power. | 7 |
5cecb4ff-f8df-4f8a-a4cd-6dad1b117cfc | 4 | 0 | 0 | 11 | 4 | 2 | 0 | 8 | Nitrogen enters a turbine at a pressure of $4$ bar (absolute) and a temperature of $30^{\circ}\mathrm{C}$, and leaves at a pressure of $1$ bar (absolute) and a temperature of $–50^{\circ}\mathrm{C}$. Kinetic energies at entry and exit are negligible and nitrogen can be modelled as a perfect gas under these conditions.
| If the process is reversible, and its path on the $T–s$ diagram is a straight line (not necessarily isentropic), evaluate the heat transfer and the shaft work per unit mass of nitrogen. (*Hint: recall the significance of the area under a reversible process path on a $T–s$ diagram.*)
\nA different process between the same end states is adiabatic; evaluate the shaft work per unit mass of nitrogen and the change in specific entropy of the gas in this process, and determine whether it is reversible or irreversible.
| 2 | 0.666667 | 2 | \n | Draw out the process on a $T-S$ diagram, bearing in mind that the entropy change is yet unknown:
  
***
The area under a $T-s$ diagram (for a **reversible** process path) corresponds to the specific heat transfer **rate**. 
***
This can be written mathematically as the area of a trapezium:
  
$\frac{\dot{q}_{12}}{\dot{m}} = \frac{T_1 + T_2}{2} \times \Delta s$
 
***
Therefore, the only unknown (apart from $\frac{\dot{q}_{12}}{\dot{m}}$) is $\Delta s$.
***
Since the pressure and temperature are both known, this can be calculated using the following equation. 
  
$\Delta s = c_{\mathrm{p}}\mathrm{ln}(\frac{T_2}{T_1}) - R\mathrm{ln}(\frac{P_2}{P_1})$
***
NOTE: every time $c_{\mathrm{v}},c_{\mathrm{p}}, R$ and $\gamma$ are used in a question, the values can be found in the data and formula book (P60).
Substituting in numbers:
  
$\Delta s = 1.04\mathrm{ln}(\frac{223.15}{303.15}) - 0.297\mathrm{ln}(\frac{1}{4}) = 0.093~\mathrm{kJ/kg.K}$
***
Therefore:
  
$\frac{\dot{q}_{12}}{\dot{m}} = \frac{223.15 + 303.15}{2} \times 0.093 = 24.5 ~\mathrm{kJ/kg}$
***
The work transfer rate can be calculated using the SFEE:
***
$\frac{\dot{q}_{12}}{\dot{m}} - \frac{\dot{w}_{12}}{\dot{m}} = h_2 - h_1 = c_{\mathrm{p}}(T_2 - T_1)$
  
Note that the kinetic and potential energy changes have been neglected. Also note that $c_{\mathrm{p}}$ is used for a steady flow process.
***
 Rearranging and substituting in numbers:
  
$\frac{\dot{w}_{12}}{\dot{m}} = 24.5 - 1.04(223.15 - 303.15) = 107.7~\mathrm{kJ/kg}$
\nSince there is no heat transfer, the SFEE can be used immediately with the heat transfer term as 0:
  
$\frac{\dot{w}_{12}}{\dot{m}} = c_{\mathrm{p}}(T_2 - T_1)$
***
Substituting numbers:
  
$\frac{\dot{w}_{12}}{\dot{m}} = 1.04(303.15 - 223.15) = 83.2~ \mathrm{kJ/kg}$
***
Since the start and end of the process are at the same states as in part (a), the entropy change is the same:
  
$\Delta s = 0.0932 ~\mathrm{kJ/kg.K}$
***
Hence, since there is a positive entropy change, the process must be irreversible. 
  
(Remember, out of adiabatic, reversible and isentropic, if two are true, the third must be true. In this case, adiabatic is true but isentropic is not. Therefore, reversible cannot be true. Note that since the process is irreversible, the area under the $T-s$ path is not equal to the heat transfer).
| Nitrogen enters a turbine at a pressure of $4$ bar (absolute) and a temperature of $30^{\circ}\mathrm{C}$, and leaves at a pressure of $1$ bar (absolute) and a temperature of $–50^{\circ}\mathrm{C}$. Kinetic energies at entry and exit are negligible and nitrogen can be modelled as a perfect gas under these conditions.
If the process is reversible, and its path on the $T–s$ diagram is a straight line (not necessarily isentropic), evaluate the heat transfer and the shaft work per unit mass of nitrogen. (*Hint: recall the significance of the area under a reversible process path on a $T–s$ diagram.*)
\nA different process between the same end states is adiabatic; evaluate the shaft work per unit mass of nitrogen and the change in specific entropy of the gas in this process, and determine whether it is reversible or irreversible.
| 139 | 6 | 17 | 17 | 280 | 0 | 87 | 2 | 0 | A different process between the same end states is adiabatic; evaluate the shaft work per unit mass of nitrogen and the change in specific entropy of the gas in this process, and determine whether it is reversible or irreversible. | 1 |
5d175063-e928-4c16-972d-c11878a6fb66 | 0 | 0 | 1 | 4 | 3 | 3 | 4 | 5 | Explain in your own words the physical meaning of shear velocity.
| Explain in your own words the physical meaning of shear velocity.
| 1 | 0.666667 | 1 | null | null | Explain in your own words the physical meaning of shear velocity.
| 11 | 0 | 0 | 0 | 0 | 0 | 11 | 0 | 0 | Explain in your own words the physical meaning of shear velocity. | 1 |
5d926bad-dc4a-4cbf-9a72-df393075ff0a | 1 | 0 | 0 | 24 | 6 | 1 | 0 | 15 | *\[Riley 6.18]* Sketch the domain of integration for the integral:
$$
\cal{I} = \int_0^1 \int_{x=y}^{1/y} \frac{y^3}{x} \exp \left[ y^2 (x^2 + x^{-2})\right]\, dx\, dy
$$
and characterise its boundaries in terms of new variables $u=xy$ and $v=y/x$. Show that the Jacobian for the change from $(x,y)$ to $(u,v)$ is equal to $(2v)^{-1}$. Hence evaluate $\cal{I}$.
***
Interestingly here, the domain has 3 sides in the $x-y$ plane but 4 sides in the $u-v$ plane. This is a bit mystifying. To understand what is going on, look at the transformation of the line $ y= $ constant to the $u,v$ plane, for constant $ =0.9,0.5,0.1,0.01 $, and this should explain what happens to the line $y=0$.
| *\[Riley 6.18]* Sketch the domain of integration for the integral:
$$
\cal{I} = \int_0^1 \int_{x=y}^{1/y} \frac{y^3}{x} \exp \left[ y^2 (x^2 + x^{-2})\right]\, dx\, dy
$$
and characterise its boundaries in terms of new variables $u=xy$ and $v=y/x$. Show that the Jacobian for the change from $(x,y)$ to $(u,v)$ is equal to $(2v)^{-1}$. Hence evaluate $\cal{I}$.
***
Interestingly here, the domain has 3 sides in the $x-y$ plane but 4 sides in the $u-v$ plane. This is a bit mystifying. To understand what is going on, look at the transformation of the line $ y= $ constant to the $u,v$ plane, for constant $ =0.9,0.5,0.1,0.01 $, and this should explain what happens to the line $y=0$.
| 1 | 1 | 3 | Changing variables in integration involves the following:
1. Changing limits
2. Changing the integrand
3. Changing the differential using the Jacobian. 
***
Sketch the region in the $xy$ plane and identify the boundaries.
***
Convert the $(x,y)$ limits to $(u,v)$. You should have found one limit to be $y=0$. To investigate this more clearly, let $y=c$, where $c$ is a constant. Find $v=v(c,u)$. As $c\to 0$, what happens to $v$ and $u$? This is best to analyse with a series of sketches in the $uv$ plane for different values of $c$. This should yield two limits of $u$ and $v$.
***
Find the Jacobian for the $(u,v)$ coordinate system. It may help you to use the reciprocal property (used in *question 14*).
***
Change the integrand, limits and differential into the $(u,v)$ system. Evaluate the integral.
| Region of integration in $xy$ plane:
Starting by changing the limits $y=1/x$ and $y=x$ to the new coordinate system $(u,v)$ (we will deal with $y=0$ later):
***
$$
\underline{y=\frac{1}{x}}\implies u=1
$$
***
$$
\underline{y=x}\implies v=1
$$
***
For the boundary $y=0$, if $x\ne 0$, we see that $u=0$ and $v=0$. More generally, let $y=c$, where $c$ is a constant. In the limit $c\to 0$:
***
Combining $u$ and $v$, we have $v=c^2/u$. As $c\to 0$, the curve tends towards $u=0,v=0$:
This defines two more limits. Therefore, the region in the $uv$ plane is given by:
***
***
Next, finding the Jacobian for the new coordinate system:
***
$$
|J|=\begin{vmatrix}\dfrac{\partial x}{\partial u} & \dfrac{\partial y}{\partial u}\\[1em] \dfrac{\partial{x}}{\partial v}&\dfrac{\partial y}{\partial v} \end{vmatrix}=\frac{1}{|J'|}, \quad |J'|= \begin{vmatrix}\dfrac{\partial u}{\partial x} & \dfrac{\partial v}{\partial x}\\[1em] \dfrac{\partial{u}}{\partial y}&\dfrac{\partial v}{\partial y} \end{vmatrix}
$$
We have taken the reciprocal of the Jacobian, because $u=u(x,y)$ and $v=v(x,y)$, which will make the partial derivatives easier to find.
***
$$
|J'|=\begin{vmatrix}y & -y/x^2\\ x&1/x\end{vmatrix} = \frac{y}{x}-\left(-\frac{y}{x}\right) = 2v
$$
Thus:
$$
|J| = \frac{1}{2v}
$$
***
Changing the integrand, differential and limits, The integral becomes:
***
$$
\begin{aligned}
\cal{I} &= \iint_R{v(uv)\exp\left[{uv\left(\frac{u}{v}+\frac{v}{u}\right)}\right]\frac{1}{2v}\,du\,dv}\\
&=\frac{1}{2}\int_{v=0}^{1}\int_{u=0}^{1}{uv\,\exp(u^2+v^2)\,du\,dv}
\end{aligned}
$$
Since the integral limits are constants, we can split the double integral into two separate integrals in terms of $u$ and $v$:
***
$$
\begin{aligned}
\cal{I}&=\frac{1}{2}\int_{u=0}^{1}{ue^{u^2}}\,du\int_{v=0}^{1}{ve^{v^2}\,dv}\\
&=\frac{1}{2}\left[\frac{1}{2}e^{u^2}\right]_0^1\left[\frac{1}{2}e^{v^2}\right]_0^1\\
&=\frac{1}{8}(e-1)^2
\end{aligned}
$$
| *\[Riley 6.18]* Sketch the domain of integration for the integral:
$$
\cal{I} = \int_0^1 \int_{x=y}^{1/y} \frac{y^3}{x} \exp \left[ y^2 (x^2 + x^{-2})\right]\, dx\, dy
$$
and characterise its boundaries in terms of new variables $u=xy$ and $v=y/x$. Show that the Jacobian for the change from $(x,y)$ to $(u,v)$ is equal to $(2v)^{-1}$. Hence evaluate $\cal{I}$.
***
Interestingly here, the domain has 3 sides in the $x-y$ plane but 4 sides in the $u-v$ plane. This is a bit mystifying. To understand what is going on, look at the transformation of the line $ y= $ constant to the $u,v$ plane, for constant $ =0.9,0.5,0.1,0.01 $, and this should explain what happens to the line $y=0$.
| 102 | 13 | 29 | 29 | 172 | 16 | 102 | 13 | 0 | Show that the Jacobian for the change from $(x,y)$ to $(u,v)$ is equal to $(2v)^{-1}$. To understand what is going on, look at the transformation of the line $ y= $ constant to the $u=xy$0 plane, for constant $u=xy$1, and this should explain what happens to the line $u=xy$2. | 2 |
5e72706e-d114-47ad-a37d-d01f74ec6e6b | 0 | 0 | 4 | 14 | 4 | 2 | 9 | 4 | Solve Problem 10.4 by considering a cylindrical frame of reference attached to the ground, with origin in the centre of the tank and with axial direction $\vec{e}_z$. | Determine an expression for the velocity field\nDetermine the deformation tensor.\nDetermine the spin tensor\nDetermine the vorticity | 4 | 0.333333 | 3 | \n\n\n | \n\n\n | Solve Problem 10.4 by considering a cylindrical frame of reference attached to the ground, with origin in the centre of the tank and with axial direction $\vec{e}_z$.Determine an expression for the velocity field\nDetermine the deformation tensor.\nDetermine the spin tensor\nDetermine the vorticity | 42 | 1 | 0 | 0 | 1 | 0 | 15 | 0 | 0 | Solve Problem 10.4 by considering a cylindrical frame of reference attached to the ground, with origin in the centre of the tank and with axial direction $\vec{e}_z$.Determine an expression for the velocity field Determine the deformation tensor. Determine the spin tensor Determine the vorticity | 2 |
5e9131cd-ed1c-40a2-9bb7-bf14211657a1 | 4 | 0 | 0 | 11 | 4 | 2 | 3 | 9 | A mixture of $1 ~\mathrm{kg} ~ \mathrm{CO}$ and $1~ \mathrm{kg} ~ \mathrm{H}_2$ in a sealed, rigid container is at a pressure of $2$ bar and a temperature of $20^{\circ}\mathrm{C}$. The container is heated until the pressure is $4$ bar. Find:
| The final temperature of the mixture.
\nThe change in specific enthalpy of the mixture.
\nThe change in specific entropy of the mixture.
\nThe magnitude of the heat transfer.
| 4 | 0.666667 | 2 | \n\n\n | Since the mixture can be treated as a perfect gas, the ideal gas equation can be used:
  
$PV = n\bar{R}T$
***
where $n$ is the total number of moles in the mixture and can be calculated as follows:
  
$n =\Sigma (\frac{m_\mathrm{i}}{M_\mathrm{i}})$
***
Substituting in numbers:
  
$n = \frac{1}{28.01} + \frac{1}{2.016} = 0.532~\mathrm{kmol}$
***
Substituting this into the ideal gas equation and rearranging for the initial volume:
  
$V = \frac{0.532\times8314\times(20+273.15)}{2\times10^5} = 6.48~\mathrm{m^3}$
***
Since the container is rigid, the volume and composition of the mixture remain constant. Substituting the final value for pressure in to the ideal gas equation and rearranging for the final temperature:
  
$T = \frac{4\times10^5\times 6.48}{0.532\times 8314} = 586~\mathrm{K}$
\n$\Delta h = c_\mathrm{p}(T_2 - T_1)$
***
Find $c_\mathrm{p}$ for the mixture:
  
$c_\mathrm{p} =\Sigma( \frac{m_\mathrm{i}}{m}c_\mathrm{p,i})$
***
Substituting in numbers:
  
$c_\mathrm{p} = (\frac{1}{2}\times1.04) +( \frac{1}{2}\times14.31) = 7.675~\mathrm{kJ/kg.K}$
***
Substituting this into the equation for $\Delta h$:
  
$\Delta h = 7.675(586-(20+273.15) = 2250~\mathrm{kJ/kg}$
\nThe change in entropy can be calculated using the following equation:
  
$\Delta s = c_\mathrm{p}(\frac{T_2}{T_1}) -R\mathrm{ln}(\frac{P_2}{P_1})$
***
The values for $T_2,T_1,P_2,P_1$ and $c_\mathrm{p}$ are all known. 
***
$R$ can be calculated as follows:
  
$R = \Sigma (\frac{m_\mathrm{i}}{m}R_\mathrm{i})$
***
Substituting in numbers:
  
$R = (\frac{1}{2}\times0.297)+(\frac{1}{2}\times4.12) = 2.2085~\mathrm{kJ/kg.K}$
***
Substituting this into the equation for $\Delta s$:
  
$\Delta s = 7.675\mathrm{ln}(\frac{586}{15+273.15}) - 2.2085\mathrm{ln}(\frac{4}{2}) = 3.79~\mathrm{kJ/kg.K}$
\nApply the first law equation, bearing in mind that there is no shaft work:
  
$Q = \Delta u = mc_\mathrm{v}(T_2-T_1)$
***
$c_\mathrm{v}$ can be calculated as follows:
  
$c_\mathrm{v} = \Sigma (\frac{m_\mathrm{i}}{m}c_\mathrm{v,i})$
***
Substituting in numbers:
  
$c_\mathrm{v} = (\frac{1}{2}\times 0.74) + (\frac{1}{2}\times 10.18) = 5.46~\mathrm{kJ/kg.K}$
***
Substituting this into the first law equation:
  
$Q = 2\times5.46(586-(20+273.15)) = 3200\mathrm{kJ}$
| A mixture of $1 ~\mathrm{kg} ~ \mathrm{CO}$ and $1~ \mathrm{kg} ~ \mathrm{H}_2$ in a sealed, rigid container is at a pressure of $2$ bar and a temperature of $20^{\circ}\mathrm{C}$. The container is heated until the pressure is $4$ bar. Find:
The final temperature of the mixture.
\nThe change in specific enthalpy of the mixture.
\nThe change in specific entropy of the mixture.
\nThe magnitude of the heat transfer.
| 63 | 5 | 25 | 25 | 213 | 0 | 28 | 0 | 0 | Find: The final temperature of the mixture. | 1 |
5eabf4ab-0d86-49d4-a486-8e28845a586a | 1 | 0 | 0 | 13 | 4 | 2 | 0 | 1 | The Imperial Racing Green car below is initially horizontal, has a mass of $2.1~\mathrm{tonnes}$ and its centre of mass is $1.5~\mathrm{m}$ above and $2.0~\mathrm{m}$ forward of the point where the rear wheel touches the ground. The moment of inertia of the car about its centre of mass is $3000~\mathrm{kgm^2}$.
The car accelerates from rest as a result of the rear wheel exerting a horizontal tractive force of $41.2~\mathrm{kN}$.
| Assuming there is no slip, determine the initial angular acceleration of the car if it does a wheelie.
*(Take rotation in the anti-clockwise direction to be positive)*
| 1 | 0.666667 | 3 | Think also about the acceleration of the centre of gravity of the car in all directions. Again, pick a suitable point around which to take moments (which point does the car rotate about whilst doing a wheelie?).
***
You will need to refer to the vector equation in your notes to relate the accelerations of different parts of the car. Make sure you impose the correct initial conditions for the relevant variables.
***
Consider the conditions necessary for the car to pop a ‘wheelie’ – what would be the reaction forces at the wheels in that case? What value must the angular acceleration have in order to prove that the car will do a wheelie?
| Free body diagram and kinematic diagram:
***
Resolving forces in the x-axis:
$$
\sum F_x = ma_{G_x}
$$
$$
F_B = ma_{G_x}~~\mathrm{(Equation 1)}
$$
***
Resolving forces in the y-axis:
$$
\sum F_y = ma_{G_y}
$$
$$
R_A + R_B -mg = ma_{G_y}~~(\mathrm{Equation~2})
$$
***
Taking moments about the centre of mass:
$$
\sum M_G = I_G\ddot{\theta}
$$
$$
R_Ax_{AG} - R_Bx_{BG} + F_By_{BG} = I_G\ddot{\theta}~~(\mathrm{Equation~3})
$$
***
But for a wheelie:
$$
R_A = 0
$$
***
Hence Equation 2 becomes:
$$
R_B -mg = ma_{G_y}
$$
***
Rearranging:
$$
R_B = mg + ma_{G_y}~~(\mathrm{Equation~4})
$$
***
And Equation 3 becomes:
$$
- R_Bx_{BG} + F_By_{BG} = I_G\ddot{\theta}~~(\mathrm{Equation~5})
$$
***
Substituting for $R_B$ from Equation 4 into Equation 5:
$$
- (mg + ma_{G_y})x_{BG} + F_By_{BG} = I_G\ddot{\theta}~~(\mathrm{Equation~6})
$$
***
Applying the kinematic equation for acceleration:
$$
a_B = a_G + \ddot\theta \times r - \ddot\theta^2r
$$
***
$$
\begin{pmatrix} a_B \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} a_{G_x} \\ a_{G_y} \\ 0 \end{pmatrix} + \begin{pmatrix} 0 \\ 0 \\ \ddot\theta \end{pmatrix} \times \begin{pmatrix} -x_{BG} \\ -y_{BG} \\ 0 \end{pmatrix} - 0
$$
***
The **i** component is:
$$
a_B = a_{G_x} + y_{BG}\ddot\theta~~(\mathrm{Equation~7})
$$
***
The **j** component is:
$$
0 = a_{G_y} - x_{BG}\ddot\theta
$$
***
Rearranging:
$$
a_{G_y} = x_{BG}\ddot\theta~~(\mathrm{Equation~8})
$$
***
Substituting $a_{G_y}$ from Equation 8 into Equation 6:
$$
- (mg + m(x_{BG}\ddot\theta))x_{BG} + F_By_{BG} = I_G\ddot{\theta}
$$
***
Rearranging:
$$
- mgx_{BG} + F_By_{BG} = (I_G + mx_{BG}^2)\ddot{\theta}
$$
$$
\ddot\theta = \cfrac{F_By_{BG} - mgx_{BG}}{I_G + mx_{BG}^2}
$$
***
Substituting in the values of the parameters (with consideration of units) gives:
$$
\ddot\theta = \cfrac{(41.2\times1000\times1.5) - (2100\times9.81\times2.0)}{3000 + 2100\times2.0^2}
$$
$$
\text{Acceleration} = \ddot\theta = 1.81~\mathrm{rad/s^2}
$$
| The Imperial Racing Green car below is initially horizontal, has a mass of $2.1~\mathrm{tonnes}$ and its centre of mass is $1.5~\mathrm{m}$ above and $2.0~\mathrm{m}$ forward of the point where the rear wheel touches the ground. The moment of inertia of the car about its centre of mass is $3000~\mathrm{kgm^2}$.
The car accelerates from rest as a result of the rear wheel exerting a horizontal tractive force of $41.2~\mathrm{kN}$.
Assuming there is no slip, determine the initial angular acceleration of the car if it does a wheelie.
*(Take rotation in the anti-clockwise direction to be positive)*
| 98 | 5 | 23 | 23 | 115 | 0 | 27 | 0 | 1 | Assuming there is no slip, determine the initial angular acceleration of the car if it does a wheelie. Take rotation in the anti-clockwise direction to be positive | 2 |
5ecfd50a-13f5-4c4a-8029-eb5dbd6318a6 | 0 | 1 | 0 | 17 | 6 | 1 | 2 | 5 | [In a previous year, students were challenged to work with their tutorial groups to invent new questions to appear on the problem sheets. Here is one of the most interesting questions they came up with.]
  
A person of mass $m = 60\,$kg, standing in a stationary train carriage, starts walking towards the front of the train, accelerating from initial velocity $u = 0\,$m$\cdot$s$^{-1}$ to final velocity $v_w = 1\,$m$\cdot$s$^{-1}$ relative to the carriage. As the person accelerates, their muscles do work , converting stored chemical energy into kinetic energy. According to the work-energy theorem (which states that the work done equals the change in KE):
$$
\text{work done on person}
= \text{KE}_{\text{final}} - \text{KE}_{\text{initial}}
= \frac{1}{2}m v_w^2 - 0 = 30\,\text{J}.
$$
  
Suppose now that the train is travelling with an initial speed $u =
20\,$m$\cdot$s$^{-1}$. When the same person accelerates from rest in the carriage to a final velocity of $1\,$m$\cdot$s$^{-1}$ relative to the carriage,
$$
\text{work done on person}
= \text{KE}_{\text{final}} - \text{KE}_{\text{initial}}
= \frac{1}{2} m (u + v_w)^2 - \frac{1}{2} m
u^2 = 1230\,\text{J},
$$
which is $41$ times higher than the result when the train is at rest. Does the person use $41$ times more energy? If not, what was wrong with the approach taken in this question? Explain your reasoning in detail.
| [In a previous year, students were challenged to work with their tutorial groups to invent new questions to appear on the problem sheets. Here is one of the most interesting questions they came up with.]
  
A person of mass $m = 60\,$kg, standing in a stationary train carriage, starts walking towards the front of the train, accelerating from initial velocity $u = 0\,$m$\cdot$s$^{-1}$ to final velocity $v_w = 1\,$m$\cdot$s$^{-1}$ relative to the carriage. As the person accelerates, their muscles do work , converting stored chemical energy into kinetic energy. According to the work-energy theorem (which states that the work done equals the change in KE):
$$
\text{work done on person}
= \text{KE}_{\text{final}} - \text{KE}_{\text{initial}}
= \frac{1}{2}m v_w^2 - 0 = 30\,\text{J}.
$$
  
Suppose now that the train is travelling with an initial speed $u =
20\,$m$\cdot$s$^{-1}$. When the same person accelerates from rest in the carriage to a final velocity of $1\,$m$\cdot$s$^{-1}$ relative to the carriage,
$$
\text{work done on person}
= \text{KE}_{\text{final}} - \text{KE}_{\text{initial}}
= \frac{1}{2} m (u + v_w)^2 - \frac{1}{2} m
u^2 = 1230\,\text{J},
$$
which is $41$ times higher than the result when the train is at rest. Does the person use $41$ times more energy? If not, what was wrong with the approach taken in this question? Explain your reasoning in detail.
| 1 | 1 | 3 | You might well be able to guess that the statement is false. Can you think why this is? (see next hint if unsure) ...
***
... The problem is that the question only calculates how the work done changes the KE of the walker, ignoring the fact that the train also speeds up or slows down. What law is this a consequence of? Let the change in velocity of the train be $v_t$. 
***
Working in the platform frame, what is the final velocity of the walker, and the final velocity of the train?
***
Invoking the law from the second hint, write an expression (let this be equation 1) relating:
* The mass of the person, $m$
* The mass of the train $M$
* The velocities $u, v_w$ and $v_t$. 
***
Next, what is the change of kinetic energy $\Delta(\text{KE})$ of the system (use the velocities from hint 3)? Let this be equation 2.
***
Simplify equation 2 using equation 1. You should see that $\Delta(\text{KE})$ is not dependent on the initial velocity $u$ of the train.
***
To analyse your equation further, use equation 1 to find $v_t$. This should give you the $\Delta(\text{KE})$ as a function of $v_\text{w}$, $m$ and $M$ only...
***
... Consider that $m \ll M$. In this limit, do you obtain the correct result from working in the frame of reference of the train and ignoring the change in KE of the train?
| No, the walker does not use $41$ times more energy.
***
When viewed from the inertial frame of reference with velocity equal to the initial velocity of the train carriage, the person’s journey looks exactly the same in both cases. Since both frames are inertial, Newton’s laws apply equally well in both. The amount of chemical energy used by the person’s muscles should therefore be independent of the train’s initial velocity.
***
The problem is that the question only calculates how the work done changes the KE of the walker, ignoring the fact that the train also speeds up or slows down. The total work done by the walker’s muscles is the sum of the change in KE of the walker and the change in KE of the train.
***
Let us analyse this more carefully. The train and walker are initially moving at the same velocity $u$ (which might be $0$m$\cdot$s$^{-1}$ or might be $20$m$\cdot$s$^{-1}$) as measured in the station frame. The final velocity of the train of mass $M$ $(\gg m)$ is $u + v_t$ and the final velocity of the walker is $u + v_t + v_w$, where $v_w = 1$m$\cdot$s$^{-1}$ is the final velocity of the walker relative to the train carriage. The signs of $v_t$ and $v_w$ are opposite.
***
The total momentum of the system consisting of walker $+$ train is conserved because no external forces are acting on it, so
***
$$
\begin{aligned}
&\qquad
&
M(u + v_t) + m(u + v_t + v_w)
&= (M + m)u \\
\Rightarrow
&
&
M v_t + m(v_t + v_w)
&= 0 .
\end{aligned}
$$
***
The total change in KE is
$$
\begin{aligned}
\Delta(\text{KE})
&=
\frac{1}{2} M (u + v_t)^2 + \frac{1}{2} m (u + v_t + v_w)^2
- \frac{1}{2} M u^2 - \frac{1}{2} m u^2 \\
&= M u v_t + m u (v_t + v_w) +
\frac{1}{2} M v_t^2 + \frac{1}{2} m (v_t + v_w)^2 .
\end{aligned}
$$
***
The first two terms on the right-hand side sum to zero by the momentum conservation equation above, leaving
$$
\begin{aligned}
\Delta(\text{KE})
&= \frac{1}{2} M v_t^2 + \frac{1}{2} m ( v_t + v_w)^2 .
\end{aligned}
$$
As expected, this is independent of the initial velocity $u$. The total work done by the walker’s muscles does not depend on the initial velocity of the train carriage.
***
We can get a little further by rearranging the momentum conservation equation to give
$$
\begin{aligned}
v_t &= -\frac{m v_w}{M+m} , &
v_w + v_t &= \frac{M v_{w}}{M+m} ,
\end{aligned}
$$
and using these results to rewrite the change in total KE as:
***
$$
\Delta(\text{KE})
= \frac{1}{2}M \left ( \frac{m v_w}{M+m} \right )^2
+ \frac{1}{2} m \left ( \frac{M v_w}{M+m} \right )^2
= \frac{1}{2} \frac{Mm}{M + m} v_w^2 .
$$
***
In the limit when $m \ll M$, this is approximately $\frac{1}{2} m
v_w^2$, which is the result you get working in the frame of reference of the train and ignoring the change in KE of the train, as in the question.
***
When the train and walker, both initially moving at $u = 20$m$\cdot$s$^{-1}$, are seen from the station platform, the train slows down by:
***
$$
v_t = -\frac{m v_w}{M+m} ,
$$
which is very small if $m/M \ll 1$. At the same time, the walker speeds up by
***
$$
v_w + v_t = \frac{M v_w}{M+m},
$$
which is only very slightly less than $v_m = 1\,$m$\cdot$ s$^{-1}$. The reduction in the train’s speed is small, but its mass is large and the reduction in its KE offsets most of the increase in the walker’s KE.
| [In a previous year, students were challenged to work with their tutorial groups to invent new questions to appear on the problem sheets. Here is one of the most interesting questions they came up with.]
  
A person of mass $m = 60\,$kg, standing in a stationary train carriage, starts walking towards the front of the train, accelerating from initial velocity $u = 0\,$m$\cdot$s$^{-1}$ to final velocity $v_w = 1\,$m$\cdot$s$^{-1}$ relative to the carriage. As the person accelerates, their muscles do work , converting stored chemical energy into kinetic energy. According to the work-energy theorem (which states that the work done equals the change in KE):
$$
\text{work done on person}
= \text{KE}_{\text{final}} - \text{KE}_{\text{initial}}
= \frac{1}{2}m v_w^2 - 0 = 30\,\text{J}.
$$
  
Suppose now that the train is travelling with an initial speed $u =
20\,$m$\cdot$s$^{-1}$. When the same person accelerates from rest in the carriage to a final velocity of $1\,$m$\cdot$s$^{-1}$ relative to the carriage,
$$
\text{work done on person}
= \text{KE}_{\text{final}} - \text{KE}_{\text{initial}}
= \frac{1}{2} m (u + v_w)^2 - \frac{1}{2} m
u^2 = 1230\,\text{J},
$$
which is $41$ times higher than the result when the train is at rest. Does the person use $41$ times more energy? If not, what was wrong with the approach taken in this question? Explain your reasoning in detail.
| 192 | 17 | 35 | 35 | 460 | 14 | 192 | 17 | 0 | According to the work-energy theorem which states that the work done equals the change in KE: $ \text{work done on person} = \text{KE}_{\text{final}} - \text{KE}_{\text{initial}} = \frac{1}{2}m v_w^2 - 0 = 30\,\text{J}. When the same person accelerates from rest in the carriage to a final velocity of $u = 0\,$1m$\cdot$s$^{-1}$ relative to the carriage, $u = 0\,$4 which is $u = 0\,$5 times higher than the result when the train is at rest. Does the person use $u = 0\,$5 times more energy? If not, what was wrong with the approach taken in this question? Explain your reasoning in detail. | 5 |
5eef3bdd-0639-4b55-aef4-81fbb780fb52 | 3 | 0 | 2 | 4 | 3 | 3 | 3 | 2 | A wide river is dredged to make it more navigable. This releases a significant mass $M$ of contaminated soil into the river. The dredged river-section is sufficiently large to model this problem as 1-dimensional (in $z$). Assume that the suspended soil behaves as a passive scalar that is released instantaneously at the bed and that the flow conditions are those usually encountered for steady uniform open channel flows.
| Starting from the three-dimensional case, show that the above assumptions allow us to reduce the governing equations for the contaminant to one dimension. Give the corresponding boundary conditions.
\nThe eddy viscosity in the channel can be described by (see sediment transport lectures):
$$
\nu_T = u_* \kappa z\left(1-\dfrac{z}{h}\right).
$$
Obtain an expression for the depth-averaged value for $D_T(z)$ by assuming that the turbulent Schmidt number $Sc_T = 1$.
\nUsing the average value of $\langle D_T \rangle$ and modelling only the bed boundary condition, give the solution to this problem.
\nComment on how you expect the analytical approximation to differ from the solution to the governing equations of this problem.
| 4 | 1 | 2 | \n\n\n | \nSince $Sc_T = 1$, $ D_T = \nu_T = u_* \kappa z\left(1-\dfrac{z}{h}\right) $. 
We then take the integral of the expression of $D_T$ and averaging it over $h$ with
$$
\displaystyle \langle D_T \rangle = \dfrac{1}{h}\int_{0}^{h} {u_* \kappa z(1-\dfrac{z}{h})}\mathrm{d}z.
$$
***
This can be evaluated further until a final expression
$$
\langle D_T \rangle = \dfrac{1}{h} u_* \kappa \left[\left(\dfrac{1}{2}h^2-\dfrac{1}{3}h^2\right) - 0\right]
$$
$$
\implies \langle D_T \rangle = \dfrac{\kappa}{6} u_* h.
$$
\nIgnoring the top boundary condition, and remembering that $D_{T}$ is not dependent on $z$, the governing equation becomes
$$
\dfrac{\partial \bar{C}}{\partial t} = \langle D_{T} \rangle \dfrac{\partial^2 \bar{C}}{\partial z^2},
$$
with the bottom boundary condition given in equation $q_z = -D_{T} \dfrac{\partial \bar{C}}{\partial z} \bigg|_h = -D_{T} \dfrac{\partial \bar{C}}{\partial z} \bigg|_0 =0$. Since the release occurs at $z=0$, the mirror source coincides with it and we can use the known solution for the 1-D diffusion equation with a release strength of $2M$. The solution is
$$
C(z,t) = \dfrac{2M}{\sqrt{2 \pi} \sigma} e^{\dfrac{-z^2}{2 \sigma^2}}\quad \text{ with }\sigma = \sqrt{\dfrac{\kappa}{3} u_* h t}.
$$
\n | A wide river is dredged to make it more navigable. This releases a significant mass $M$ of contaminated soil into the river. The dredged river-section is sufficiently large to model this problem as 1-dimensional (in $z$). Assume that the suspended soil behaves as a passive scalar that is released instantaneously at the bed and that the flow conditions are those usually encountered for steady uniform open channel flows.
Starting from the three-dimensional case, show that the above assumptions allow us to reduce the governing equations for the contaminant to one dimension. Give the corresponding boundary conditions.
\nThe eddy viscosity in the channel can be described by (see sediment transport lectures):
$$
\nu_T = u_* \kappa z\left(1-\dfrac{z}{h}\right).
$$
Obtain an expression for the depth-averaged value for $D_T(z)$ by assuming that the turbulent Schmidt number $Sc_T = 1$.
\nUsing the average value of $\langle D_T \rangle$ and modelling only the bed boundary condition, give the solution to this problem.
\nComment on how you expect the analytical approximation to differ from the solution to the governing equations of this problem.
| 170 | 6 | 14 | 14 | 101 | 0 | 100 | 4 | 1 | Assume that the suspended soil behaves as a passive scalar that is released instantaneously at the bed and that the flow conditions are those usually encountered for steady uniform open channel flows. Starting from the three-dimensional case, show that the above assumptions allow us to reduce the governing equations for the contaminant to one dimension. Give the corresponding boundary conditions. The eddy viscosity in the channel can be described by see sediment transport lectures: $ u_T = u_* \kappa z\left(1-\dfrac{z}{h}\right). $ Obtain an expression for the depth-averaged value for $D_T(z)$ by assuming that the turbulent Schmidt number $Sc_T = 1$. Using the average value of $\langle D_T \rangle$ and modelling only the bed boundary condition, give the solution to this problem. | 6 |
5f1a9fd0-9e55-422f-be15-10c651210951 | 0 | 10 | 0 | 2 | 1 | 2 | 0 | 9 | Are the following flows steady? Explain your answer.
| Blood flow in the heart is
\nFlow in a river is
\nWater flow from a tap
\nFlow in a vein
| 4 | 0.333333 | 0 | \n\n\n | \n\n\n | Are the following flows steady? Explain your answer.
Blood flow in the heart is
\nFlow in a river is
\nWater flow from a tap
\nFlow in a vein
| 28 | 0 | 0 | 0 | 1 | 0 | 20 | 0 | 0 | Are the following flows steady? Explain your answer. | 2 |
5f89db98-8d5c-4e61-9c36-a7726b260a1f | 4 | 1 | 0 | 4 | 3 | 3 | 6 | 3 | A river stream has a flow rate of $q = 20.98\ \mathrm{m^2 /s}$ and a river depth of $h = 8.23\ \mathrm{m}$. The sediment of the bed has a typical sediment size of $d_{50} = 0.28\ \mathrm{mm}$.
For relatively large particles in large water depths, sediment suspension occurs in a small layer close to the bed. In such cases, we can assume that $z<<h$. Using this assumption the sediment concentration profile is simplified to
$$
\overline{C} = C_a \left( \dfrac{z}{z_a} \right)^{-b}.
$$
You may wish to use [the Shields diagram](https://bb.imperial.ac.uk/webapps/blackboard/execute/content/file?cmd=view\&content_id=_2543151_1\&course_id=_30255_1).
| Using this simplified expression, calculate the sediment concentration at the following vertical elevations with respect to the bed: $z_1 = 0.0020\ \mathrm{m}$, $z_2 = 2.0\ \mathrm{m}$, and $z_3 = 4.0\ \mathrm{m}$.
\nCompare your values with the original Rouse profile. Is the assumption of suspended sediment transport occurring in a small layer close to the bed valid?
\nObtain the total sediment transport rate.
| 3 | 1 | 2 | \n\n | Assuming a rough turbulent flow $z_0 = \dfrac{k_s}{30} =\dfrac{2.5 d_{50}}{30} = 2.333 \times 10^{-5}\ \mathrm{m}$, the friction coefficient is computed as
$$
C_f = \left[ \dfrac{1}{\kappa} \left( \ln \left( \dfrac{h}{z_0} \right) -1 \right) \right]^{-2}
= \left[ \dfrac{1}{0.41} \left( \ln \left( \dfrac{8.23}{2.333 \times 10^{-5}} \right) -1 \right) \right]^{-2}
= 0.001213,
$$
and the shear velocity $u_*= \sqrt{C_f} U = 0.08877\ \mathrm{m/s}$.
***
We can check now if the flow has a rough or smooth turbulence by finding $\mathrm{Re}_k$, which is
$$
\mathrm{Re}_k = \dfrac{k_s u_*}{\nu} = \dfrac{7 \times 10^{-4} \cdot 0.08877}{10^{-6}} = 62.14,
$$
which is much larger than 10, and so the flow is rough turbulent.
***
We then compute the settling velocity for the given sediment size using
$$
w_s = \sqrt{\dfrac{4g d_{50}}{3 C_D} \left( \dfrac{\rho_s}{\rho_w} -1 \right)},
$$
where $C_D = 1.4 + \dfrac{36}{\text{Re}_p}$ and $\mathrm{Re}_p = \dfrac{w_s d_{50}}{\nu}$.
We thus proceed iteratively, starting with a initial guess value of $w_s = 0.1\ \mathrm{m/s}$. We have (1) $w_s = 0.04743\ \mathrm{m/s}$, (2) $w_s = 0.03834\ \mathrm{m/s}$... and a final value $w_s = 0.03424\ \mathrm{m/s}$ after 9 iterations.
We can verify that $u_* > w_s$ and there will be sediment suspension.
***
The Rouse number is given as
$$
b = \dfrac{w_s}{\kappa u_*} = \dfrac{0.3424}{0.41 \cdot 0.08877} = 0.9407.
$$
The Shields parameter is obtained from
$$
\theta = \dfrac{u_*^2}{(s-1)gd_{50}} = \dfrac{0.08877^2}{(2.65-1) \cdot 9.81 \cdot 2.8 \times 10^{-3}} = 1.739.
$$
And so, the reference concentration becomes
$$
C_a = \dfrac{0.331 (\theta - 0.045)^{1.75}}{1 + 0.72 (\theta - 0.045)^{1.75}} = 0.2962\ \mathrm{m^3/m^3},
$$
with $z_a = k_s = 7 \times 10^{-4}\ \mathrm{m}$.
***
We compute the suspended sediment concentration using the simplified relation
$$
\overline{C} = C_a \left( \dfrac{z}{z_a} \right)^{-b},
$$
and obtain the values:
* $\overline{C}(z_1 = 0.0020\ \mathrm{m}) = 1.103 \times 10^{-1}\ \mathrm{m^3/m^3}$
* $\overline{C}(z_2 = 2.0\ \mathrm{m}) = 1.661 \times 10^{-4}\ \mathrm{m^3/m^3}$
* $\overline{C}(z_3 = 4.0\ \mathrm{m}) = 8.654 \times 10^{-5}\ \mathrm{m^3/m^3}$
\n\nThe total sediment transport rate is obtained as $q_t = q_b + q_s$ where $q_b$ is the bed-load sediment transport rate and $q_s$ is the suspended sediment transport rate. The suspended sediment concentration rate is obtained as
$$
\displaystyle q_s = \int_{z_a}^{h} \overline{u}(z) \overline{C}(z) = 4.792 \times 10^{-3}\ \mathrm{m^2/s},
$$
and the bed load sediment transport rate is obtained from Ribberink (1998). The critical Shields parameter is obtained from the Shields diagram using the shear Reynolds number, which leads to
$$
\mathrm{Re}_* = \dfrac{u_* d_{50}}{\nu} = \dfrac{0.08777 \cdot 2.8\times 10^{-4}}{10^{-6}} = 24.85.
$$
The Shields diagram provides a $\theta_{\mathrm{cr}} \approx 0.035$.
***
The non-dimensional bed-load sediment transport rate is thus
$$
\Phi_b = 10.4(\theta-\theta_{\mathrm{cr}})^{1.67} = 10.4 \cdot (1.739 - 0.035)^{1.67} = 25.315,
$$
and the bed-load sediment transport rate is
$$
q_b = \Phi_b\sqrt{(s-1) \cdot g \cdot d_{50}^3} =4.772 \times 10^{-4}\ \mathrm{m^2/s} \text{ or } \mathrm{m^3/s/m}.
$$
The total sediment transport rate is
$$
q_t =q_b + q_s = 4.772 \times 10^{-4} + 4.792 \times 10^{-3}=5.270 \times 10^{-3}\ \mathrm{m^2 /s}.
$$
Thus, the suspended load is dominant in this case.
| A river stream has a flow rate of $q = 20.98\ \mathrm{m^2 /s}$ and a river depth of $h = 8.23\ \mathrm{m}$. The sediment of the bed has a typical sediment size of $d_{50} = 0.28\ \mathrm{mm}$.
For relatively large particles in large water depths, sediment suspension occurs in a small layer close to the bed. In such cases, we can assume that $z<<h$. Using this assumption the sediment concentration profile is simplified to
$$
\overline{C} = C_a \left( \dfrac{z}{z_a} \right)^{-b}.
$$
You may wish to use [the Shields diagram](https://bb.imperial.ac.uk/webapps/blackboard/execute/content/file?cmd=view\&content_id=_2543151_1\&course_id=_30255_1).
Using this simplified expression, calculate the sediment concentration at the following vertical elevations with respect to the bed: $z_1 = 0.0020\ \mathrm{m}$, $z_2 = 2.0\ \mathrm{m}$, and $z_3 = 4.0\ \mathrm{m}$.
\nCompare your values with the original Rouse profile. Is the assumption of suspended sediment transport occurring in a small layer close to the bed valid?
\nObtain the total sediment transport rate.
| 130 | 8 | 30 | 30 | 266 | 0 | 56 | 3 | 0 | In such cases, we can assume that $z<<h$. Using this assumption the sediment concentration profile is simplified to $ \overline{C} = C_a \left( \dfrac{z}{z_a} \right)^{-b}. $ You may wish to use . Using this simplified expression, calculate the sediment concentration at the following vertical elevations with respect to the bed: $z_1 = 0.0020\ \mathrm{m}$, $z_2 = 2.0\ \mathrm{m}$, and $z_3 = 4.0\ \mathrm{m}$. Compare your values with the original Rouse profile. Is the assumption of suspended sediment transport occurring in a small layer close to the bed valid? Obtain the total sediment transport rate. | 7 |
5fca4d66-12e3-4564-8582-c6cb93a7fa9c | 2 | 0 | 0 | 16 | 6 | 1 | 6 | 3 | Are the following exact differentials? If so - of what functions?
Yes - Input the function
No - Type 'none'
| $$
e^y \mathrm{d}x+ x(e^y + 1) \mathrm{d}y
$$
\n$$
(e^y + ye^x) \mathrm{d}x + (e^x + xe^y + 1) \, \mathrm{d}y
$$
| 2 | 0.333333 | 2 | The equation $P(x,y)\mathrm{d}x + Q(x,y)\mathrm{d}y$ is an exact differential when
$$
\begin{aligned} {\partial P\over\partial y} = {\partial Q\over\partial x} \end{aligned}
$$
is satisfied.
Now you can try identifying $P(x,y)$ and $Q(x,y)$, and then can you see if this is satisfied?
\nThe equation $P(x,y)\mathrm{d}x + Q(x,y)\mathrm{d}y$ is an exact differential when
$$
\begin{aligned} {\partial P\over\partial y} = {\partial Q\over\partial x} \end{aligned}
$$
is satisfied.
Now you can try identifying $P(x,y)$ and $Q(x,y)$, and then can you see if this is satisfied?
***
You should find that $(e^y + ye^x) \mathrm{d}x + (e^x + xe^y + 1) \, \mathrm{d}y$ is an exact differential.
For finding what function it is an exact differential of, remember that $\displaystyle P = {\partial u\over \partial x}$ and $\displaystyle Q={\partial u\over\partial y}$.
***
You can integrate $P$ with respect to $x$ to find $u(x,y)$. Remember that the integration constant can be a function of $y$ as you are integrating a partial derivative, so you can denote it as $A(y)$.
You can integrate $Q$ with respect to $y$ to find $u(x,y)$. Remember that the integration constant can be a function of $x$ as you are integrating a partial derivative, so you can denote it as $B(x)$.
Compare these to find $A(y), B(x)$ so that you can find $u(x,y)$.
| The equation $P(x,y)\mathrm{d}x + Q(x,y)\mathrm{d}y$ is an exact differential when
$$
\begin{aligned} {\partial P\over\partial y} = {\partial Q\over\partial x} \end{aligned}
$$
is satisfied.
Now you can try identifying $P(x,y)$ and $Q(x,y)$, and then can you see if this is satisfied?
***
$$
\begin{aligned}
P(x,y) &= e^y \\
Q(x,y) &= x(e^y+1)
\end{aligned}
$$
Can you see if it is an exact differential now?
***
$$
\begin{aligned}
{\partial Q\over\partial x} &= e^y+1 \\
{\partial P\over\partial y} &= e^y
\end{aligned}
$$
Is it an exact differential?
***
These are not identical, and so the condition is not satisfied. $e^y\mathrm{d}x+x(e^y+1)\mathrm{d}y$ is not an exact differential.
\nThe equation $P(x,y)\mathrm{d}x + Q(x,y)\mathrm{d}y$ is an exact differential when
$$
\begin{aligned} {\partial P\over\partial y} = {\partial Q\over\partial x} \end{aligned}
$$
is satisfied.
Now you can try identifying $P(x,y)$ and $Q(x,y)$, and then can you see if this is satisfied?
***
$$
P(x,y) = e^y+ye^x \\
Q(x,y) = e^x+xe^y+1
$$
Can you find out if it is an exact differential now?
***
$$
\begin{aligned}
{\partial Q\over\partial x} &= e^x+e^y \\
{\partial P\over\partial y} &= e^y+e^x
\end{aligned}
$$
Is it an exact differential?
***
These are identical, so $(e^y + ye^x) \mathrm{d}x + (e^x + xe^y + 1) \, \mathrm{d}y$ is an exact differential.
For finding what function it is an exact differential of, remember that $\displaystyle P = {\partial u\over \partial x}$ and $\displaystyle Q={\partial u\over\partial y}$.
***
You can integrate $P$ with respect to $x$ to find $u(x,y)$. Remember that the integration constant can be a function of $y$ as you are integrating a partial derivative, so you can denote it as $A(y)$.
You can integrate $Q$ with respect to $y$ to find $u(x,y)$. Remember that the integration constant can be a function of $x$ as you are integrating a partial derivative, so you can denote it as $B(x)$.
Compare these to find $A(y), B(x)$ so that you can find $u(x,y)$.
***
$$
\begin{aligned}
P = {\partial u\over \partial x} &= e^y+ye^x
\end{aligned}
$$
Integrating this with respect to $x$,
$$
\begin{aligned}
u(x,y) &= xe^y + ye^x + A(y)
\end{aligned}
$$
where $A$ is a function of only $y$.
$$
\begin{aligned}
Q = {\partial u\over \partial y} &= e^x+xe^y+1
\\
u(x,y) &= ye^x + xe^y + y + B(x)
\end{aligned}
$$
where $B$ is a function of $x$ only.
Compare these to find $A(y), B(x)$ so that you can find $u(x,y)$.
***
Comparing these,
$$
u(x,y) = xe^y + ye^x + A(y) = ye^x + xe^y + y + B(x)
$$
You can see that $A(y) - y = B(x)$, so $B(x)$ must be a constant as there are no terms in $x$ on the LHS. Then $A(y) = y+\mathrm{constant}$.
What is $u(x,y)$?
***
Finally,
$$
u(x,y) = xe^y+ye^x + y+\mathrm{constant}
$$
| Are the following exact differentials? If so - of what functions?
Yes - Input the function
No - Type 'none'
$$
e^y \mathrm{d}x+ x(e^y + 1) \mathrm{d}y
$$
\n$$
(e^y + ye^x) \mathrm{d}x + (e^x + xe^y + 1) \, \mathrm{d}y
$$
| 23 | 2 | 45 | 45 | 306 | 23 | 3 | 2 | 0 | Are the following exact differentials? If so - of what functions? | 2 |
5ff4fab4-25d8-4f7b-97b7-e9e89b8969e8 | 4 | 0 | 1 | 8 | 4 | 2 | 2 | 1 | A cylindrical pressure vessel is $0.6\text{ m}$ diameter, with hemispherical ends, and is made of steel with a yield strength of $250 \text{ MPa}$, $13 \text{ mm}$ thick. It is pressurised to $100 \text{ bar}.$
| Calculate the safety factor for the hemispherical ends, using both Tresca and Von Mises criteria.
\nCalculate the safety factor for the cylindrical part, using both criteria.
\nExplain, by drawing a yield locus, why the answers are the same in (a), but different in (b).
| 3 | 0.333333 | 2 | \n\n | For this pressure vessel with hemispherical ends, we know that:
* $\text{Diameter } (d)=0.6\text{ m}$
* $\text{Thickness } (t)=13\text{ mm}$
* $\text{Pressure } (P)=100\text{ bar}$
* $\text{Yield Strength } (\sigma_y)=250\text{ MPa}$
   
***
The stress in the hemispherical ends is the same in all directions:
   
$$
\sigma_\theta=\frac{PR}{2t}=\frac{100\times10^5\times0.3}{2\times0.013}=115.4\text{ MPa}
$$
   
***
The safety factor using the Tresca failure criteria:
   
$$
\text{SF}_{\text{Tresca}}=\frac{\sigma_y}{|\sigma_{\text{max}}-\sigma_{\text{min}}|} =\frac{250}{115.4-0} =\boxed{2.167}
$$
   
***
The safety factor using the Von Mises failure criteria:
   
$$
\begin{align*}
\text{SF}_{\text{Von Mises}}&=\frac{\sigma_y}{\sqrt{(\sigma_1^2+\sigma_2^2-\sigma_1\sigma_2)}} \\
\hspace{10pt} \\
&=\frac{250}{\sqrt{(115.4)^2+(115.4)^2-(115.4)(115.4)}}\\
\hspace{10pt} \\
&=\boxed{2.167}
\end{align*}
$$
\nExamining the cylinder itself, the hoop, axial, and radial stresses are calculated as such:
   
$$
\sigma_\theta=\frac{PR}{t}=230.8\text{ MPa} \hspace{40pt} \sigma_z=\frac{PR}{2t}=115.4 \text{ MPa} \hspace{40pt} \sigma_r\approx0
$$
   
***
The safety factor using the Tresca failure criteria:
   
$$
\text{SF}_{\text{Tresca}}=\frac{250}{230.8-0} =\boxed{1.0833}
$$
   
***
The safety factor using the Von Mises failure criteria:
   
$$
\text{SF}_{\text{Von Mises}}=\frac{250}{\sqrt{(230.8)^2+(115.4)^2-(230.8)(115.4)}}=\boxed{1.251}
$$
\n | A cylindrical pressure vessel is $0.6\text{ m}$ diameter, with hemispherical ends, and is made of steel with a yield strength of $250 \text{ MPa}$, $13 \text{ mm}$ thick. It is pressurised to $100 \text{ bar}.$
Calculate the safety factor for the hemispherical ends, using both Tresca and Von Mises criteria.
\nCalculate the safety factor for the cylindrical part, using both criteria.
\nExplain, by drawing a yield locus, why the answers are the same in (a), but different in (b).
| 73 | 4 | 10 | 10 | 112 | 0 | 44 | 0 | 0 | It is pressurised to $100 \text{ bar}.$ Calculate the safety factor for the hemispherical ends, using both Tresca and Von Mises criteria. Calculate the safety factor for the cylindrical part, using both criteria. Explain, by drawing a yield locus, why the answers are the same in a, but different in b. | 3 |
6124b15c-8d77-49b9-8d46-129ffdf45e53 | 4 | 0 | 0 | 10 | 4 | 2 | 7 | 0 | A crack of length $10\text{ mm}$ has been found at the center of a large thin plate, with an end load of $150\text{ kN}$. The plate is $100\text{ mm}$ wide and $5\text{ mm}$ thick. Assuming that the material is **perfectly linear elastic.** 
 
$$
\footnotesize
\text{Figure Q2.1: Defect in the centre of a large wide plate}
$$
| Calculate the remote (applied) stress on the plate.
\nAssuming that $Y=\sqrt\pi$, calculate the stress intensity factor, $K$
\nGiven that the materials fracture toughness is $80\text{ MPa}\sqrt{\text{m}}$, calculate the critical crack length in the material for fast (unstable brittle) fracture to occur.
\nCalculate the local stress at a distance of $10~\mu \text{m}$ directly ahead of the crack tip and comment on whether this stress is realistic. 
| 4 | 0.666667 | 2 | \n\n\n | $$
\text{Remote Stress}=\frac{F}{A}
= \frac{150\times10^3}{(0.1)(0.005)}=\boxed{300\text{ MPa}}
$$
\nSince the crack is in the center of the plate, the crack length $(a)$ is equal to half of the crack length.
   
$$
a=10\text{ mm}
$$
   
***
Therefore $K$ can be calculated as below:
   
$$
K=Y\sigma \sqrt{a}=\sqrt\pi\times300\times\sqrt{0.01}=\boxed{53.2\text{ MPa}\sqrt{\text{m}}}
$$
\nGiven that the fracture toughness of the material is provided, the critical crack length can be calculated.
   
$$
\begin{align*}
K_\text{IC}&=Y\sigma_c\sqrt{a_c}\\
a_c&=\left(\frac{K_\text{IC}}{Y\sigma_c}\right)^2\\
&=\left(\frac{80\times10^6}{\sqrt{\pi}\times300\times10^6}\right)^2\\
a_c&=\boxed{22.6\text{ mm}}
\end{align*}
$$
\nThe local stress is calculated as below:
$$
\begin{align*}
\sigma_{yy}&=\frac{K}{\sqrt2\pi r_y} \\
&=\frac{53.2\times10^6}{\sqrt2\times\pi\times10\times10^{-6}}\\
&=\boxed{6.7\text{ GPa}}
\end{align*}
$$
   
This is an unrealistic stress and is too high. 
| A crack of length $10\text{ mm}$ has been found at the center of a large thin plate, with an end load of $150\text{ kN}$. The plate is $100\text{ mm}$ wide and $5\text{ mm}$ thick. Assuming that the material is **perfectly linear elastic.** 
 
$$
\footnotesize
\text{Figure Q2.1: Defect in the centre of a large wide plate}
$$
Calculate the remote (applied) stress on the plate.
\nAssuming that $Y=\sqrt\pi$, calculate the stress intensity factor, $K$
\nGiven that the materials fracture toughness is $80\text{ MPa}\sqrt{\text{m}}$, calculate the critical crack length in the material for fast (unstable brittle) fracture to occur.
\nCalculate the local stress at a distance of $10~\mu \text{m}$ directly ahead of the crack tip and comment on whether this stress is realistic. 
| 106 | 9 | 7 | 7 | 80 | 0 | 65 | 4 | 0 | A crack of length $10\text{ mm}$ has been found at the center of a large thin plate, with an end load of $150\text{ kN}$. Assuming that the material is perfectly linear elastic. $ \footnotesize \text{Figure Q2.1: Defect in the centre of a large wide plate} $ Calculate the remote applied stress on the plate. Assuming that $Y=\sqrt\pi$, calculate the stress intensity factor, $K$ Given that the materials fracture toughness is $80\text{ MPa}\sqrt{\text{m}}$, calculate the critical crack length in the material for fast unstable brittle fracture to occur. Calculate the local stress at a distance of $10~\mu \text{m}$ directly ahead of the crack tip and comment on whether this stress is realistic. | 5 |
61305f9a-eb06-4522-b6dc-8742d4f2ad37 | 3 | 0 | 1 | 14 | 4 | 2 | 4 | 4 | The resistance of a sea-going ship is due to wave-making and viscous drag, and it may be expressed in functional form as $F_D = f(U,l,B,\rho,\mu,g)$, where $F_D$ is the drag force, $U$ is the ship speed, $l$ is its length, $B$ is its width, $\rho$ and $\mu$ are the sea water density and viscosity, and $g$ is the acceleration due to gravity. | Find the non-dimensional groups that describe the problem.
\nIf we are to test a model of the ship to determine the drag, what are the requirements for dynamic similarity?\nWe are going to test a $1/25$ scale model of a $100\,\mathrm{m}$ long ship. If the maximum velocity of the full-scale ship is $10\,\mathrm{m/s}$, what should the maximum speed of the model be?\nWe are going to test a $1/25$ scale model of a $100\,\mathrm{m}$ long ship. What should the kinematic viscosity ($\nu = \mu/\rho$) of the model test fluid be compared to the kinematic viscosity of sea water? | 4 | 0.5 | 2 | \n\n\n | \n\n\n | The resistance of a sea-going ship is due to wave-making and viscous drag, and it may be expressed in functional form as $F_D = f(U,l,B,\rho,\mu,g)$, where $F_D$ is the drag force, $U$ is the ship speed, $l$ is its length, $B$ is its width, $\rho$ and $\mu$ are the sea water density and viscosity, and $g$ is the acceleration due to gravity.Find the non-dimensional groups that describe the problem.
\nIf we are to test a model of the ship to determine the drag, what are the requirements for dynamic similarity?\nWe are going to test a $1/25$ scale model of a $100\,\mathrm{m}$ long ship. If the maximum velocity of the full-scale ship is $10\,\mathrm{m/s}$, what should the maximum speed of the model be?\nWe are going to test a $1/25$ scale model of a $100\,\mathrm{m}$ long ship. What should the kinematic viscosity ($\nu = \mu/\rho$) of the model test fluid be compared to the kinematic viscosity of sea water? | 157 | 14 | 0 | 0 | 1 | 0 | 97 | 6 | 0 | The resistance of a sea-going ship is due to wave-making and viscous drag, and it may be expressed in functional form as $F_D = f(U,l,B,\rho,\mu,g)$, where $F_D$ is the drag force, $U$ is the ship speed, $l$ is its length, $B$ is its width, $\rho$ and $\mu$ are the sea water density and viscosity, and $g$ is the acceleration due to gravity.Find the non-dimensional groups that describe the problem. If we are to test a model of the ship to determine the drag, what are the requirements for dynamic similarity? If the maximum velocity of the full-scale ship is $F_D$0, what should the maximum speed of the model be? What should the kinematic viscosity $F_D$3 of the model test fluid be compared to the kinematic viscosity of sea water? | 4 |
613e706e-06a3-4f0b-a712-d1ae8794752b | 1 | 0 | 0 | 15 | 5 | 0 | 0 | 4 | The Q-cycle is a part of the electron transport chain in mitochondria, plastids, and many prokaryotes. The specific version found in mitochondria oxidise ubiquinol ($\mathrm{UQH_{2}}$) to ubiquinone ($\mathrm{UQ}$) by removing two electrons (and two protons) and passing them to the next component of the electron transport chain: cytochrome-*c*. ($\mathrm{cyt\text{-}c}$)
$\mathrm{UQH_{2}}$ has two electrons to pass on, and $\mathrm{cyt\text{-}c}$ can only take up one.
The way the Q-cycle does this is quite complicated: one of the electrons from the oxidation of a molecule of $\mathrm{UQH_{2}}$ is passed to $\mathrm{cyt\text{-}c}$ but the other electron is passed *back* to a different $\mathrm{UQ}$ quinone molecule, which it half-reduces to a $\mathrm{UQH\cdot}$ free-radical. A *second* $\mathrm{UQH_{2}}$ is then oxidised, passing on one of its electrons to $\mathrm{cyt\text{-}c}$, and the other to the $\mathrm{UQH\cdot}$, fully reducing it to $\mathrm{UQH_{2}}$.
In essence, the protein complex is performing the process below:
$$
\mathrm{2UQH_{2}+1UQ+2cyt\text{-}c_{ox}\rightarrow 2UQ+1UQH_{2}+2cyt\text{-}c_{red}}
$$
For every two $\mathrm{UQH_{2}}$ molecules we oxidise, we get one $\mathrm{UQH_{2}}$ back! That one regenerated $\mathrm{UQH_{2}}$ molecule can then re-enter the Q cycle again (note the halving of all the terms):
$$
\mathrm{1UQH_{2}+\frac{1}{2}UQ+1cyt\text{-}c_{ox} \rightarrow UQ+ \frac{1}{2}UQH_{2}+1cyt\text{-}c_{red}}
$$
And the ‘half a molecule of $\mathrm{UQH_{2}}$ regenerated by this can re-re-enter the Q-cycle too:
$$
\mathrm{\frac{1}{2}UQH_{2}+\frac{1}{4}UQ+\frac{1}{2}cyt\text{-}c_{ox}\rightarrow \frac{1}{2}UQ+\frac{1}{4}UQH_{2}+\frac{1}{2}cyt\text{-}c_{red}}
$$
And so on.
In reality, there are no fractional molecules of $\mathrm{UQ}$ or $\mathrm{UQH_{2}}$; each time we run the Q-cycle, we simply halve the number of $\mathrm{UQH_{2}}$ molecules in the membrane.
| How many molecules of $\mathrm{cyt\text{-}c}$ are reduced by the complete oxidation of a single $\mathrm{UQH_{2}}$ molecule to $\mathrm{UQ}$? There are two ways to get the answer, and one of them requires an infinite sum. See if you can see both paths to the solution.
| 1 | 1 | 1 | null | The answer can be worked out from the chemistry without recourse to the maths: if we cancel the terms that appear on both sides of the chemical formula:
$$
\mathrm{2\,UQH_2 + 1\,UQ + 2\,cyt\text{-}c_{ox}\rightarrow 2\,UQ + 1\,UQH_2 + 2\,cyt\text{-}c_{red}}
$$
we get:
$$
\mathrm{UQH_2 + 2\,cyt\text{-}c_{ox} \rightarrow UQ + 2\,cyt\text{-}c_{red}}
$$
The $\mathrm{UQH_2}$ has two electrons to give up, and $cyt_c$ can only take one each. However complex the internal reactions, we’re only passing two electrons from $\mathrm{1\,UQH_2}$ to $\mathrm{2\,cyt\text{-}c_{red}}$.
Alternatively, you can see this as the sum of an infinite series. Oxidation of $\mathrm{1\,UQH_2}$ gives us $\mathrm{1\,cyt\text{-}c_{red}}$ plus one half of a molecule of $\mathrm{UQH_2}$ back. Oxidation of that $\mathrm{\frac{1}{2}\,UQH_2}$ gives us another half of a $\mathrm{cyt\text{-}c_{red}}$, plus one quarter of a $\mathrm{UQH_2}$ back. Oxidation of that $\mathrm{\frac{1}{4}\,UQH_2}$ gives us another quarter of a $\mathrm{cyt\text{-}c_{red}}$ plus one eighth of a $\mathrm{UQH_2}$ back. And so on. The total amount of $\mathrm{cyt\text{-}c_{red}}$ is therefore 
$$
1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots = 2
$$
the exact same geometric series we discussed earlier. In more detail, the first round of the Q-cycle does this:
$$
\mathrm{2\,UQH_2 + 1\,UQ + 2\,cyt\text{-}c_{ox} \rightarrow 2\,UQ + 1\,UQH_2 + 2\,cyt\text{-}c_{red}}
$$
We oxidise $\mathrm{2\,UQH_2}$, but get $\mathrm{1\,UQH_2}$ back. So, we have oxidised net $\mathrm{1\,UQH_2}$, and produced $\mathrm{2\,cyt\text{-}c_{red}}$ from it. The $\mathrm{1\,UQH_2}$ we get back is now oxidised too in a second round of the Q-cycle:
$$
\mathrm{1\,UQH_2 + \frac{1}{2}\,UQ + 1\,cyt\text{-}c_{ox} \rightarrow 1\,UQ + \frac{1}{2}\,UQH_2 + 1\,cyt\text{-}c_{red}}
$$
We oxidise $\mathrm{1\,UQH_2}$, but get $\mathrm{\frac{1}{2}\,UQH_2}$ back. So, we have oxidised net $\mathrm{\frac{1}{2}\,UQH_2}$, and produced $\mathrm{1\,cyt\text{-}c_{red}}$ from it. The $\mathrm{\frac{1}{2}\,UQH_2}$ we get back is now oxidised too in a third round of the Q-cycle:. 
$$
\mathrm{\frac{1}{2}\,UQH_2 + \frac{1}{4}\,UQ + \frac{1}{2}\,cyt\text{-}c_{ox} \rightarrow \frac{1}{2}\,UQ + \frac{1}{4}\,UQH_2 + \frac{1}{2}\,cyt\text{-}c_{red}}
$$
We oxidise $\mathrm{\frac{1}{2}\,UQH_2}$, but get $\mathrm{\frac{1}{4}\,UQH_2}$ back. So, we have oxidised net $\mathrm{\frac{1}{4}\,UQH_2}$, and produced $\mathrm{\frac{1}{2}\,cyt\text{-}c_{red}}$ from it. The $\mathrm{\frac{1}{4}\,UQH_2}$ we get back is now oxidised too in a fourth round of the Q-cycle: 
$$
\mathrm{\frac{1}{4}\,UQH_2 + \frac{1}{8}\,UQ + \frac{1}{4}\,cyt\text{-}c_{ox} \rightarrow \frac{1}{4}\,UQ + \frac{1}{8}\,UQH_2 + \frac{1}{4}\,cyt\text{-}c_{red}}
$$
We oxidise $\mathrm{\frac{1}{4}\,UQH_2}$, but get $\mathrm{\frac{1}{8}\,UQH_2}$ back. So, we have oxidised net $\mathrm{\frac{1}{8}\,UQH_2}$, and produced $\mathrm{\frac{1}{4}\,cyt\text{-}c_{red}}$ from it. The $\mathrm{\frac{1}{8}\,UQH_2}$ we get back is now oxidised too in a fifth round of the Q-cycle, and so on and so on.
How much $\mathrm{UQH_2}$ have we oxidised in total if we continue this forever? The sum of all the net oxidations of $\mathrm{UQH_2}$ listed above:
$$
1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} \ldots = 2
$$
And how much $\mathrm{cyt\text{-}c_{red}}$ have we made? The sum of all the $\mathrm{cyt\text{-}c_{red}}$ on the right of the equations:
$$
2 + 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} \ldots = 4
$$
| The Q-cycle is a part of the electron transport chain in mitochondria, plastids, and many prokaryotes. The specific version found in mitochondria oxidise ubiquinol ($\mathrm{UQH_{2}}$) to ubiquinone ($\mathrm{UQ}$) by removing two electrons (and two protons) and passing them to the next component of the electron transport chain: cytochrome-*c*. ($\mathrm{cyt\text{-}c}$)
$\mathrm{UQH_{2}}$ has two electrons to pass on, and $\mathrm{cyt\text{-}c}$ can only take up one.
The way the Q-cycle does this is quite complicated: one of the electrons from the oxidation of a molecule of $\mathrm{UQH_{2}}$ is passed to $\mathrm{cyt\text{-}c}$ but the other electron is passed *back* to a different $\mathrm{UQ}$ quinone molecule, which it half-reduces to a $\mathrm{UQH\cdot}$ free-radical. A *second* $\mathrm{UQH_{2}}$ is then oxidised, passing on one of its electrons to $\mathrm{cyt\text{-}c}$, and the other to the $\mathrm{UQH\cdot}$, fully reducing it to $\mathrm{UQH_{2}}$.
In essence, the protein complex is performing the process below:
$$
\mathrm{2UQH_{2}+1UQ+2cyt\text{-}c_{ox}\rightarrow 2UQ+1UQH_{2}+2cyt\text{-}c_{red}}
$$
For every two $\mathrm{UQH_{2}}$ molecules we oxidise, we get one $\mathrm{UQH_{2}}$ back! That one regenerated $\mathrm{UQH_{2}}$ molecule can then re-enter the Q cycle again (note the halving of all the terms):
$$
\mathrm{1UQH_{2}+\frac{1}{2}UQ+1cyt\text{-}c_{ox} \rightarrow UQ+ \frac{1}{2}UQH_{2}+1cyt\text{-}c_{red}}
$$
And the ‘half a molecule of $\mathrm{UQH_{2}}$ regenerated by this can re-re-enter the Q-cycle too:
$$
\mathrm{\frac{1}{2}UQH_{2}+\frac{1}{4}UQ+\frac{1}{2}cyt\text{-}c_{ox}\rightarrow \frac{1}{2}UQ+\frac{1}{4}UQH_{2}+\frac{1}{2}cyt\text{-}c_{red}}
$$
And so on.
In reality, there are no fractional molecules of $\mathrm{UQ}$ or $\mathrm{UQH_{2}}$; each time we run the Q-cycle, we simply halve the number of $\mathrm{UQH_{2}}$ molecules in the membrane.
How many molecules of $\mathrm{cyt\text{-}c}$ are reduced by the complete oxidation of a single $\mathrm{UQH_{2}}$ molecule to $\mathrm{UQ}$? There are two ways to get the answer, and one of them requires an infinite sum. See if you can see both paths to the solution.
| 279 | 26 | 47 | 47 | 355 | 0 | 45 | 3 | 1 | The specific version found in mitochondria oxidise ubiquinol $\mathrm{UQH_{2}}$ to ubiquinone $\mathrm{UQ}$ by removing two electrons and two protons and passing them to the next component of the electron transport chain: cytochrome-c. $\mathrm{cyt\text{-}c}$ $\mathrm{UQH_{2}}$ has two electrons to pass on, and $\mathrm{cyt\text{-}c}$ can only take up one. How many molecules of $\mathrm{cyt\text{-}c}$ are reduced by the complete oxidation of a single $\mathrm{UQH_{2}}$ molecule to $\mathrm{UQ}$? See if you can see both paths to the solution. | 3 |
614376cf-25d5-4712-b4d1-2260204101f7 | 5 | 0 | 0 | 17 | 6 | 1 | 3 | 7 | A ball is dropped onto a table from an initial height $h_0$. If it lands with speed $v$, it bounces up with speed $ev$, where the coefficient of restitution $e$ is a positive constant less than 1.
| Write down expressions for the speed $v_0$ of the ball when it first hits the table and the time $t_0$ it takes to drop, both in terms of $h_0$. 
\nFind the height $h_1$ reached by the ball after its first bounce and the corresponding drop time $t_1$. Hence write down expressions for the height $h_n$ and drop time $t_n$ of the $n^{\text{th}}$ bounce, in terms of $h_0$ and $t_0$ respectively.
\nHow long does the ball take to stop completely? Evaluate this time if the initial height $h_0 = 5\,$m and the coefficient of restitution $e = 0.5$. Take $g = 10\,\text{m}\cdot\text{s}^{-2}$.
\[**Hint:** you will need the formula for the sum of a geometric series: $\displaystyle a + ar + ar^2 + ar^3 + \ldots =
\frac{a}{1-r}, \qquad |r|<1.]$
| 3 | 0.666667 | 2 | Use conservation of energy or SUVAT equations.
\nRecall from the question that if the ball hits the ground with speed $v$, it bounces up with speed $ev$.
***
As in part (a), use conservation of energy/SUVAT equations to find $h_1$ and $t_1$.
***
Write $h_1$ and $t_1$ in terms of $h_0$ and $t_0$ respectively.
***
From here, you should be able to see the general pattern for $h_n$ and $t_n$. It may help to calculate $h_2$ and $t_2$ to confirm this to yourself.
\nWhat is the duration of the $n^\text{th}$ bounce? Use your $t_n$ formula from part (b). Note that this excludes $t_0$, the time for the zeroth bounce. 
***
Write down the total time $T$ taken for the ball to stop bouncing as a series... 
***
... Express this using the geometric-series formula, as suggested in the hint in the question.
***
After finding $t_0$ (refer to part (a)), find $T$.
| Using conservation of energy,
***
The speed $v_0$ of the ball when it first hits the ground is related to the initial height $h_0$ by $mgh_0 = \frac{1}{2}mv_0^2$. This gives:
***
$$
v_0 = \sqrt{2gh_0}.
$$
(Could also have used the SUVAT equation $v^2 = u^2 + 2as$.) 
***
Since $v_0 = g t_0$,
***
$$
t_0 = \sqrt{\frac{2h_0}{g}}.
$$
\nThe upward speed after the first bounce is $v_1 = ev_0$. The height reached after the first bounce is:
***
$$
h_1 = \frac{v_1^2}{2g} = \frac{e^2 v_0^2}{2g}
= e^2 h_0.
$$
***
The time taken to reach the top of the first bounce is
***
$$
t_1 = \sqrt{\frac{2h_1}{g}} = et_0.
$$
From this, we can establish the relationship for the $n^{\text{th}}$ bounce:
***
The height $h_n$ and drop time $t_n$ for the $n^{\text{th}}$ bounce are:
$$
h_n = e^{2n} h_0, \qquad t_n = e^n t_0.
$$
\nThe initial drop takes time $t_0$. The duration of the $n^{\text{th}}$ bounce is $2t_n$, not $t_n$, because the ball has to rise and then drop again. 
***
The total time $T$ for the ball to stop bouncing is the sum of the durations of the initial drop and all subsequent bounces:
***
$$
\begin{aligned}
T &= t_0 + 2 e t_0 + 2 e^2 t_0 + 2 e^3 t_0 + \ldots \\
&= 2 t_0 (1 + e + e^2 + e^3 + \ldots) - t_0 \\
\end{aligned}
$$
Each bounce takes less and less time, so the series converges.
***
$T$ can be condensed using the geometric series formula given in the question,
***
$$
\begin{aligned}
T &= \frac{2 t_0}{1 - e} - t_0\\
&= \frac{1 + e}{1 - e} \, t_0.
\end{aligned}
$$
***
If $h_0 = 5$m, $e = 0.5$, and $g = 10\text{m}\cdot\text{s}^{-2}$, then $t_0$ is given by:
$$
t_0 = \sqrt{\frac{2 \times 5}{10}} = 1\,\text{s},
$$
and the time $T$ taken to stop bouncing is:
***
$$
T = \frac{1 + 0.5}{1 - 0.5} \, t_0 = 3\,\text{s}.
$$
| A ball is dropped onto a table from an initial height $h_0$. If it lands with speed $v$, it bounces up with speed $ev$, where the coefficient of restitution $e$ is a positive constant less than 1.
Write down expressions for the speed $v_0$ of the ball when it first hits the table and the time $t_0$ it takes to drop, both in terms of $h_0$. 
\nFind the height $h_1$ reached by the ball after its first bounce and the corresponding drop time $t_1$. Hence write down expressions for the height $h_n$ and drop time $t_n$ of the $n^{\text{th}}$ bounce, in terms of $h_0$ and $t_0$ respectively.
\nHow long does the ball take to stop completely? Evaluate this time if the initial height $h_0 = 5\,$m and the coefficient of restitution $e = 0.5$. Take $g = 10\,\text{m}\cdot\text{s}^{-2}$.
\[**Hint:** you will need the formula for the sum of a geometric series: $\displaystyle a + ar + ar^2 + ar^3 + \ldots =
\frac{a}{1-r}, \qquad |r|<1.]$
| 154 | 18 | 30 | 30 | 215 | 18 | 114 | 14 | 0 | Write down expressions for the speed $v_0$ of the ball when it first hits the table and the time $t_0$ it takes to drop, both in terms of $h_0$. Find the height $h_1$ reached by the ball after its first bounce and the corresponding drop time $t_1$. Hence write down expressions for the height $h_n$ and drop time $v$0 of the $v$1 bounce, in terms of $h_0$ and $t_0$ respectively. How long does the ball take to stop completely? Evaluate this time if the initial height $v$4m and the coefficient of restitution $v$5. Take $v$6. | 6 |
61444717-1285-46e9-8574-5e6ee1d6cdc8 | 0 | 2 | 2 | 21 | 6 | 1 | 2 | 4 | In this problem, we will consider some practical applications of the Fourier transform in optics. You may use the result from lectures that the amplitude $U$ of a diffraction pattern is given by:
$$
U(l) = \frac{\lambda}{\sqrt{2\pi}}A(2\pi l/\lambda) \; ,
$$
where $l=\sin \theta$ with $\theta$ the scattering angle, $\lambda$ the wavelength of the light, and $A$ is the Fourier Transform of the aperture function $a(x)$ (where $x$ is the displacement along the aperture). Do not worry about normalisation factors. 
| Consider two infinitely narrow slits that are separated by a distance $d$. Calculate the amplitude of the diffraction pattern as a function of $l$. Then, calculate the corresponding intensity that would be recorded by a detector.
***
In the answer box below, input the form of the intensity diffraction pattern that you have derived. See the solutions for a more detailed breakdown of how the form is obtained.
\nFind an expression for $U(l)$ with $N$ slits on each side of $x=0$, each separated by $d$ (a diffraction grating). Which intensity pattern will be recorded in the limit $N \to \infty$? 
***
Select your derived answer to this in the answer boxes below. See the solutions for a more detailed breakdown of how the form is obtained.
\nRevert to just two slits now, but make the slits have a Gaussian absorption profile with width $\sigma$. Find an expression for the amplitude and intensity again and comment on the effect. You can use $\sigma \ll d$. Make a sketch of the intensity pattern.
\nIt would seem that from the recorded pattern of intensities of the scattered light, we would be able to find the shape of the slits through an inverse Fourier transform. Discuss why this is not the case and see of you can come up with some ideas for how to achieve this anyway.
| 4 | 1 | 4 | First, what is $a(x)$ (the *aperture function*) for two infinitely narrow slits...?
***
... Consider that an infinitely narrow slit can be represented by a delta function.
***
Now, your goal is to find $A(\omega)$, the Fourier transform of $a(x)$ (**Note:** you will let $\omega = 2\pi l/\lambda$ later on), in order to find $U(l)$... 
***
... To find $A(\omega)$, start by using Fourier transform properties (see **section 4.3**) to simplify the Fourier Transform...
***
... What is the Fourier Transform of a delta function (see **section 4.2.2**)?
***
Let $\omega = 2\pi l / \lambda$. 
***
How can you find intensity from amplitude? Apply this to your result to find the intensity pattern.
\nThe aperture function $a(x)$ is a summation of $2N+1$
delta functions separated by distance $d$. Can you write $A(x)$?
***
Then, apply a Fourier Transform to $a(x)$ to find $U(l)$...
***
... First, apply a Fourier Transform property (see **section 4.3**) to extract $\delta(x)$. 
***
... You will need to consider the Fourier Transform of a delta function (see **section 4.2.2**).
***
What is the position of the maxima in the interference pattern? 
***
What happens if you let $N\rightarrow\infty$?
\nRather than the aperture function being two $\delta$ functions at $x=\pm d/2$ as in part (a), we now have two Gaussian centred at $x\pm d/2$, with width $\sigma$...
***
... Considering the general form of a Gaussian (ignoring constants in-front), can you write down $a(x)$?
***
Now, evaluate the Fourier Transform $A(\omega)$ to find $U(l)$...
***
... First apply a property of Fourier transforms to remove the shifting factor...
***
... Then, consider the Fourier transform of a Gaussian. 
***
After finding $U(l)$, find the intensity pattern. 
***
... What is the shape of the intensity pattern (either sketch, or use a graphing software)?
\nIf we start from the intensity pattern, can you think why it would be difficult to retrieve the aperture function...?
***
... What information is lost by finding the intensity from $U(l)$?
| In the lectures, we derived that the Fourier transform of the aperture function, $a(x)$, gives the form of an interference pattern. 
***
If $x=0$ is the centre of aperture, the slits are at $x=+d/2$ and $x=-d/2$:
***
So, we can write the aperture function as the sum of two delta functions,
$$
a(x) = \delta(x-d/2) + \delta(x+d/2) \; .
$$
***
We need to solve for $A(\omega)$, the Fourier transform of $a(x)$. 
***
Using translation properties (see **section 4.3**), we can simplify $A(\omega)$:
***
$$
\begin{aligned}
&A(\omega)=\mathcal{F}[a(x)]= \\
&= \mathcal{F}[\delta(x-d/2) + \delta(x+d/2)] \\
&= \mathcal{F}[\delta(x-d/2)]+\mathcal{F}[\delta(x+d/2)]\\
&= e^{i\omega d/2}\mathcal{F}[\delta(x)] + e^{-i\omega d/2}\mathcal{F}[\delta(x)]
\end{aligned}
$$
***
Then, we use that the Fourier transform of a delta function is a constant (see **section 4.2.2**):
$$
\mathcal{F}[\delta(x)] =\frac{1}{\sqrt{2\pi}}
$$
Inserting this into $A(\omega)$, we have:
***
$$
\begin{aligned}
A(\omega) &= \frac{1}{\sqrt{2\pi}} (e^{i\omega d/2} + e^{-i\omega d/2}) \\
\end{aligned}
$$
Then, using the identity:
$$
\cos\phi = \frac{e^{i\phi} + e^{-i\phi}}{2}
$$
***
$$
\begin{aligned}
&A(\omega)= \frac{2}{\sqrt{2\pi}} \cos(\omega d/2) \\
&= A_0 \cos\left( \frac{\pi l d}{\lambda} \right) \; .
\end{aligned}
$$
with $A_0$ a normalisation constant, and $\omega = 2\pi l /\lambda$. 
***
The intensity is the amplitude squared so:
***
$$
I = A_0^2 \cos^2 \left( \frac{\pi l d}{\lambda} \right) \; .
$$
The diffraction pattern is thus a set of lines that have a constant separation of $\lambda/d$ (corresponding to shifts in $\pi$ inside the cosine).
\nHere, we adapt the approach from part (a) to look at $M = 2N+1$ narrow slits (delta functions) separated by a distance $d$. We can place the slits symmetrically around $x=0$, giving the aperture function $a(x)$:
***
$$
a(x) = \sum_{n=-N}^{n=N}\delta(x-nd)
$$
Then, we must find $A(\omega)=\mathcal{F}[a(x)]$. You should not be worried by the summation, because each term in the summation can be treated independently. 
***
First, applying the translation property of the Fourier Transform (see **section 4.3**):
***
$$
A(\omega) = \mathcal{F}[a(x)] = \mathcal{F}\left[\sum_{n=-N}^{n=N}\delta(x-nd)\right] = \sum_{n=-N}^{n=N}e^{i\omega nd}\mathcal{F}[\delta(x)]
$$
***
The Fourier transform of a $\delta$ function is a constant, which we will write as $A_0$ (the value of $A_0$ is irrelevant to the form of the diffraction pattern):
$$
U(l) = A_0 \sum_{n=-N}^N{e^{i\omega n d}} \;
$$
***
The intensity is the square of this result:
***
$$
I = A_0^2 \left( \sum_{n=-N}^N{e^{i\omega n d}} \right)^2 \; .
$$
***
Looking at the sum, we see that the individual parts all add up coherently (meaning that they have the same phase). Maxima occur for $\omega d = 2\pi m$, where $m \in \mathbf{Z}$, or:
$$
l = \frac{m \lambda}{d} \; .
$$
***
As with part (a), we get lines separated by $\lambda/d$. As $N \to \infty$ the cancellation for other values of $l$ gets better and better and the pattern approaches a set of delta functions, each separated by $l/d$. This function is called the *Dirac comb* and is in fact identical to the function that we find the Fourier transform of.
\nThis is identical to the first part but we have to insert the Fourier transform of the Gaussian rather than of the delta function. The aperture function is the sum of two shifted Gaussians at $x\pm d/2$: 
***
$$
a(x) = \exp\left[-\frac{(x+d/2)^2}{2\sigma^2}\right]+\exp\left[-\frac{(x-d/2)^2}{2\sigma^2}\right]
$$
***
Now, using the Fourier Transform translation property to find $A(\omega)$ (**section 4.3**): 
***
$$
\begin{aligned}
&A(\omega)=\mathcal{F}\left[\exp\left(-\frac{(x+d/2)^2}{2\sigma^2}\right)+\exp\left(-\frac{(x-d/2)^2}{2\sigma^2}\right)\right]\\
&=\mathcal{F}\left[\exp\left(-\frac{x^2}{2\sigma^2}\right)\right]\times\left[\exp\left(\frac{i\omega d}{2}\right)+\exp\left(\frac{-i\omega d}{2}\right)\right]
\end{aligned}
$$
Hence, applying the Fourier Transform of a Gaussian (see **section 4.2.1**), and ignoring constants, we have:
***
$$
\begin{aligned}
U(l) &= A_0 (e^{i\omega d/2} + e^{-i\omega d/2})
\exp\left(\frac{-\sigma^2 \omega^2}{2} \right) \\
&= \cos(\omega d/2) \exp\left(\frac{-\sigma^2 \omega^2}{2} \right) \; , \\
\end{aligned}
$$
***
and:
$$
\begin{aligned}
I(l) &= \cos^2\left(\frac{\pi l d}{\lambda}\right)
\exp\left(\frac{-4 \sigma^2 \pi^2 l^2}{\lambda^2} \right) \; .
\end{aligned}
$$
The intensity will be an interference pattern with separation $l/d$ between peaks as before, but the intensity will be modulated by a broad Gaussian of width:
***
$$
\sigma' = \frac{\lambda}{2\sqrt{2}\pi\sigma} \; .
$$
Given the initial assumption we see that $\sigma' \gg \lambda/d$.
***
  
\nThe basic problem is that we record the intensity rather than the magnitude and phase in each direction. So a straight forward inverse Fourier transform will not work. 
***
A method that can be used is to parameterise the aperture, and then use the Fourier transform of that parametrisation in a fit to the observed data. Information can also be obtained from an inverse Fourier transform of the intensity but it leads to a solution with all the possible vectors connecting different parts of the aperture and thus not the direct information. The Wikipedia article on the *Phase problem* is informative.
| In this problem, we will consider some practical applications of the Fourier transform in optics. You may use the result from lectures that the amplitude $U$ of a diffraction pattern is given by:
$$
U(l) = \frac{\lambda}{\sqrt{2\pi}}A(2\pi l/\lambda) \; ,
$$
where $l=\sin \theta$ with $\theta$ the scattering angle, $\lambda$ the wavelength of the light, and $A$ is the Fourier Transform of the aperture function $a(x)$ (where $x$ is the displacement along the aperture). Do not worry about normalisation factors. 
Consider two infinitely narrow slits that are separated by a distance $d$. Calculate the amplitude of the diffraction pattern as a function of $l$. Then, calculate the corresponding intensity that would be recorded by a detector.
***
In the answer box below, input the form of the intensity diffraction pattern that you have derived. See the solutions for a more detailed breakdown of how the form is obtained.
\nFind an expression for $U(l)$ with $N$ slits on each side of $x=0$, each separated by $d$ (a diffraction grating). Which intensity pattern will be recorded in the limit $N \to \infty$? 
***
Select your derived answer to this in the answer boxes below. See the solutions for a more detailed breakdown of how the form is obtained.
\nRevert to just two slits now, but make the slits have a Gaussian absorption profile with width $\sigma$. Find an expression for the amplitude and intensity again and comment on the effect. You can use $\sigma \ll d$. Make a sketch of the intensity pattern.
\nIt would seem that from the recorded pattern of intensities of the scattered light, we would be able to find the shape of the slits through an inverse Fourier transform. Discuss why this is not the case and see of you can come up with some ideas for how to achieve this anyway.
| 298 | 17 | 8 | 8 | 250 | 24 | 226 | 9 | 0 | You may use the result from lectures that the amplitude $U$ of a diffraction pattern is given by: $ U(l) = \frac{\lambda}{\sqrt{2\pi}}A(2\pi l/\lambda) \; , $ where $l=\sin \theta$ with $\theta$ the scattering angle, $\lambda$ the wavelength of the light, and $A$ is the Fourier Transform of the aperture function $a(x)$ where $x$ is the displacement along the aperture. Consider two infinitely narrow slits that are separated by a distance $d$. Calculate the amplitude of the diffraction pattern as a function of $l$. Then, calculate the corresponding intensity that would be recorded by a detector. In the answer box below, input the form of the intensity diffraction pattern that you have derived. See the solutions for a more detailed breakdown of how the form is obtained. Find an expression for $ U(l) = \frac{\lambda}{\sqrt{2\pi}}A(2\pi l/\lambda) \; , $0 with $ U(l) = \frac{\lambda}{\sqrt{2\pi}}A(2\pi l/\lambda) \; , $1 slits on each side of $ U(l) = \frac{\lambda}{\sqrt{2\pi}}A(2\pi l/\lambda) \; , $2, each separated by $d$ a diffraction grating. Which intensity pattern will be recorded in the limit $ U(l) = \frac{\lambda}{\sqrt{2\pi}}A(2\pi l/\lambda) \; , $4? Select your derived answer to this in the answer boxes below. See the solutions for a more detailed breakdown of how the form is obtained. Find an expression for the amplitude and intensity again and comment on the effect. You can use $ U(l) = \frac{\lambda}{\sqrt{2\pi}}A(2\pi l/\lambda) \; , $6. Make a sketch of the intensity pattern. It would seem that from the recorded pattern of intensities of the scattered light, we would be able to find the shape of the slits through an inverse Fourier transform. Discuss why this is not the case and see of you can come up with some ideas for how to achieve this anyway. | 15 |
61591033-37c0-4c98-bf3c-29b4c44617ec | 2 | 0 | 0 | 0 | 0 | 2 | 5 | 3 | For a specific design Mach number $M$, the geometry of an inlet can make the reflected shock wave cancel. Consider the geometry shown in the figure below (not to scale) for a supersonic diffuser to reduce the Mach number from $M=3.5$. The deflection angle is $\theta = 10^{\circ}$. 
| Find the downstream height $h$. 
\nFind the downstream Mach number $M_{3}$. 
| 2 | 1 | 3 | If the angle the reflected shock makes with the wall, $ \phi $, and the length $ AB $ are known , $ h $ can be determined geometrically. How can these two quantities be determined? Hint: use oblique shock tables for $ \phi $ and simple geometry for $ AB $. 
***
$ \phi = \beta_{2} - \theta $. Determine $ \beta_{2} $ and $ \theta $ using the oblique shock tables, and thus find $ \phi $. 
***
Use the law of sines to find $ AB $. 
\nFirst you need the Mach number behind the first shock, with that and the deflection angle $\theta$, you can find the Mach number behind the second shock
| The first oblique shock is reflected off the opposite wall. As the flow needs to be turned parallel to the wall, the effective deflection angle at the wall is again $\theta$.
  
***
The strength of the second shock is diminished, as the Mach number behind the first shock is going to be lower, to the shock angle of the second shock, $\beta_2$, is different than $\beta_1$. The reflected shock makes an angle with the wall $\phi=\beta_2-\theta$.
***
First let's compute the angle $ \phi $ using oblique shock tables.
 
 For $ M_{1} = 3.5 $ and $ \theta_{1} = 10^{\circ} $, $ \beta_{1} = 24.384^{\circ} $and $ M_{2} = 2.904 $. 
***
Remember that $ \theta_{2} = \theta_{1} = 10^{\circ} $.
 
Again using oblique shock tables with $M_2=2.904$ and $\theta=10^{\circ}$, $ \beta_{2} = 28.099^{\circ} $. 
 
 $\phi = \beta_{2} - \theta = 28.099^{\circ} -10^{\circ} = 18.099^{\circ}$  
***
Now it is a matter of trigonometry and triangles. 
 
$ \widehat{ACB} = \beta_{1} - \theta = 14.384^{\circ}; \ \widehat{CBA} = \beta_{2} = 28.099^{\circ}; \ \widehat{BAC} = 180^{\circ} -14.384^{\circ} - 28.099^{\circ} = 137.517^{\circ} $
 
$\mathrm{sin} \beta_{1} = \frac{1}{AC}$ , so $\ AC = \frac{1}{\mathrm{sin(24.384)}} = 2.422 \ \mathrm{m}$
 
***
By the law of sines $ \frac{AB}{\mathrm{sin(14.384)}} = \frac{AC}{\mathrm{sin(28.096)}} $, thus $ AB = 1.278 \ \mathrm{m} $. 
 
***
Finally $ \mathrm{sin} \phi=\frac{h}{AB} $, so
 
 $ h = 1.278 \cdot \mathrm{sin(18.096)} = 0.397 \ \mathrm{m} $
\nFirst you need to know the Mach number behind the first shock, which has been calculated in part (a).
 
***
With $M_2=2.904$ and $\theta=10^{\circ}$, we can find the Mach number behind the second shock. 
 
Using oblique shock tables, $ M_{3} = 2.427 $
| For a specific design Mach number $M$, the geometry of an inlet can make the reflected shock wave cancel. Consider the geometry shown in the figure below (not to scale) for a supersonic diffuser to reduce the Mach number from $M=3.5$. The deflection angle is $\theta = 10^{\circ}$. 
Find the downstream height $h$. 
\nFind the downstream Mach number $M_{3}$. 
| 63 | 5 | 24 | 24 | 208 | 11 | 13 | 2 | 1 | Consider the geometry shown in the figure below not to scale for a supersonic diffuser to reduce the Mach number from $M=3.5$. Find the downstream height $h$. Find the downstream Mach number $M_{3}$. | 3 |
61598941-66ef-425d-9d76-03b3226e4d80 | 0 | 1 | 1 | 4 | 3 | 3 | 7 | 0 | The attached sketch illustrates a bed form called antidune, occurring when the Froude number is close or above 1 (supercritical). In the antidune system the water surface is in phase with the bed configuration with a larger water depth at the antidune crest than at the antidune trough.
| Discuss, using the Exner equation, the expected migration of the antidune after an appropriate time interval.
\nWould it be possible to obtain upstream migration (against the flow) of an antidune system if the total sediment transport is downstream? Discuss your answer.
| 2 | 0.666667 | 2 | \n | \n | The attached sketch illustrates a bed form called antidune, occurring when the Froude number is close or above 1 (supercritical). In the antidune system the water surface is in phase with the bed configuration with a larger water depth at the antidune crest than at the antidune trough.
Discuss, using the Exner equation, the expected migration of the antidune after an appropriate time interval.
\nWould it be possible to obtain upstream migration (against the flow) of an antidune system if the total sediment transport is downstream? Discuss your answer.
| 90 | 0 | 0 | 0 | 1 | 0 | 41 | 0 | 1 | Discuss, using the Exner equation, the expected migration of the antidune after an appropriate time interval. Would it be possible to obtain upstream migration against the flow of an antidune system if the total sediment transport is downstream? Discuss your answer. | 3 |
6187b997-5513-41f9-a9eb-3d0d80a2fb02 | 2 | 0 | 1 | 17 | 6 | 1 | 1 | 6 | A particle of mass $m$ moves in the $x$ direction under the action of a conservative force. Its potential energy is given by
$$
U(x) = \frac{cx}{x^2 + a^2}
$$
where $a$ and $c$ are constants.
| Find the position of **stable** equilibrium and the value of $U(x)$ at that point. Sketch $U(x)$.
\nIf the particle starts from the position of stable equilibrium with velocity $v$, find the range of velocities for which it:
* (i) escapes to $+\infty$
* (ii) escapes to $-\infty$
* (iii) oscillates
(**Note:** This question doesn't have a response area to type your answer because Lambda Feedback does not yet accept inequality inputs.)
\nDerive an expression for the period of small oscillations about the position of stable equilibrium.
| 3 | 1 | 4 | What is the relationship between $F(x)$ and $U(x)$? 
***
Equilibrium occurs when $F(x)=0$. Therefore, solve for $x$ and plug this back into $U(x)$.
***
Therefore, solve for $x$ and plug this back into $U(x)$. You should obtain **two** equilibrium points...
***
... Which equilibrium point is stable? There are several ways to deduce this:
* Deduce the nature of each equilibrium point just by making a sketch of $U(x)$ (see next hint if you need to), or
* You may be able to see just by looking at the values of $x$ and $U(x)$, or
* Find the sign of $d^2 U /dx^2$ at each equilibrium point. 
***
To sketch $U(x)$, consider $U(x)$ at:
* The stationary points
* $x=0$
* The limits $x\to \pm \infty$
Also, it may help you to consider the symmetry of $U(x)$ (is it even or odd)? 
\nRefer to your graph from part (a), and the point of stable equilibrium...
***
... In each case, your goal is to find the minimum kinetic energy that would be required for the motion (from this, you can derive the velocity range).
***
... Recall that, for a conservative force, the total energy is the sum of the kinetic and potential energy.
***
The total energy therefore needs to be larger than the value of the potential energy function for motion in that region to be *viable*.
***
You should also consider in what *direction* the particle is moving. This is most relevant in (ii).
\nTo obtain simple harmonic motion, we require $U(x)$ to be quadratic. The derivative of this gives a force that is proportional to $x$, which satisfies simple harmonic motion.
***
Therefore, what is the formula of a quadratic with its minimum at the stable equilibrium that you found in part (a)? You should include a constant, $s$, in front of the $x^2$ term (this is analogous to the spring constant). Your quadratic approximates to $U(x)$ for small $x$. (**Note:** You may wish to find the quadratic this via a Taylor Series expansion if you have met this before).
***
Now, you must solve for $s$. Differentiate your quadratic approximation $U(x)$ twice. What is $s$?
***
Hence make use of the potential function in the question (you can start with your $U'(x)$ from part (a)) to solve for $s$.
***
Use the standard expression for the period of simple harmonic motion with your 'spring constant' $s$.
| The force $F(x)$ corresponding to the given potential function $U(x)$ is:
***
$$
F(x) = -\frac{dU(x)}{dx}
$$
Apply the quotient rule:
***
$$
F(x) = -\frac{dU(x)}{dx} = \frac{2cx^2 - c(x^2 + a^2)}{(x^2 +
a^2)^2} = \frac{c(x^2 - a^2)}{(x^2 + a^2)^2}.
$$
***
At the position of equilibrium the force must be zero, implying that $x = \pm a$ are the only positions of equilibrium.
***
The corresponding values of the potential are $U(\pm a) = \pm\frac{c}{2a}$. 
You might also notice (useful for the sketch) that:
* For $x=0$, $U=0$
* For large positive $x$, $U$ is positive
* For large negative $x$, $U$ is negative.
* There are no asymptotes (other than $U=0$)
***
From this, we deduce that $x = -a$ is a minimum (a position of **stable equilibrium**) and $x = +a$ is a maximum (a position of unstable equilibrium). You can also verify that $x=-a$ is the stable equilibrium point by finding the sign of $d^2 U / dx^2$ (should be $>0$). Sketch:
***
\nThe particle oscillates if its initial KE is less than $c/2a$. The range of velocities is then:
***
$$
-\sqrt{\frac{c}{ma}} < v < \sqrt{\frac{c}{ma}}.
$$
, The particle can escape to $-\infty$ in two ways. If its initial velocity is negative, all that is required is for the initial KE to be greater than $c/2a$, implying $v < - \sqrt{c/ma}$. 
***
If its initial velocity is positive, the initial KE must lie between $c/2a$ and $c/a$. In that case the particle does not have enough energy to climb the barrier to its right and bounces back, but does have enough KE to climb the barrier to its left and travel off to $x = -\infty$. The range of $v$ is then:
***
$$
v < -\sqrt{\frac{c}{ma}} \quad \text{or} \quad
\sqrt{\frac{c}{ma}} < v < \sqrt{\frac{2c}{ma}} .
$$
, Graph from part (a):
 The particle starts from the point of stable equilibrium $(-a,-c/2a)$
, The particle escapes to $+\infty$ if it is initially moving to the right and its initial kinetic energy is greater than $c/a$, which is the height of the potential barrier it has to surmount.
***
$$
\frac{1}{2}mv^2 > \frac{c}{a} \quad \Rightarrow \quad
v > \sqrt{\frac{2c}{ma}}.
$$
\nTo find the period of small oscillations about the minimum at $x = -a$, approximate the potential function as a quadratic near that point:
***
$$
U(x) \approx U(-a) + \frac{1}{2} s (x - (-a))^2 \quad \text{for small
values of } |x - (-a)|.
$$
This quadratic approximation for $U(x)$ looks just like the potential of a simple harmonic oscillator with spring constant $s$. 
***
Differentiating the approximation for $U$ twice with respect to $x$ shows that $U''(x) \approx s$ when $|x - (-a)|$ is small. This means that we can find $s$ by differentiating $U(x)$ twice with respect to $x$ and then setting $x = -a$.
***
Returning to the exact expression for the force $F = -dU/dx$ from part (a) above and differentiating it once more gives:
***
$$
U''(x) = \frac{d}{dx} \left ( \frac{c(a^2 - x^2)}{(x^2 +
a^2)^2} \right )
= \frac{-2cx(x^2 + a^2) - 4c(a^2 - x^2)x}{(x^2 + a^2)^3}.
$$
***
Setting $x = -a$ then gives the spring constant:
$$
s = U''(-a) = \frac{c}{2a^3}.
$$
The time period of small oscillations is given by the usual formula for simple harmonic motion:
$$
T = 2\pi\sqrt{\frac{m}{s}} = 2\pi \sqrt{\frac{2a^3 m}{c}}.
$$
,
, You may have knowledge of the Taylor series, which provides another method to find $U(x)$. 
***
Performing a Taylor series expansion of a general $U(x)$ about $x=-a$ up to terms in $x^2$ (a quadratic approximation):
***
$$
U(x) \approx U(-a) + U'(-a)(x-(-a)) + \frac{1}{2!}U''(-a)(x-(-a))^2
$$
***
The first derivative evaluates to $0$ at $x=-a$ since this is a minimum. We will then let $U''(-a)=s$:
***
$$
U(x) \approx U(-a) + \frac{1}{2} s (x +a)^2
$$
For small values of $|x+a|$. 
***
This quadratic approximation for $U(x)$ looks just like the potential of a simple harmonic oscillator with spring constant $s$ (because $F=-dU/dx$ scales as $-x$, demonstrating that the oscillations are simple harmonic). From here, we can solve for $s=U''(-a)$ as in the standard method. 
| A particle of mass $m$ moves in the $x$ direction under the action of a conservative force. Its potential energy is given by
$$
U(x) = \frac{cx}{x^2 + a^2}
$$
where $a$ and $c$ are constants.
Find the position of **stable** equilibrium and the value of $U(x)$ at that point. Sketch $U(x)$.
\nIf the particle starts from the position of stable equilibrium with velocity $v$, find the range of velocities for which it:
* (i) escapes to $+\infty$
* (ii) escapes to $-\infty$
* (iii) oscillates
(**Note:** This question doesn't have a response area to type your answer because Lambda Feedback does not yet accept inequality inputs.)
\nDerive an expression for the period of small oscillations about the position of stable equilibrium.
| 118 | 10 | 51 | 51 | 413 | 28 | 88 | 5 | 0 | Its potential energy is given by $ U(x) = \frac{cx}{x^2 + a^2} $ where $a$ and $c$ are constants. Find the position of stable equilibrium and the value of $U(x)$ at that point. Sketch $U(x)$. If the particle starts from the position of stable equilibrium with velocity $v$, find the range of velocities for which it: i escapes to $+\infty$ ii escapes to $-\infty$ iii oscillates Note: This question doesn't have a response area to type your answer because Lambda Feedback does not yet accept inequality inputs. Derive an expression for the period of small oscillations about the position of stable equilibrium. | 5 |
619b472c-839f-4fd4-9e7a-c581134f37b6 | 2 | 0 | 0 | 2 | 1 | 2 | 2 | 3 | The pressure force in a hydrostatic fluid is a vector given by $F_{pressure}=-\nabla p$.
| In cartesian co-ordinates, what is the $x$-component of the pressure?
\nWhat would we expect the x-component of the pressure to equal? Why?
| 2 | 0.333333 | 0 | The vector describing the pressure force is $ \nabla p=\frac{\delta p}{\delta x}+\frac{\delta p}{\delta y}+\frac{\delta p}{\delta z} $. What is the x-component of this vector?
***
The x-component is $\frac{\delta p}{\delta x}$. Think about the forces acting on the fluid and which direction they act in. How will this relate to this pressure component.
\n | The vector describing the pressure force is $ -\nabla p=-(\frac{\delta p}{\delta x}+\frac{\delta p}{\delta y}+\frac{\delta p}{\delta z}) $. The x-component of this vector is $-\frac{\delta p}{\delta x}$
\n
In a hydrostatic fluid the only forces acting are due to pressure and gravity. The convention for gravity is that it acts in the z direction, and has no x-component.
 So the only force acting in the x-direction is $\frac{\delta p}{\delta x}$ and as the fluid is in equilibrium the sum of forces acting in this direction must be zero so that $\frac{\delta p}{\delta x}=0$
| The pressure force in a hydrostatic fluid is a vector given by $F_{pressure}=-\nabla p$.
In cartesian co-ordinates, what is the $x$-component of the pressure?
\nWhat would we expect the x-component of the pressure to equal? Why?
| 37 | 2 | 4 | 4 | 78 | 2 | 23 | 1 | 0 | The pressure force in a hydrostatic fluid is a vector given by $F_{pressure}=- abla p$. In cartesian co-ordinates, what is the $x$-component of the pressure? What would we expect the x-component of the pressure to equal? Why? | 4 |
627119a8-7b2f-406c-a366-45b4cc230be6 | 1 | 1 | 1 | 17 | 6 | 1 | 6 | 1 | The figure below shows two equal masses attached to either end of a light stiff rod, which is mounted in the middle on a vertical axle.
The rod and masses rotate around the axle with fixed angular velocity $\omega$ at a fixed angle $\theta$. The upper half of the rod joining the two masses traces out the curved surface of a cone with its point at the bottom, while the lower half traces out the curved surface of a cone with its point at the top. The bearing is at the centre of mass. Ignore the effects of gravity.
| Does the axle exert a net force on the rotating rod? 
\nWhen the rod rotates fast enough, the bearing breaks. Why?
\nFind vector expressions for the angular momentum of each mass and add the two vectors to get the total angular momentum $\vec{\boldsymbol{L}}_{\text{tot}}$ of the contraption. Show that the torque $\vec{\boldsymbol{G}}$ exerted on the rod and masses by the axle is given by
$$
\vec{\boldsymbol{G}} = -m\omega^2 R^2 \sin(2\theta) \, \boldsymbol{\hat{\phi}} .
$$
It is easiest to work in a coordinate system with the origin at the bearing and the $z$ axis along the axle. The three unit vectors are:
* $\boldsymbol{\hat{k}}$: which points along the $z$ (rotation) axis;
* $\boldsymbol{\hat{r}}$: which lies in the $xy$ plane and points from the rotation axis to the upper mass;
* $\boldsymbol{\hat{\phi}}$: which always points in the direction in which the upper mass is moving.
Note that $\boldsymbol{\hat{r}}$ and $\boldsymbol{\hat{\phi}}$ rotate as the mass rotates.
| 3 | 0.666667 | 3 | Remember that we are ignoring gravity in this question.
***
The rod provides a force to keep the masses in orbit.
***
Consider N3. 
\nWhat happens to the required torque as $\omega$ increases?
***
Consider that the bearing has a limit to the torques it can exert.
\nBefore starting this question, use the right-hand rule to find the cross products:
* $ \boldsymbol{\hat{k}} \times \boldsymbol{\hat{r}} $
* $ \boldsymbol{\hat{r}} \times \boldsymbol{\hat{\phi}} $
* $ \boldsymbol{\hat{\phi}} \times \boldsymbol{\hat{k}} $
These results will be useful for later on in the question.
***
Also, what is the direction of the $\boldsymbol{\vec{\omega}}$ vector? ...
***
... This is defined as the direction of the rotation axis. 
***
The angular momentum is $\vec{\boldsymbol{L}}_{\pm} = \vec{\boldsymbol{r}}_{\pm}\times m \vec{\boldsymbol{v}}_{\pm}$, where $\pm$ denotes the two components of the total angular momentum from each of the masses.
***
For each angular momentum components, what is $\vec{\boldsymbol{r}}$? ...
***
... Express this in terms of $\boldsymbol{\hat{r}}$ and $\boldsymbol{\hat{k}}$. 
***
Next, what is $\boldsymbol{\vec{v}}$ for each of the masses? ...
***
... You can find $\boldsymbol{\vec{v}}$ in two ways:
1. Recognise that the motion of the masses is circular and use the formula for circular motion. In what direction vector is circular motion?
2. Recognise that the radius of motion is constant, and so the equation $\boldsymbol{\vec{v}}=\boldsymbol{\vec{\omega}}\times\boldsymbol{\vec{r}}$ applies. 
***
Hence find $\boldsymbol{\vec{L}}$ for each mass and sum the results...
***
... Use the unit vector cross products you found at the start. 
***
The *torque* is the rate of change of angular momentum. 
***
In your angular momentum formula, what value(s) (including vectors) are changing with respect to time? What is the rate of change of these values (refer to lectures if unsure)? Hence find $\boldsymbol{\vec{G}}$.
| The axle supplies the centripetal forces required to keep the two masses orbiting in circles, but these are always equal and opposite (N3). 
***
The nett force is zero.
\nThe axle generates the centripetal forces required to keep the masses in orbit by applying a torque to the rod via the bearing. 
***
Since the bearing is small (invisible on the diagram in the question!) and the magnitude of the torque is force $\times$ distance, this requires the presence of large forces where the axle passes through the bearing. If the rod is rotating fast enough, these forces will break the bearing or axle. 
***
Another way to see this is to realise that the rotating rod would "like" to be horizontal. The axle has to exert forces on the bearing to prevent this from happening.
\nBy looking at the diagram and applying the right-hand rule, it is straightforward to check that $\boldsymbol{\hat{k}} \times \boldsymbol{\hat{r}} = \boldsymbol{\hat{\phi}}$, $\boldsymbol{\hat{r}}
\times \boldsymbol{\hat{\phi}} = \boldsymbol{\hat{k}}$, and $\boldsymbol{\hat{\phi}} \times \boldsymbol{\hat{k}} = \boldsymbol{\hat{r}}$. 
***
The angular velocity vector is $\boldsymbol{\vec{\omega}} = \omega \boldsymbol{\hat{k}}$. 
***
The angular momentum of the masses is found by calculating $\vec{\boldsymbol{L}}_{\pm} = \vec{\boldsymbol{r}}_{\pm}\times m \vec{\boldsymbol{v}}_{\pm}$ where $\pm$ denotes the upper and lower components respectively. 
***
The position vectors $\vec{\boldsymbol{r}}_{\pm}$ of the upper and lower masses are:
***
$$
\vec{\boldsymbol{r}}_{\pm} = \pm R\cos\theta \,\boldsymbol{\hat{k}} \pm R\sin\theta \,\boldsymbol{\hat{r}}
$$
***
and their velocities are:
$$
\begin{aligned}
\hspace{10pt} \vec{\boldsymbol{v}}_{\pm} &= \omega\boldsymbol{\hat{k}} \times \vec{\boldsymbol{r}}_{\pm} = \omega \boldsymbol{\hat{k}} \times (\pm R\cos\theta\, \boldsymbol{\hat{k}} \pm R\sin\theta\, \boldsymbol{\hat{r}})
\end{aligned}
$$
***
$$
\begin{aligned}
&= \pm \omega R \sin\theta \, \boldsymbol{\hat{\phi}} .
\end{aligned}
$$
This can also be found by recognising that the motion is circular, with a radius $r= R\sin\theta$ of circular motion, and applying the formula $v=\omega r$. Their angular momentum vectors are then:
***
$$
\begin{aligned}
\hspace{13pt} \vec{\boldsymbol{L}}_{\pm} &= \vec{\boldsymbol{r}}_{\pm}\times m \vec{\boldsymbol{v}}_{\pm} \\
&= (\pm R\cos\theta \, \boldsymbol{\hat{k}} \pm R\sin\theta \, \boldsymbol{\hat{r}}) \times
\left ( \pm m \omega R \sin\theta \, \boldsymbol{\hat{\phi}} \right ) \\
\end{aligned}
$$
***
$$
\begin{aligned}
&= -m\omega R^2 \sin\theta\cos\theta \, \boldsymbol{\hat{r}} +
m\omega R^2 \sin^2\theta\, \boldsymbol{\hat{k}} .
\end{aligned}
$$
Notice that both masses have the same angular momentum. The total angular momentum is:
$$
\vec{\boldsymbol{L}}_{\text{tot}} = \vec{\boldsymbol{L}}_{+} + \vec{\boldsymbol{L}}_{-} = -2m\omega R^2
\sin\theta\cos\theta\, \boldsymbol{\hat{r}} + 2m \omega R^2 \sin^2\theta\, \boldsymbol{\hat{k}}.
$$
***
The $\boldsymbol{\hat{k}}$ component of the total angular momentum is constant but the $\boldsymbol{\hat{r}}$ component varies with time because $\boldsymbol{\hat{r}}$ varies with time. Using torque = rate of change of angular momentum gives:
***
$$
\vec{\boldsymbol{G}} = -2m\omega R^2 \sin\theta \cos\theta \, \frac{d\boldsymbol{\hat{r}}}{dt}
= -2m\omega^2 R^2 \sin\theta \cos\theta \, \boldsymbol{\hat{\phi}}
= -m\omega^2 R^2 \sin(2\theta) \boldsymbol{\hat{\phi}} ,
$$
where:
$$
\frac{d\boldsymbol{\hat{r}}}{dt} = \omega \boldsymbol{\hat{k}} \times \boldsymbol{\hat{r}} = \omega \boldsymbol{\hat{\phi}}.
$$
| The figure below shows two equal masses attached to either end of a light stiff rod, which is mounted in the middle on a vertical axle.
The rod and masses rotate around the axle with fixed angular velocity $\omega$ at a fixed angle $\theta$. The upper half of the rod joining the two masses traces out the curved surface of a cone with its point at the bottom, while the lower half traces out the curved surface of a cone with its point at the top. The bearing is at the centre of mass. Ignore the effects of gravity.
Does the axle exert a net force on the rotating rod? 
\nWhen the rod rotates fast enough, the bearing breaks. Why?
\nFind vector expressions for the angular momentum of each mass and add the two vectors to get the total angular momentum $\vec{\boldsymbol{L}}_{\text{tot}}$ of the contraption. Show that the torque $\vec{\boldsymbol{G}}$ exerted on the rod and masses by the axle is given by
$$
\vec{\boldsymbol{G}} = -m\omega^2 R^2 \sin(2\theta) \, \boldsymbol{\hat{\phi}} .
$$
It is easiest to work in a coordinate system with the origin at the bearing and the $z$ axis along the axle. The three unit vectors are:
* $\boldsymbol{\hat{k}}$: which points along the $z$ (rotation) axis;
* $\boldsymbol{\hat{r}}$: which lies in the $xy$ plane and points from the rotation axis to the upper mass;
* $\boldsymbol{\hat{\phi}}$: which always points in the direction in which the upper mass is moving.
Note that $\boldsymbol{\hat{r}}$ and $\boldsymbol{\hat{\phi}}$ rotate as the mass rotates.
| 247 | 13 | 21 | 21 | 297 | 15 | 146 | 11 | 1 | Ignore the effects of gravity. Does the axle exert a net force on the rotating rod? When the rod rotates fast enough, the bearing breaks. Why? Find vector expressions for the angular momentum of each mass and add the two vectors to get the total angular momentum $\vec{\boldsymbol{L}}_{\text{tot}}$ of the contraption. Show that the torque $\vec{\boldsymbol{G}}$ exerted on the rod and masses by the axle is given by $ \vec{\boldsymbol{G}} = -m\omega^2 R^2 \sin(2\theta) \, \boldsymbol{\hat{\phi}} . Note that $\boldsymbol{\hat{r}}$ and $\theta$0 rotate as the mass rotates. | 7 |
6281e7a1-0b80-40fc-9b0f-718dc0d5bbb5 | 0 | 3 | 0 | 21 | 6 | 1 | 0 | 1 | The following functions are orthogonal on the interval (**True** / **False**):
|  $\sin(\pi x/L)$ and $\cos(2\pi x/L)$ over $-L\leq x\leq L$
\n$\sin(\pi x/L)$ and $\cos(2\pi x/L)$ over $0\leq x\leq 2L$
\n$\sin(\pi x/L)$ and $\cos(2\pi x/L)$ over $0\leq x\leq L$
| 3 | 0.666667 | 2 | If the two functions are orthogonal, the *inner product* of the functions must be 0 (see section **2.3** of the notes). This is like the dot product of two orthogonal vectors. 
**Note:** we are not testing for Kronecker-delta here because there is no $n,m$ dependence in the functions.
***
After setting-up the inner product integral (start with the first interval $-L\le x \le L$), there are two methods to determine if the inner product evaluates to 0:
***
(**Method 1**): Evaluate the integral directly by using the formulae:
$$
\begin{aligned}
\cos{\theta} = \frac{e^{i\theta}+e^{-i\theta}}{2}\\
\sin{\theta} = \frac{e^{i\theta}-e^{-i\theta}}{2}
\end{aligned}
$$
***
(**Method 2**): Determine whether the $\sin$ and $\cos$ functions are even or odd in the interval.
It may help you to set $L=\pi$ (or some other number) for the purpose of sketching the function...
***
(**Method 2**): ... Then, use the properties of even/odd functions to evaluate the integral. 
\nUse the hints from part (a).
\nUse the hints from part (a).
| First, performing the substitution $k=\pi x/L$ for simplification:
$$
\begin{aligned}
&= \frac{L}{\pi}\int_{-\pi}^{\pi}{\sin(k) \cos(2k) dk} \\
\end{aligned}
$$
***
Then, applying the formulae for $\sin$ and $\cos$ in imaginary terms:
$$
\begin{aligned}
\sin{\theta} = \frac{e^{i\theta}-e^{-i\theta}}{2}\\
\cos{\theta} = \frac{e^{i\theta}+e^{-i\theta}}{2}\\
\end{aligned}
$$
***
$$
\begin{aligned}
&= \frac{L}{\pi}\int_{-\pi}^{\pi}{
\left(
\frac{e^{ik}-e^{-ik}}{2i}\frac{e^{2ik}+e^{-2ik}}{2}
\right) dk} \\
&= \frac{L}{\pi}\frac{1}{4i}\int_{-\pi}^{\pi}{((e^{3ik} - e^{-3ik}) - (e^{ik} - e^{-ik}) )dk} \\
&= 0 \; .
\end{aligned}
$$
The last equality follows as the complex exponentials are integrated over 3 and 1 full period, respectively, and thus average out to zero, meaning that the functions are orthogonal. You can check this by performing the integral explicitly.
So the answer is **True**.
, We need to evaluate the inner product and show if it is zero or not. Hence we evaluate the integral,
***
$$
\begin{aligned}
\braket{\sin(\pi x/L) | \cos(2\pi x/L)}
& = \int_{-L}^{L}{\sin(\pi x/L) \cos(2\pi x/L) dx} \\
\end{aligned}
$$
This integral can be evaluated by two methods:
, Plotting the two functions in the interval $-L<x<L$, we have:
***
***
About the line $x=0$, $\sin(\pi x/L)$ is odd and $\cos(2\pi x/L)$ is even. 
***
The product of the functions is therefore odd.
***
Therefore, the integral over the interval is 0; and hence the functions are orthogonal. 
\n
, Changing the sketch from part (a) to the interval $[0,2L]$, we now consider whether the functions are even/odd about $x=L$ (the half-way point):
***
***
***
$\sin(\pi x/L)$ is odd over the interval, whereas $\cos(2\pi x/L)$ is even.
***
The product of the functions is therefore odd.
***
The integral therefore evaluates to 0, so the functions are orthogonal in the interval.
, An integral of trigonometric functions over an integral number of periods doesn't depend on where the integral stops and starts, so the integral still vanishes as in part (a). Answer is true.
\nThis is the same integral as in the first part but now with limits that are not across a full period; so we cannot immediately conclude that the integral evaluates to 0. 
***
Each term in the integral is not zero, but we need to evaluate it to make sure that there is not an accidental cancellation. The integral to evaluate is:
***
$$
\begin{aligned}
I=\frac{L}{\pi}\frac{1}{4i}\int_{0}^{\pi}{((e^{3ik} - e^{-3ik}) - (e^{ik} - e^{-ik}) )dk}
= -\frac{2}{3}\frac{L}{\pi} \; ,
\end{aligned}
$$
***
Therefore, the functions are **not** orthogonal in this interval. 
, Over the interval $0<x<L$, we measure whether the functions are odd about the line $x=L/2$:
***
***
Both $\sin(\pi x/L)$ and $\cos(2\pi x/L)$ are even about $x=L/2$.
***
The product of the functions is therefore even. 
***
The integral does not evaluate to 0, so the functions are not orthogonal. 
,
| The following functions are orthogonal on the interval (**True** / **False**):
 $\sin(\pi x/L)$ and $\cos(2\pi x/L)$ over $-L\leq x\leq L$
\n$\sin(\pi x/L)$ and $\cos(2\pi x/L)$ over $0\leq x\leq 2L$
\n$\sin(\pi x/L)$ and $\cos(2\pi x/L)$ over $0\leq x\leq L$
| 29 | 9 | 11 | 11 | 388 | 6 | 18 | 9 | 0 | The following functions are orthogonal on the interval True / False: $\sin(\pi x/L)$ and $\cos(2\pi x/L)$ over $-L\leq x\leq L$ $\sin(\pi x/L)$ and $\cos(2\pi x/L)$ over $0\leq x\leq 2L$ $\sin(\pi x/L)$ and $\cos(2\pi x/L)$ over $0\leq x\leq L$ | 1 |
62a5b748-6653-44bb-9eea-3eac5a486698 | 0 | 1 | 1 | 11 | 4 | 2 | 4 | 0 | Refrigerators and heat pumps are both examples of reversed heat engines.
| What is the difference between a refrigerator and a heat pump? 
\nIf a refrigerator and a heat pump operate on the same cycle, which has the higher $COP$?
| 2 | 0.333333 | 0 | \n | \nFor a refrigerator:
  
$COP_\mathrm{R} = \frac{\dot{Q}_\mathrm{in}}{|\dot{W}_\mathrm{net}|}$
  
where $\dot{Q}_\mathrm{in}$ is the heat transferred from the cold space into the refrigerant.
***
For a heat pump:
  
$COP_\mathrm{HP} = \frac{|\dot{Q}_\mathrm{out}|}{|\dot{W}_\mathrm{net}|}$
  
where $|\dot{Q}_\mathrm{out}|$ is the heat transferred out of the refrigerant to the hot space.
***
Using the first law equation:
***
$|\dot{W}_\mathrm{net}| +\dot{Q}_\mathrm{in} = |\dot{Q}_\mathrm{out}|$
***
Dividing through by $|\dot{W}_\mathrm{net}|$:
  
$1+COP_\mathrm{R} = COP_\mathrm{HP}$
***
Hence:
  
$COP_\mathrm{HP} > COP_\mathrm{R}$
| Refrigerators and heat pumps are both examples of reversed heat engines.
What is the difference between a refrigerator and a heat pump? 
\nIf a refrigerator and a heat pump operate on the same cycle, which has the higher $COP$?
| 40 | 1 | 8 | 8 | 61 | 0 | 29 | 1 | 0 | What is the difference between a refrigerator and a heat pump? If a refrigerator and a heat pump operate on the same cycle, which has the higher $COP$? | 2 |
637204d2-f4c7-494e-844c-6b169616db05 | 3 | 0 | 0 | 24 | 6 | 1 | 0 | 11 | **\[Boas 5.4.1 (a),(b)]** For the disk $\rho\leq a$, find by integration using polar coordinates:
| The area of the disk.
\nThe centroid of one quadrant \[take the positive quadrant] of the disk.
***
The centroid is the Cartesian point $(X,Y)$ such that $X = \frac{1}{A} \iint_R xdxdy$ and $Y = \frac{1}{A} \iint_R ydxdy$ where $R$ is the region of integration and $A$ is the area of that region. 
| 2 | 0.666667 | 2 | $$
A = \iint_R{dA}
$$
Can you express $dA$ in polar coordinates? ...
***
... This should be of the form $|J|d\rho d\phi$, where $|J|$ is the *Jacobian* in polar coordinates.
***
Sketch the disk. What are the limits of $\rho$ and $\phi$?
***
Hence perform the double integral, keeping the variable you are not integrating with respect to constant in the inner integral. 
***
Does the area you found match the expected area of a circle?
\nChange the integrals $X$ and/or $Y$ to polar coordinates (you may wish to evaluate only one integral and use symmetry for the other)...
***
... Remember that:
$$
dx\,dy = |J|d\rho\,d\phi
$$
and remember to convert the integrands into polar coordinates.
***
You are considering the disc in a single quadrant. How does this change the limits of the integrals in comparison to part (a)?
***
Also remember that the arae $A$ is in the positive quadrant only.
| Sketch of the disk:
***
The area of the disk is the sum of the area elements:
$$
A = \iint_{R}{dA}
$$
This integral could be performed in Cartesian coordinates, but it is much easier to exploit the circular symmetry of the problem. In polar coordinates, the area element is given by:
***
$$
dA = |J|d\rho \, d\phi =\rho\,d\rho\, d\phi
$$
The limits of the disk in polar coordinates are:
***
$$
\begin{aligned}
&\rho: 0\to a \\
&\phi: 0\to 2\pi
\end{aligned}
$$
The integral is therefore:
***
$$
A = \iint_{R}dA = \int_{\phi=0}^{2\pi}\int_{\rho=0}^{a}\rho\,d\rho\, d\phi
$$
Since the inner integral has no dependence on $\phi$ (in both the limits and the integrand), the integrals can be evaluated concurrently: 
***
$$
A = 2\pi \left[\frac{\rho^2}{2}\right]_{\rho=0}^{a} = \pi a^2
$$
This is the area of a circle.
\nFirst, we change $X$ and $Y$ into polar coordinates. Recall the Jacobian:
***
$$
dx \,dy = \rho \,d\rho\,d\phi
$$
and the conversion of $x$ and $y$ into polar coordinates:
***
$$
x= \rho \cos\phi\quad \quad y=\rho \sin\phi
$$
The limits of $\phi$ in each integral are reduced to the positive quadrant, as well as the area:
***
$$
\begin{aligned}
\phi: 0 \to\, &\pi/2 \qquad \rho: 0\to a\\
&A = \pi a^2 /4
\end{aligned}
$$
Starting with $X$:
***
$$
\begin{aligned}
X &= \frac{4}{\pi a^2}\int_{\phi=0}^{\pi/2}\int_{\rho=0}^{a}{\rho \cos\phi\,\rho\,d\rho\,d\phi}\\
&=\frac{4}{\pi a^2}\int_{\phi=0}^{\pi/2}\cos\phi\,d\phi \int_{\rho=0}^{a}\rho^2 \,d\rho
\end{aligned}
$$
Since none of the limits are a function of a variable, it is possible to separate out the integrals as shown above (You could also evaluate the integrals one after the other rather than following this method). 
***
$$
\begin{aligned}
&X = \frac{4}{\pi a^2}[\sin{\phi}]_0^{\pi/2}\left[\frac{\rho^3}{3}\right]_0^a\\
&=\frac{4a}{3\pi}
\end{aligned}
$$
***
Similarly, by symmetry,
$$
Y=\frac{4a}{3\pi}
$$
| **\[Boas 5.4.1 (a),(b)]** For the disk $\rho\leq a$, find by integration using polar coordinates:
The area of the disk.
\nThe centroid of one quadrant \[take the positive quadrant] of the disk.
***
The centroid is the Cartesian point $(X,Y)$ such that $X = \frac{1}{A} \iint_R xdxdy$ and $Y = \frac{1}{A} \iint_R ydxdy$ where $R$ is the region of integration and $A$ is the area of that region. 
| 59 | 6 | 18 | 18 | 195 | 10 | 45 | 5 | 0 | Boas 5.4.1 a,b For the disk $\rho\leq a$, find by integration using polar coordinates: The area of the disk. | 1 |
638eb85e-0b37-46ad-b4c7-021442fe87f1 | 2 | 0 | 0 | 14 | 4 | 2 | 5 | 4 | Newton had already formulated that the propagation of sound in air followed from the oscillatory motion of the fluid particles (which he viewed as ``the law of the oscillating pendulum''). While he did not state it explicitly, it appears that he considered such a process to be isothermal. | Rederive the speed of sound for an ideal gas, but assume the successive compression/expansion of the fluid particle to be isothermal. Give your answer in terms of the gas temperature. 
\nWhat is the relative error between Newton's sound speed for ideal gases and the one obtained in class? Is it a function of the temperature? Give a numerical value in the case of air.
  
Note: error is defined as 
$$
\mathrm{error}=\frac{\text{approximate value }- \text{ actual value}}{\text{actual value}}
$$
| 2 | 0.5 | 2 | \n | \n | Newton had already formulated that the propagation of sound in air followed from the oscillatory motion of the fluid particles (which he viewed as ``the law of the oscillating pendulum''). While he did not state it explicitly, it appears that he considered such a process to be isothermal.Rederive the speed of sound for an ideal gas, but assume the successive compression/expansion of the fluid particle to be isothermal. Give your answer in terms of the gas temperature. 
\nWhat is the relative error between Newton's sound speed for ideal gases and the one obtained in class? Is it a function of the temperature? Give a numerical value in the case of air.
  
Note: error is defined as 
$$
\mathrm{error}=\frac{\text{approximate value }- \text{ actual value}}{\text{actual value}}
$$
| 118 | 1 | 0 | 0 | 1 | 0 | 71 | 1 | 0 | Newton had already formulated that the propagation of sound in air followed from the oscillatory motion of the fluid particles which he viewed as ``the law of the oscillating pendulum''. While he did not state it explicitly, it appears that he considered such a process to be isothermal.Rederive the speed of sound for an ideal gas, but assume the successive compression/expansion of the fluid particle to be isothermal. Give your answer in terms of the gas temperature. What is the relative error between Newton's sound speed for ideal gases and the one obtained in class? Is it a function of the temperature? Give a numerical value in the case of air. | 6 |
6396919f-7055-4a64-9fea-9aafa18f0068 | 4 | 0 | 0 | 0 | 0 | 2 | 4 | 0 | Determine the following quantities downstream of the continuous $10^{\circ}$ convex corner shown below, with the flow upstream of the corner being at Mach 2: 
| The Mach number, $M_2$.
\nThe pressure, $p_2$, as a function of the upstream pressure $p_1$.
\nFind the angles $\mu_{1}$ and $\mu_{2}$ that the forward and rearward Mach lines form with respect to their local flow.
| 3 | 0.666667 | 3 | Use tables to determine $\nu_{1}$, knowing $M_{1}$.
***
Find $\nu_{2}$, carefully considering whether angles are clockwise or anticlockwise, and whether waves are left-running or right-running.
***
Use tables to find $M_{2}$. 
\nKnowing the flow is isentropic, express $\frac{p_{2}}{p_{1}}$ as a function of knowable pressure ratios
***
Use tables to find any unknown pressure ratios. 
\nUse the angle-Mach number relation
| $M_{1}=2$ and $\theta_{1}=0^{\circ}$. 
   
From 1D compressible flow tables, for $M_{1}=2$, $\nu_{1}=26.38^{\circ}$. 
 
***
$\theta_{2}$ is clockwise, so it is negative: $\theta_{2}=-10^{\circ}$
   
The waves are are L-R, so we take the positive sign in the expansion expression:
   
$\theta_{1}+\nu_{1}=\theta_{2} + \nu_{2}$
   
From which we can solve for $\nu_2$:
 
 $\nu_{2} = \nu_{1} - \theta_{2}=26.38+10=36.38^{\circ}$
***
   
From the 1D tables again, for $\nu_{2}=36.38^{\circ}$, $M_{2}=2.385$
\nWe can get the pressure $p_2$ by using the following product of pressure ratios:
  
$\frac{p_{2}}{p_{1}}=\frac{p_{2}}{p_{02}} \cdot \frac{p_{02}}{p_{01}} \cdot \frac{p_{01}}{p_{1}}$
   
The flow is isentropic, so $p_{01} = p_{02}$
   
***
We first get the isentropic pressure ratio for the velocity before the expansion from the isentropic tables:  
That is, for $M_1=2.0$: 
  
$\frac{p_{01}}{p_1}=7.8244$
***
  
We can also get the isentropic pressure ratio for the velocity behind the expansion, which is $M_{2}=2.365$:
 
 $\frac{p_{02}}{p_{2}}=14.287$
***
   
Therefore $\frac{p_{2}}{p_{1}}=\frac{1}{14.35} \cdot 7.8244=0.548$
   
And $p_2=0.548 p_1$.
\nWe can obtain both Mach angles using the expression that relates them to their corresponding Mach number of the flow:
$\mu_{1}=sin^{-1}\big(\frac{1}{M_{1}}\big)=30^{\circ}$
  
$\mu_{2}=sin^{-1}\big(\frac{1}{M_{2}}\big)=24.8^{\circ}$
| Determine the following quantities downstream of the continuous $10^{\circ}$ convex corner shown below, with the flow upstream of the corner being at Mach 2: 
The Mach number, $M_2$.
\nThe pressure, $p_2$, as a function of the upstream pressure $p_1$.
\nFind the angles $\mu_{1}$ and $\mu_{2}$ that the forward and rearward Mach lines form with respect to their local flow.
| 63 | 6 | 22 | 22 | 182 | 5 | 38 | 5 | 1 | Determine the following quantities downstream of the continuous $10^{\circ}$ convex corner shown below, with the flow upstream of the corner being at Mach 2: The Mach number, $M_2$. Find the angles $\mu_{1}$ and $\mu_{2}$ that the forward and rearward Mach lines form with respect to their local flow. | 2 |
63b21872-5801-47d8-8f9c-6445c6d37812 | 0 | 0 | 1 | 17 | 6 | 1 | 2 | 3 | In lectures, you saw how to use Newton’s second law for two particles,
$$
\begin{aligned}
\frac{d\boldsymbol{\vec{p}}_1}{dt} &= \boldsymbol{\vec{f}}_1^{\text{ ext}} + \boldsymbol{\vec{f}}_{2\text{ on }1}
,
&
\frac{d\boldsymbol{\vec{p}}_2}{dt} &= \boldsymbol{\vec{f}}_2^{\text{ ext}} + \boldsymbol{\vec{f}}_{1\text{ on }2}
,
\end{aligned}
$$
to show that the rate of change of the total momentum, $\boldsymbol{\vec{P}} = \boldsymbol{\vec{p}}_1 +
\boldsymbol{\vec{p}}_2$, is equal to the total external force, $\boldsymbol{\vec{F}}^{\text{ ext}} =
\boldsymbol{\vec{f}}_1^{\text{ ext}} + \boldsymbol{\vec{f}}_2^{\text{ ext}}$. The internal forces cancel out. Generalise this derivation to the case of $N$ particles.
| In lectures, you saw how to use Newton’s second law for two particles,
$$
\begin{aligned}
\frac{d\boldsymbol{\vec{p}}_1}{dt} &= \boldsymbol{\vec{f}}_1^{\text{ ext}} + \boldsymbol{\vec{f}}_{2\text{ on }1}
,
&
\frac{d\boldsymbol{\vec{p}}_2}{dt} &= \boldsymbol{\vec{f}}_2^{\text{ ext}} + \boldsymbol{\vec{f}}_{1\text{ on }2}
,
\end{aligned}
$$
to show that the rate of change of the total momentum, $\boldsymbol{\vec{P}} = \boldsymbol{\vec{p}}_1 +
\boldsymbol{\vec{p}}_2$, is equal to the total external force, $\boldsymbol{\vec{F}}^{\text{ ext}} =
\boldsymbol{\vec{f}}_1^{\text{ ext}} + \boldsymbol{\vec{f}}_2^{\text{ ext}}$. The internal forces cancel out. Generalise this derivation to the case of $N$ particles.
| 1 | 0.666667 | 1 | Each particle $i$ feels an external force and an internal force from (in principle) *every other particle*. Express $d\boldsymbol{\vec{p}_i}/dt$...
***
... The internal forces should be expressed as a summation over every particle $j \ne i$.
***
Sum up the forces to find the total force acting on the system.
***
Why do the internal forces cancel out to leave only the external forces?
| Label the particles $i = 1, 2, \ldots, N$. N2 applies to all of them:
***
$$
\begin{aligned}
\frac{d\boldsymbol{\vec{p}}_1}{dt}
&= \boldsymbol{\vec{f}}_1^{\text{ ext}} +
\sum_{j\,(\neq 1)}\boldsymbol{\vec{f}}_{j\text{ on }1},
& \text{(particle $1$)}\\
& \ldots & \\
& \ldots & \\
\frac{d\boldsymbol{\vec{p}}_i}{dt}
&= \boldsymbol{\vec{f}}_i^{\text{ ext}} +
\sum_{j\,(\neq i)} \boldsymbol{\vec{f}}_{j\text{ on }i},
& \text{(particle $i$)}\\
& \ldots \\
& \ldots \\
\frac{d\boldsymbol{\vec{p}}_N}{dt}
&= \boldsymbol{\vec{f}}_N^{\text{ ext}} +
\sum_{j\,(\neq N)}\boldsymbol{\vec{f}}_{j\text{\,on }N}.
& \text{(particle $N$)}
\end{aligned}
$$
***
Adding these equations gives:
***
$$
\sum_{i} \frac{d\boldsymbol{\vec{p}}_i}{dt}
= \sum_{i} \boldsymbol{\vec{f}}_i^{\text{ ext}} + \sum_{i} \sum_{j\,(\neq i)}
\boldsymbol{\vec{f}}_{j\text{\,on\,}i} .
$$
***
The double sum includes two terms for every pair of particles. For example, for particles $7$ and $234$, there is one term with $i = 7$ and $j = 234$ and another term with $j = 7$ and $i = 234$. The sum of these two terms is
$$
\boldsymbol{\vec{f}}_{234\text{\,on\,}7} + \boldsymbol{\vec{f}}_{7\text{\,on\,}234} = \vec{0}
$$
by N. 
***
The double summation gives zero and we get
$$
\frac{d\boldsymbol{\vec{p}}}{dt} = \boldsymbol{\vec{F}}^{\text{ ext}} ,
$$
where $\boldsymbol{\vec{p}} \equiv \sum_i \boldsymbol{\vec{p}}_i$ and $\boldsymbol{\vec{F}}^{\text{ ext}} \equiv \sum_i
\boldsymbol{\vec{f}}^{\text{ ext}}_i$.
***
When you push on an object (a shopping trolley, say), you are exerting forces on enormous numbers of surface atoms, all of which in turn exert forces on the atoms within the trolley, but the rate of change of the momentum of the trolley as a whole is exactly as if it were a point particle feeling the total applied force. This mathematical miracle explains why N2 is just as useful in our macroscopic world as it is in the world of point particles.
| In lectures, you saw how to use Newton’s second law for two particles,
$$
\begin{aligned}
\frac{d\boldsymbol{\vec{p}}_1}{dt} &= \boldsymbol{\vec{f}}_1^{\text{ ext}} + \boldsymbol{\vec{f}}_{2\text{ on }1}
,
&
\frac{d\boldsymbol{\vec{p}}_2}{dt} &= \boldsymbol{\vec{f}}_2^{\text{ ext}} + \boldsymbol{\vec{f}}_{1\text{ on }2}
,
\end{aligned}
$$
to show that the rate of change of the total momentum, $\boldsymbol{\vec{P}} = \boldsymbol{\vec{p}}_1 +
\boldsymbol{\vec{p}}_2$, is equal to the total external force, $\boldsymbol{\vec{F}}^{\text{ ext}} =
\boldsymbol{\vec{f}}_1^{\text{ ext}} + \boldsymbol{\vec{f}}_2^{\text{ ext}}$. The internal forces cancel out. Generalise this derivation to the case of $N$ particles.
| 50 | 4 | 13 | 13 | 166 | 3 | 50 | 4 | 0 | In lectures, you saw how to use Newton’s second law for two particles, $ \begin{aligned} \frac{d\boldsymbol{\vec{p}}_1}{dt} &= \boldsymbol{\vec{f}}_1^{\text{ ext}} + \boldsymbol{\vec{f}}_{2\text{ on }1} , & \frac{d\boldsymbol{\vec{p}}_2}{dt} &= \boldsymbol{\vec{f}}_2^{\text{ ext}} + \boldsymbol{\vec{f}}_{1\text{ on }2} , \end{aligned} $ to show that the rate of change of the total momentum, $\boldsymbol{\vec{P}} = \boldsymbol{\vec{p}}_1 + \boldsymbol{\vec{p}}_2$, is equal to the total external force, $\boldsymbol{\vec{F}}^{\text{ ext}} = \boldsymbol{\vec{f}}_1^{\text{ ext}} + \boldsymbol{\vec{f}}_2^{\text{ ext}}$. | 1 |
63d7bf5e-3b9e-4583-bf45-a2c0665a35ab | 2 | 5 | 2 | 2 | 1 | 2 | 7 | 13 | In this question you will derive the flow profile of an incompressible Newtonian fluid that is flowing with volume flux $Q$ in a channel of width $h$ created by the gap between two plates, each of length $L$ and width $W$, at $y=0$ and $y=h$.
  
We will make the following assumptions:
* Two-dimensional flow (no dependence on $z$ and no flow in the third direction).
* Steady flow
* Fully developed flow (the velocity does not depend on $x$).
* Gravity can be neglected.
Secondly, we will do the calculation in terms of the velocity components and the pressure.
| What do each of the assumptions imply about the velocity components $u(x,y,z,t)$, $v(x,y,z,t)$ and $w(x,y,z,t)$ and the pressure $p(x,y,z,t)$?
\nShow that the continuity equation implies that $v$ is constant.
\nWrite down the boundary conditions at the top and bottom of the channel. What does this tell you about $v$?
\nUse the Navier--Stokes equations to derive equations for $\frac{\delta p}{\delta x}$ and $\frac{\delta p}{\delta y}$. Use these to show that $p$ only depends on $x$ and also that
$\frac{\delta p}{\delta x}=\mu\frac{\delta ^2u}{\delta y^2}.$
\n Solve the equation you found in Part 8.14d) and apply the boundary conditions on $u$ to find $u$ as a function of $\frac{\delta p}{\delta x}$, $\mu$, $y$ and $h$.
\nHence find $\frac{\delta p}{\delta x}$ in terms of $Q$. What do you notice about $\frac{\delta p}{\delta x}$, and what does that tell you about $p$? Eliminate $\frac{\delta p}{\delta x}$ from the expression for $u$.
| 6 | 1 | 4 | \nWrite out the expanded form of the continuity equation *i.e.* the one used throughout Question 8.2.
***
Since $u$ and $v$ don't depend on $x$ what does the equation $\frac{\delta u}{\delta x}+\frac{\delta v}{\delta y}=0$ simplify to? 
***
What does the solution of the simplified PDE suggest about $v$?
\nPreviously we saw that $v$ is constant in space and time. We also know that $v=0$ at the plates due to no slip boundary conditions. What does this suggest about the value of $v$ everywhere else?
\nNeglecting gravity and assuming two-dimensional flow, we have an equation where we can cancel terms knowing that $u$ only depends on $y$.
$\rho\cancel{(\frac{\delta u}{\delta t}+u\frac{\delta u}{\delta x}+v\frac{\delta u}{\delta y})}=-\frac{\delta p}{\delta x}+\mu(\cancel{\frac{\delta ^2u}{\delta x^2}}+\frac{\delta ^2u}{\delta y^2}),$ 
We can repeat this for the equation for $v$, knowing it doesn't depend on $y$ or $x$.
***
From this you should get $\frac{\delta p}{\delta x}=\mu\frac{\delta ^2u}{\delta y^2}$ and $\frac{\delta p}{\delta y}=0$.
Now you can see that $p$ therefore only depends on $x$
\nSince $\frac{\delta p}{\delta x}$ does not depend on $y$, we have $\frac{\delta ^2u}{\delta y^2}=\frac1{\mu}\frac{\delta p}{\delta x}$ which we can treat as an ODE and integrate to get constants of integrations.
***
Apply boundary conditions to the equation found ($u=\frac1{2\mu}\frac{\delta p}{\delta x}y^2+Ay+B$) to find values for $A$ and $B$. You can substitute these back into the original equation and rearrange to get the final answer.
\nThe flux is given by
$Q=\int_{\sf Area}u\,dA$, where we know $u$ and the area of integration is $h$ in the $y$ and $W$ in the $z$ direction. Knowing this carry out the double integral for the flux, treating $\frac{\delta p}{\delta x}$ as a constant.
***
You should have gotten $Q=-\frac{h^3W}{12\mu}\frac{\delta p}{\delta x}$, which we can rearrange to make $\frac{\delta p}{\delta x}$ the subject. We know this expression will be constant so $p$ would decrease linearly across the channel. We can substitute the rearranged form back into $u$ to get our final expression for $u$.
| We have
* The assumption of two-dimensional flow means that $u$, $v$ and $p$ do not depend on $z$ and that $w=0$. Thus we have $u(x,y,t)$, $v(x,y,t)$, $p(x,y,t)$.
* The assumption of steady flow means that none of the variables depend on $t$. Thus $u(x,y)$, $v(x,y)$, $p(x,y)$,
* The assumption of fully developed flow means that the velocity does not depend on $x$. Thus $u(y)$, $v(y)$, but it doesn't tell us anything about $p$.
* Gravity can be neglected tells us that we can leave the term $\rho{\bf g}$ out of the Navier--Stokes equation.
Hence we have that $u$ and $v$ are functions of $y$ only, while $w=0$ and $p$ is a function of $x$ and $y$.
\nSince $u$ and $v$ do not depend on $x$, we have
$\frac{\delta u}{\delta x}+\frac{\delta v}{\delta y}=\frac{\delta v}{\delta y}=0.$
Hence $v$ must be constant, since it does not depend on any variables.
\n
\nNeglecting gravity and assuming two-dimensional flow, we have
$\rho(\frac{\delta u}{\delta t}+u\frac{\delta u}{\delta x}+v\frac{\delta u}{\delta y})=-\frac{\delta p}{\delta x}+\mu(\frac{\delta ^2u}{\delta x^2}+\frac{^\delta 2u}{\delta y^2}),$
$\rho(\frac{\delta v}{\delta t}+u\frac{\delta v}{\delta x}+v\frac{\delta v}{\delta y})=-\frac{\delta p}{\delta y}+\mu(\frac{\delta ^2v}{\delta x^2}+\frac{\delta ^2v}{\delta y^2}).$
Since $v=0$ and $u$ only depends on $y$, these become
$\rho(0+0+0)=-\frac{\delta p}{\delta x}+\mu(0+\frac{\delta ^2u}{\delta y^2})\quad\Rightarrow\quad\frac{\delta p}{\delta x}=\mu\frac{\delta ^2u}{\delta y^2},$
$\rho(0+0+0)=-\frac{\delta p}{\delta y}+\mu(0+0)\quad\Rightarrow\quad\frac{\delta p}{\delta y}=0.$
Hence $ p $ only depends on $x$.
\nSince $\frac{\delta p}{\delta x}$ does not depend on $y$, we have
$\frac{\delta ^2u}{\delta y^2}=\frac1{\mu}\frac{\delta p}{\delta x} \quad\Rightarrow\quad u=\frac1{2\mu}\frac{\delta p}{\delta x}y^2+Ay+B,$
where $A$ and $B$ are constants. Imposing the boundary condition $u=0$ at $y=0$ and at $y=h$ implies
$B=0$
$\frac1{2\mu}\frac{\delta p}{\delta x}h^2+Ah+B=0\quad\Rightarrow\quad A=-\frac{h}{2\mu}\frac{\delta p}{\delta x}.$
Hence
$u=\frac1{2\mu}\frac{\delta p}{\delta x}(y^2-hy)$
$=-\frac1{2\mu}\frac{\delta p}{\delta x}y(h-y).$
\nThe flux is given by
$Q=\int_{\sf Area}u\,dA$
$=\int_0^W\int_0^h-\frac{1}{2\mu}\frac{\delta p}{\delta x}y(h-y)\,dy\,dz$
$=-\frac1{2\mu}\frac{\delta p}{\delta x}W\frac{h^3}{6}$
$=-\frac{h^3W}{12\mu}\frac{\delta p}{\delta x}$
and hence
$\frac{\delta p}{\delta x}=-\frac{12\mu Q}{h^3W}.$
This is constant, and therefore $p$ decreases linearly along the channel.
Thus the velocity is
$u=\frac{6Q}{h^3W}y(h-y),\quad v=0.$
| In this question you will derive the flow profile of an incompressible Newtonian fluid that is flowing with volume flux $Q$ in a channel of width $h$ created by the gap between two plates, each of length $L$ and width $W$, at $y=0$ and $y=h$.
  
We will make the following assumptions:
* Two-dimensional flow (no dependence on $z$ and no flow in the third direction).
* Steady flow
* Fully developed flow (the velocity does not depend on $x$).
* Gravity can be neglected.
Secondly, we will do the calculation in terms of the velocity components and the pressure.
What do each of the assumptions imply about the velocity components $u(x,y,z,t)$, $v(x,y,z,t)$ and $w(x,y,z,t)$ and the pressure $p(x,y,z,t)$?
\nShow that the continuity equation implies that $v$ is constant.
\nWrite down the boundary conditions at the top and bottom of the channel. What does this tell you about $v$?
\nUse the Navier--Stokes equations to derive equations for $\frac{\delta p}{\delta x}$ and $\frac{\delta p}{\delta y}$. Use these to show that $p$ only depends on $x$ and also that
$\frac{\delta p}{\delta x}=\mu\frac{\delta ^2u}{\delta y^2}.$
\n Solve the equation you found in Part 8.14d) and apply the boundary conditions on $u$ to find $u$ as a function of $\frac{\delta p}{\delta x}$, $\mu$, $y$ and $h$.
\nHence find $\frac{\delta p}{\delta x}$ in terms of $Q$. What do you notice about $\frac{\delta p}{\delta x}$, and what does that tell you about $p$? Eliminate $\frac{\delta p}{\delta x}$ from the expression for $u$.
| 244 | 31 | 57 | 57 | 248 | 36 | 140 | 23 | 1 | In this question you will derive the flow profile of an incompressible Newtonian fluid that is flowing with volume flux $Q$ in a channel of width $h$ created by the gap between two plates, each of length $L$ and width $W$, at $y=0$ and $y=h$. We will make the following assumptions: Two-dimensional flow no dependence on $z$ and no flow in the third direction. What do each of the assumptions imply about the velocity components $u(x,y,z,t)$, $v(x,y,z,t)$ and $h$0 and the pressure $h$1? Show that the continuity equation implies that $h$2 is constant. Write down the boundary conditions at the top and bottom of the channel. What does this tell you about $h$2? Use the Navier--Stokes equations to derive equations for $h$4 and $h$5. Use these to show that $h$6 only depends on $x$ and also that $h$8 Solve the equation you found in Part 8.14d and apply the boundary conditions on $h$9 to find $h$9 as a function of $h$4, $L$2, $L$3 and $h$. Hence find $h$4 in terms of $Q$. What do you notice about $h$4, and what does that tell you about $h$6? Eliminate $h$4 from the expression for $h$9. | 11 |
644c4af4-fec2-4fcd-93d7-b775dd08d41b | 4 | 0 | 0 | 0 | 0 | 2 | 2 | 3 | The figure below shows a convergent-divergent nozzle. It is observed that the flow (air, $\gamma=1.4$) exits the nozzle as a perfectly expanded (isentropic) supersonic flow. 
   
  
| Determine the mass flow rate through the nozzle.
\nDetermine the Mach number at the nozzle exit ($M_{exit}$)
\nDetermine the pressure at the nozzle exit ($p_{exit}$).
\nThe pressure at the nozzle exit is gradually increased from its original value. For what range of values of $p_{exit}$ would the mass flow rate through the nozzle remain unchanged? 
| 4 | 1 | 3 | Mass flow rate is constant throughout the nozzle, so we can analyze it at any point. At what section is there enough information to do this? Hint: the only quantities remaining to be determined are $ \rho^{*} $ and $ v^{*} $. 
***
To determine $ \rho^{*} $, use the ideal gas law. To determine the any quantities not given in the ideal gas law expression, use tables. 
***
Rearrange the expression for Mach number (which is known) for $ v^{*} $. To find $ a^{*} $, use tables. $ v^{*} $, and then $ \dot{m} $, can now be calculated directly. 
\nWhat is the ratio of exit area to throat area?
***
How can the ratio determined at the previous step be used, with the help of tables, to find the exit Mach number? 
\nThe flow is perfectly expanded supersonic flow, therefore isentropic. Use this information, along with tables at the relevant Mach number. 
\nUse the area ratio to determine the exit Mach number. Will it be subsonic or supersonic? 
***
Use tables at the previously determined exit Mach number to determine the exit pressure. 
***
What do we know about mass flow rate in choked flow? For what Mach number at the throat does the flow become choked? 
| We will analyse the mass flow rate at the throat, where we know that $ M^{*} = 1 $, and $A^{*} = 0.1 \ \mathrm{m^{2}}$. The expression for mass flow rate here is
   
 $ \dot{m} = \rho^{*} \cdot A^{*} \cdot v^{*} $.
 
 Thus the quantities we need to determine are $ \rho^{*} $ and $ v^{*} $. 
   
***
To determine $ \rho^{*} $, first we must get $ \rho_{0} $, which we can calculate from the equation of state:
   
$\rho_{0} = \frac{P_{0}}{R T_{0}} = 3.604 \ \mathrm{kg/m^{3}}$. 
***
   
From the tables for $ M = 1 $, $\frac{\rho_{0}}{\rho} = 1.5774$ (see highlighted section of image below).
  
Thus $ \rho^{*} = \frac{3.604}{1.577} = 2.285 \ \mathrm{kg/m^{3}} $ 
  
***
   
The velocity at the throat is $ v^{*} = M^{*} \cdot a^{*} = 1 \cdot \sqrt{\gamma R T^{*}} $. 
   
From tables at $ M = 1 $, $ \frac{T_{0}}{T} = 1.200 $ (again see highlighted row), thus $ T^{*} = \frac{290}{1.2} = 241.67 \ \mathrm{K} $. 
   
$ v^{*} = \sqrt{1.4 \cdot 287 \cdot 241.67} = 311.4 \ \mathrm{m/s} $. 
  
***
   
Finally we can now compute the mass flow rate:
  
$ \dot{m} = 2.285 \cdot 0.1 \cdot 311.5 = 71.2 \ \mathrm{kg/s} $. 
\nThe exit area is $ A_{e} = 0.2 \ \mathrm{m^{2}} $. The ratio $ \frac{A_{e}}{A^{*}} $ is therefore $ 2 $. 
   
Use tables and interpolate to obtain $ M_{e} = 2.20 $: 
\nThe flow is perfectly expanded supersonic flow, therefore it is isentropic and $p_{0,exit} = p_{01}$. 
   
Use the tables for $ M_{e} = 2.20 $, and obtain $ \frac{p_{0}}{p} = 10.693 $.
 Thus $ p_{exit} = 2.806 \cdot 10^{4} \ \mathrm{Pa} $. 
\nThe mass flow rate remains unchanged as long as the flow is choked. The flow becomes choked when $ M $ becomes 1 at the throat. Remember the pressure graph. 
 
To find the range of pressures for which the flow is choked, we must find the subsonic solution with $ M = 1 $ at the throat. 
***
   
The ratio of the exit area to the throat are is $ \frac{A_{e}}{A^{*}} = \frac{0.2}{0.1} = 2 $.
   
 Now look in tables for a subsonic $ M $ to match (there will be two Mach numbers corresponding to this area ratio, a supersonic one and a subsonic one). This is the first M for which the flow becomes sonic at the throat.
   
In this case $ M_{e} = 0.306 $:
***
At $ M = 0.306 $, $ \frac{p_{0}}{p} = 1.0677 $ and since flow is isentropic, $p_{0,exit} = p_{01}$. 
   
Thus $p_{e} = \frac{p_{01}}{1.0677} = 2.810 \cdot 10^{5} \ \mathrm{Pa}$, and this is the upper limit for which we will get choked flow. 
   
The pressures lower than that, there will be supersonic flow in the divergent part of the nozzle (with possible shocks in the divergent part of the nozzle), but the flow will remain choked.
| The figure below shows a convergent-divergent nozzle. It is observed that the flow (air, $\gamma=1.4$) exits the nozzle as a perfectly expanded (isentropic) supersonic flow. 
   
  
Determine the mass flow rate through the nozzle.
\nDetermine the Mach number at the nozzle exit ($M_{exit}$)
\nDetermine the pressure at the nozzle exit ($p_{exit}$).
\nThe pressure at the nozzle exit is gradually increased from its original value. For what range of values of $p_{exit}$ would the mass flow rate through the nozzle remain unchanged? 
| 89 | 4 | 34 | 34 | 379 | 7 | 59 | 3 | 1 | Determine the mass flow rate through the nozzle. Determine the Mach number at the nozzle exit $M_{exit}$ Determine the pressure at the nozzle exit $p_{exit}$. For what range of values of $p_{exit}$ would the mass flow rate through the nozzle remain unchanged? | 3 |
6452a50d-5fb9-44c9-8934-7253f74ee23a | 2 | 0 | 0 | 19 | 6 | 1 | 4 | 0 | **(L7)**: Use Cramer's rule to find the intersection of the lines
$$
\begin{aligned}
x+ y&=1\, ,\\
x - y&= -2\, .
\end{aligned}
$$
| **(L7)**: Use Cramer's rule to find the intersection of the lines
$$
\begin{aligned}
x+ y&=1\, ,\\
x - y&= -2\, .
\end{aligned}
$$
| 1 | 0.333333 | 0 | Form the determinants $\Delta,\Delta_1,\Delta_2$ (see **section 2.4**). Hence find $x,y$ using these determinants.
| Start by evaluating the 3 determinants: $\Delta, \Delta_1, \Delta_2$ (see **section 2.4**): 
***
$$
\displaystyle \Delta \ =\ \begin{array}{|c c|}
1 & 1\\
1 & -1
\end{array} = -1-1 = -2
$$
***
$$
\displaystyle \Delta _{1} \ =\ \begin{array}{|c c|}
\hskip6pt 1 & \hskip6pt 1\\
-2 & -1
\end{array} = -1+2 = 1
$$
***
$$
\displaystyle \Delta _{2} \ =\ \begin{array}{|c c|}
1 & 1\\
1 & -2
\end{array} = -2-1 = -3
$$
***
So, $x = \Delta_{1}/\Delta = -\frac{1}{2}$, and $y = \Delta_{2}/\Delta = \frac{-3}{-2} = \frac{3}{2}$.
| **(L7)**: Use Cramer's rule to find the intersection of the lines
$$
\begin{aligned}
x+ y&=1\, ,\\
x - y&= -2\, .
\end{aligned}
$$
| 12 | 1 | 6 | 6 | 23 | 2 | 12 | 1 | 0 | L7: Use Cramer's rule to find the intersection of the lines $ \begin{aligned} x+ y&=1\, ,\\ x - y&= -2\, . \end{aligned} $ | 2 |
65990727-92f1-40a7-9db9-c12b6e215476 | 4 | 0 | 0 | 19 | 6 | 1 | 8 | 3 | **(L15)**: Consider the matrix ${\mathbf{\text{A}}}=\left(\begin{array}{cr}5&\hskip3pt -7\\1&\hskip3pt -3\end{array}\right).$
| Find the eigenvalues and normalised eigenvectors of ${\mathbf{\text{A}}}$.
\nWrite down ${\mathbf{\text{S}_A}} = (\bf{x}_1, \bf{x}_2)$, the matrix of eigenvectors for ${\mathbf{\text{A}}}$. Use ${\mathbf{\text{S}_A}}$ to diagonalise ${\mathbf{\text{A}}}$.
| 2 | 0.333333 | 2 | To solve for the eigenvalues, solve the characteristic equation $\det{(\text{A}-\lambda\mathbb{I})}=0$ (**section 3.19**).
***
Then, for each eigenvalue, solve $(\text{A}-\lambda\mathbb{I})\mathbf{\underline{x}}$. 
***
Normalise this result. 
\nWrite the matrix ${\mathbf{\text{S}}} = (\bf{x}_1, \bf{x}_2)$ of eigenvectors of $\text{A}$.
***
Referring to **section 3.21**, the diagonal matrix is $\Lambda = \text{S}^{-1}\text{A}\text{S}$...
***
... Hence invert $\text{S}$.
***
Then multiply the matrices in the required order to retrieve $\Lambda$. You expect this to be the matrix of eigenvalues.
| The eigenvalues of ${\mathbf{\text{A}}}$ are determined from the characteristic equation $\det{(\text{A}-\lambda\mathbb{I})}=0$:
***
$$
\begin{aligned}
\det({\mathbf{\text{A}}}-\lambda\mathbb{I})&=
\left|
\begin{array}{cc}
5-\lambda&\hskip3pt -7\\
1&\hskip3pt -3-\lambda
\end{array}\right|=-(5-\lambda)(3+\lambda)+7\nonumber\\
&=-(15+5\lambda-3\lambda-\lambda^2)+7\nonumber\\
&=\lambda^2-2\lambda-8= (\lambda-4)(\lambda+2) = 0\, .
\end{aligned}
$$
The eigenvalues are thus $\lambda =-2,\ 4$. 
The eigenvector corresponding to eigenvalue $\lambda_1=-2$ is obtained from solving the equation $(\text{A}-\lambda\mathbb{I})\mathbf{\underline{x}}$:
***
$$
\left(
\begin{array}{cc}
7&\hskip3pt -7\\
1&\hskip3pt -1
\end{array}\right)
\left(\begin{array}{c}
x\\ y
\end{array}\right) =
\left(\begin{array}{c}
7x-7y\\
x-y
\end{array}\right) =
\left(\begin{array}{c}
0\\ 0
\end{array}\right),
$$
which has the solution $x=y$. The associated normalised eigenvector is $ \mathbf{\hat{u}}_1=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\ 1\end{array}\right) $ .
***
The eigenvector corresponding to eigenvalue $\lambda_2=4$ is obtained from
$$
\left(
\begin{array}{cc}
1&\hskip3pt -7\\
1&\hskip3pt -7
\end{array}\right)
\left(\begin{array}{c}
x\\
y
\end{array}\right) =
\left(\begin{array}{c}
x-7y\\
x-7y
\end{array}\right) =
\left(\begin{array}{c}
0\\
0
\end{array}\right),
$$
which has the solution $x=7y$. The associated normalised eigenvector is $ \mathbf{\hat{u}}_2=\frac{1}{5\sqrt{2}}\left(\begin{array}{c}7\\\ 1\end{array}\right) $.
\nWe use eigenvectors $\vec{x}_1=\left(\begin{array}{c}1\\1\end{array}\right),\qquad\vec{x}_2=\left(\begin{array}{c}7\\1\end{array}\right)$ to construct the matrix ${\mathbf{\text{S}}}$ of eigenvectors (see **section 2.21**):
$$
{\mathbf{\text{S}}}=\left(\begin{array}{cc}1&\hskip10pt 7\\1&\hskip10pt 1\end{array}\right).
$$
**Note:** You may have written this in the opposite order, or used multiples of $\vec{x}_1$ and $\vec{x}_2$.
***
The diagonalisation of $\text{A}$ is ${{\text{S}}}^{-1}{{\text{A}}}{{\text{S}}}$ (this retrieves the matrix of eigenvalues). Starting by using Gaussian elimination (or otherwise) to find ${\mathbf{\text{S}}}^{-1}$:
***
$\left( \begin{array}{c c} 1 & 7 \\ 1 & 1 \end{array}
\right| \left. \begin{array}{c c } 1 & 0 \\ 0 & 1 \end{array} \right)
\ \rightarrow \
\left( \begin{array}{c c} 1 & \hskip8pt 7 \\ 0 & -6 \end{array}
\right| \left. \begin{array}{c c }\hskip8pt 1 & 0 \\ -1 & 1 \end{array} \right)
\ \rightarrow \
\left( \begin{array}{c c} 1 & 7 \\ 0 & 1 \end{array}
\right| \left. \begin{array}{c c } 1 & 0 \\ \tfrac{1}{6} & -\tfrac{1}{6} \end{array} \right)
\ \rightarrow \
\left( \begin{array}{c c} 1 & 0 \\ 0 & 1 \end{array}
\right| \left. \begin{array}{c c } -\tfrac{1}{6} & -\tfrac{1}{5} \\ \hskip8pt \tfrac{1}{6} & -\tfrac{1}{6} \end{array} \right)$
***
Hence, the diagonal form of ${\mathbf{\text{A}}}$ is:
***
$$
\begin{aligned}
{{\text{S}}}^{-1}{{\text{A}}}{{\text{S}}}&=-{1\over 6}
\left(
\begin{array}{rr}
1&\hskip3pt -7\\
-1&\hskip3pt 1
\end{array}\right)\left(
\begin{array}{cr}
5&\hskip3pt -7\\
1&\hskip3pt -3
\end{array}\right)
\left(
\begin{array}{cc}
1&\hskip10pt 7\\
1&\hskip10pt 1
\end{array}\right)\nonumber\\
\end{aligned}
$$
***
$$
\begin{aligned}
&=-{1\over 6}
\left(
\begin{array}{rr}
1&\hskip3pt -7\\
-1&\hskip3pt 1
\end{array}\right)\left(
\begin{array}{rc}
-2&\hskip10pt 28\\
-2&\hskip10pt 4
\end{array}\right)\nonumber\\
\end{aligned}
$$
***
$$
\begin{aligned}
&=-{1\over6}\left(
\begin{array}{cr}
12&\hskip3pt 0\hskip3pt\\
0&\hskip3pt -24
\end{array}\right)=\left(
\begin{array}{rc}
-2&\hskip10pt 0\\
0&\hskip10pt 4
\end{array}\right)\, ,
\end{aligned}
$$
Notice that the diagonal elements are the eigenvalues of ${\mathbf{\text{A}}}$, as required.
| **(L15)**: Consider the matrix ${\mathbf{\text{A}}}=\left(\begin{array}{cr}5&\hskip3pt -7\\1&\hskip3pt -3\end{array}\right).$
Find the eigenvalues and normalised eigenvectors of ${\mathbf{\text{A}}}$.
\nWrite down ${\mathbf{\text{S}_A}} = (\bf{x}_1, \bf{x}_2)$, the matrix of eigenvectors for ${\mathbf{\text{A}}}$. Use ${\mathbf{\text{S}_A}}$ to diagonalise ${\mathbf{\text{A}}}$.
| 31 | 6 | 27 | 27 | 159 | 7 | 26 | 5 | 0 | L15: Consider the matrix ${\mathbf{\text{A}}}=\left(\begin{array}{cr}5&\hskip3pt -7\\1&\hskip3pt -3\end{array}\right).$ Find the eigenvalues and normalised eigenvectors of ${\mathbf{\text{A}}}$. Write down ${\mathbf{\text{S}_A}} = (\bf{x}_1, \bf{x}_2)$, the matrix of eigenvectors for ${\mathbf{\text{A}}}$. Use ${\mathbf{\text{S}_A}}$ to diagonalise ${\mathbf{\text{A}}}$. | 3 |
65ba8a09-46a5-4797-9493-cd5a35137a9f | 2 | 0 | 0 | 17 | 6 | 1 | 1 | 0 | A block of mass $m = 2\,$kg slides on a horizontal frictionless surface. The block is initially moving at velocity $v = -1\,$m$\cdot$ s$^{-1}$. (Notice that $v$ is negative, so the block is initially moving in the $-x$ direction.) A constant force $F = +10\,$N acts for 5 s.
| Find the acceleration of the block and hence its final speed and final kinetic energy. How much kinetic energy has the block gained?
\nHow far from its initial position was the block after 5 s? Calculate the work done by the force and show that this is the same as the kinetic energy gained by the block.
| 2 | 0.333333 | 1 | Use N2 to find the acceleration of the block. 
***
Can you think of a SUVAT equation that finds final velocity from initial velocity and acceleration?
***
Kinetic energy, $K=(1/2)mv^2$. Hence find the *change* in $K$. 
\nUse a SUVAT equation to find the displacement at $t=5$.
***
Since the force is constant, the equation for work done is $W = \boldsymbol{\vec{F}}\cdot\boldsymbol{\vec{d}}$. Evaluate $W$ over the distance you have calculated.
***
As an additional task, think about this question in terms of energy. The block is initially moving in the $-x$ direction, but is eventually decelerated to zero by the force, at which point it starts accelerating in the $+x$ direction.
1. During the deceleration period, is the force doing work on the block or is the block doing work on the force? 
2. During the acceleration period, is the force doing work on the block or is the block doing work on the force?
3. What is the total work done when the block returns to the origin point?
| The acceleration $a$ of the block is $F/m = 5\,\text{m}\cdot\text{s}^{-2}$. This is positive, so the block is accelerating to the right.
***
The final velocity is given by the familiar SUVAT equation:
***
$$
v = u + at = -1 + 5 \times 5 = 24\,\text{m}\cdot\text{s}^{-1}. \\
$$
***
The initial and final kinetic energy, $K_{\text{i}}$ and $K_{\text{f}}$, are $K_{\text{i}} = \frac{1}{2} m u^2 = 1$J and $K_{\text{f}} = \frac{1}{2} m v^2 = 576$J. The kinetic energy gained is $575$ J.
\nWe can find the displacement at time $t = 5$s using the SUVAT equation:
***
$$
s = ut + \frac{1}{2}at^2 = -5 + \frac{1}{2} \times 5 \times 5^2
= 57.5\,\text{m}.
$$
***
The work done by a constant force $\boldsymbol{\vec{F}}$ on an object that moves by a displacement vector $\boldsymbol{\vec{d}}$ is $\boldsymbol{\vec{F}}\cdot\boldsymbol{\vec{d}} = Fd\cos\theta$. (The $\cos\theta$ appears because only the component of force in the direction of motion does work.) In this question the force is parallel to the direction in which the block moves, so
***
$$
\text{work} = \text{force} \times \text{distance}
= Fd = 10 \times 57.5 = 575\,\text{J},
$$
in agreement with the result of part (a).
***
\[It is worth thinking a bit harder about this question. The block is always accelerating to the right but is initially moving to the left, slowing down and coming to a stop when $v = u + at = 0$ (and hence when $t = 0.2$s) before moving back past the origin and off to the right. While it is moving to the left, the block is doing work on the applied force and its kinetic energy is decreasing. As soon as the block has turned round and is moving back towards the origin, the force is doing work on the block. By the time the block has returned to the origin, the total work done is exactly zero (ignoring friction) and the block is again moving at its initial speed, although in the opposite direction. Because the total work done during the excursion to the left is zero, only the work done afterwards has to be evaluated.]
| A block of mass $m = 2\,$kg slides on a horizontal frictionless surface. The block is initially moving at velocity $v = -1\,$m$\cdot$ s$^{-1}$. (Notice that $v$ is negative, so the block is initially moving in the $-x$ direction.) A constant force $F = +10\,$N acts for 5 s.
Find the acceleration of the block and hence its final speed and final kinetic energy. How much kinetic energy has the block gained?
\nHow far from its initial position was the block after 5 s? Calculate the work done by the force and show that this is the same as the kinetic energy gained by the block.
| 104 | 7 | 17 | 17 | 288 | 7 | 56 | 0 | 0 | Notice that $v$ is negative, so the block is initially moving in the $-x$ direction. A constant force $F = +10\,$N acts for 5 s. Find the acceleration of the block and hence its final speed and final kinetic energy. How much kinetic energy has the block gained? How far from its initial position was the block after 5 s? Calculate the work done by the force and show that this is the same as the kinetic energy gained by the block. | 5 |
65ef6b13-0e9c-4858-b3cf-fa98de6f5bba | 3 | 0 | 2 | 14 | 4 | 2 | 7 | 1 | A scale model of an aeroplane is to be tested in a wind tunnel at $M=2$. The tunnel is supplied from an air reservoir at $21^{\circ}\mathrm{C}$, and the working section is held at atmospheric pressure $100\space\mathrm{kPa}$. Assume air to be a perfect gas with $\gamma=1.4$ and $R=287\mathrm{J/(kg K)}$.
| Assuming isentropic flow, what must the reservoir pressure be?\nWhat is the air speed at the test section?\nIf the cross-sectional area at the test section is $0.2 \> \mathrm{m}^{2}$, what is the mass rate of the flow?
\nSketch the pressure distribution in the tunnel.\nIf the back-pressure is reduced below atmospheric level, sketch the pressure distribution in the tunnel and explain the changes in the flow. | 5 | 0.5 | 3 | \n\n\n\n | \n\n\n\n | A scale model of an aeroplane is to be tested in a wind tunnel at $M=2$. The tunnel is supplied from an air reservoir at $21^{\circ}\mathrm{C}$, and the working section is held at atmospheric pressure $100\space\mathrm{kPa}$. Assume air to be a perfect gas with $\gamma=1.4$ and $R=287\mathrm{J/(kg K)}$.
Assuming isentropic flow, what must the reservoir pressure be?\nWhat is the air speed at the test section?\nIf the cross-sectional area at the test section is $0.2 \> \mathrm{m}^{2}$, what is the mass rate of the flow?
\nSketch the pressure distribution in the tunnel.\nIf the back-pressure is reduced below atmospheric level, sketch the pressure distribution in the tunnel and explain the changes in the flow. | 115 | 6 | 0 | 0 | 1 | 0 | 63 | 1 | 1 | Assume air to be a perfect gas with $\gamma=1.4$ and $R=287\mathrm{J/(kg K)}$. Assuming isentropic flow, what must the reservoir pressure be? What is the air speed at the test section? If the cross-sectional area at the test section is $0.2 \> \mathrm{m}^{2}$, what is the mass rate of the flow? Sketch the pressure distribution in the tunnel. If the back-pressure is reduced below atmospheric level, sketch the pressure distribution in the tunnel and explain the changes in the flow. | 6 |
66688299-b090-4830-8280-7b06bf819c0b | 3 | 0 | 1 | 11 | 4 | 2 | 3 | 12 | An air conditioning unit comprises a cooling coil with condensate removal followed by an electrical heater. It is designed to deliver $10~ \mathrm{m^3/min}$ of conditioned air. On test the following temperatures (in $^{\circ}\mathrm{C}$) were recorded:
| | Dry Bulb | Wet Bulb |
| :----- | :------- | :------- |
| Inlet | 30.0 | 27.5 |
| Outlet | 22.0 | 14.0 |
| Draw the processes on the psychrometric chart
\nCalculate the mass flow rate of condensate.
\nCalculate the rate of heat removal at the cooling coil.
\nCalculate the heater electrical power input.
| 4 | 1 | 3 | \n\n\n | \nIt can be helpful to draw a diagram:
  
***
The specific humidity ($\omega$) at the inlet and outlet can be found using the psychrometric chart:
***
$\omega_1 = 0.0223$
  
$\omega_3 = 0.0066$
***
Apply the continuity equation to the control volume enclosed by the dotted box in the diagram:
  
$\dot{m}_\mathrm{a_1} = \dot{m}_\mathrm{a_2} = \dot{m}_\mathrm{a}$ (Eq 1)
  
$\dot{m}_\mathrm{v_1} = \dot{m}_\mathrm{w_4} + \dot{m}_\mathrm{v_2}$ (Eq 2)
***
Dividing equation (2) through by $\dot{m}_\mathrm{a}$ and bearing in mind that $\omega = \frac{\dot{m}_\mathrm{v}}{\dot{m}_\mathrm{a}}$:
  
$\omega_1 = \frac{\dot{m}_\mathrm{w_4}}{\dot{m}_\mathrm{a}} + \omega_2$ (Eq 3)
***
At the outlet (state $3$), the total flow rate is $10~\mathrm{m^3/min}$. Applying continuity between states $2$ and $3$, the total volumetric flow rate at $2$ is also $10~\mathrm{m^3/min}$.
***
To convert the flow rate to $\mathrm{kg/min}$, the volumetric flow rate can be divided by the specific volume (as determined from the psychrometric chart):
  
Mass flow rate $= \frac{10}{0.845} = 11.83~\mathrm{kg/min}$
***
Hence:
  
$\dot{m}_\mathrm{a} + \dot{m}_\mathrm{v_2} = 11.83$ (Eq 4)
***
Dividing equation (4) through by $\dot{m}_\mathrm{a}$:
  
$1+\omega_2 = \frac{11.83}{\dot{m}_\mathrm{a}}$ (Eq 5)
***
Rearranging equation (3) and equation (5) to make $\dot{m}_\mathrm{a}$ the subject:
  
$\dot{m}_\mathrm{a} = \frac{\dot{m}_\mathrm{w_4}}{\omega_1-\omega_2}$ (Eq 6)
  
$\dot{m}_\mathrm{a} = \frac{11.83}{1+\omega_2}$ (Eq 7)
***
Equating equations (6) and (7):
  
$\frac{\dot{m}_\mathrm{w_4}}{\omega_1-\omega_2} = \frac{11.83}{1+\omega_2}$ (Eq 8)
***
Looking at the psychrometric chart, $\omega_2 = \omega_3 = 0.0066$.
***
Therefore, equation (8) can be rearranged to make $\dot{m}_\mathrm{w_4}$ the subject, and numbers can be substituted in to obtain:
  
$\dot{m}_\mathrm{w_4} = \frac{11.83}{1+0.0066}\times(0.0223-0.0066) = 0.185~\mathrm{kg/min}$
\nApply the SFEE to the control volume drawn in the diagram in part (b), bearing in mind that there is no displacement work and kinetic and potential energy changes can be neglected:
***
$\dot{Q} = \dot{m}_\mathrm{a}(h_\mathrm{a_2}-h_\mathrm{a_1}) + \dot{m}_\mathrm{v2}h_\mathrm{v_2} -\dot{m}_\mathrm{v_1}h_\mathrm{v_1} +\dot{m}_\mathrm{w_4}h_\mathrm{w_4}$
***
From equation (7) in part (b):
  
$\dot{m}_\mathrm{a} = \frac{11.83}{1 + 0.0066} = 11.75~\mathrm{kg/min}$
***
Multiplying the vapour mass flow rates in the SFEE by $\frac{\dot{m}_\mathrm{a}}{\dot{m}_\mathrm{a}}$:
  
$\dot{Q} = \dot{m}_\mathrm{a}(h_\mathrm{a_2}-h_\mathrm{a_1}) + \dot{m}_\mathrm{a}\omega_2h_\mathrm{v_2} -\dot{m}_\mathrm{a}\omega_1h_\mathrm{v_1} +\dot{m}_\mathrm{w_4}h_\mathrm{w_4}$
***
Now all of the quantities on the right hand side of the SFEE are known apart from the enthalpies. These can be found as follows:
***
For air:
  
$h_\mathrm{a_2} - h_\mathrm{a_1} = c_\mathrm{p,air}(T_2-T_1)$
***
Using the dry bulb temperatures (the temperature at state $2$ can be found on the psychrometric chart):
  
$h_\mathrm{a_2} - h_\mathrm{a_1} = 1.01(8-30) = -22.22~\mathrm{kJ/kg}$
***
For the vapours:
  
$h = h_\mathrm{g}$
***
Looking up the values at the relevant temperatures (dry-bulb): 
  
$h_\mathrm{v_1} = h_\mathrm{g_1} = 2555.6~\mathrm{kJ/kg}$
  
$h_\mathrm{v_2} = h_\mathrm{g_2} = 2515.6~\mathrm{kJ/kg}$
***
For the liquid:
  
$h = h_\mathrm{f}$
***
Looking up the value at the relevant temperature (wet-bulb at state $2$, found on the the psychrometric chart to be $8^{\circ}\mathrm{C}$):
  
$h_\mathrm{w_4} = h_\mathrm{f_4} = 33.63\mathrm{kJ/kg}$
***
All of the known values can now be substituted into the SFEE:
  
$\dot{Q} = \frac{11.75}{60}[(-22.22) + (0.0066\times 2515.6) -( 0.0223\times 2555.6)] +(\frac{0.185}{60}\times 33.63) = -12.2~\mathrm{kW}$
\nApply the SFEE between states $2$ and $3$, bearing in mind that there is no shaft work and that kinetic and potential energy changes can be neglected:
 
***
$\dot{Q} = \dot{m}_\mathrm{a}(h_\mathrm{a_3}-h_\mathrm{a_2}) + \dot{m}_\mathrm{v_3}h_\mathrm{v_3} - \dot{m}_\mathrm{v_2}h_\mathrm{v_2}$
***
The mass of vapour does not change between states $2$ and $3$. Therefore:
  
$\dot{m}_\mathrm{v_3} = \dot{m}_\mathrm{v_2} =\dot{m}_\mathrm{a}\omega_2$
***
Find the enthalpies:
***
For dry air:
  
$h_\mathrm{a_3} - h_\mathrm{a_2} = c_\mathrm{p}(T_3 - T_2)$
  
where the temperatures $T_3$ and $T_2$ are the dry-bulb temperatures at states $3$ and $2$ respectively.
***
Substituting in numbers:
  
$h_\mathrm{a_3} - h_\mathrm{a_2} = 1.01(22-8) = 14.14~\mathrm{kJ/kg}$
***
For vapour:
  
$h_\mathrm{v}$ at each state $= h_\mathrm{g}$. The values can be looked up in the steam tables at the relevant temperatures (dry-bulb):
***
$h_\mathrm{v_2} = h_\mathrm{g_2} = 2515.6~\mathrm{kJ/kg}$
  
$h_\mathrm{v_3} = h_\mathrm{g_3} = 2541.1~\mathrm{kJ/kg}$
***
All of the known quantities can be substituted into the SFEE:
  
$\dot{Q} = \frac{11.75}{60}[(14.14) +0.0066(2541.1-2515.6)] = 2.80~\mathrm{kW}$
| An air conditioning unit comprises a cooling coil with condensate removal followed by an electrical heater. It is designed to deliver $10~ \mathrm{m^3/min}$ of conditioned air. On test the following temperatures (in $^{\circ}\mathrm{C}$) were recorded:
| | Dry Bulb | Wet Bulb |
| :----- | :------- | :------- |
| Inlet | 30.0 | 27.5 |
| Outlet | 22.0 | 14.0 |
Draw the processes on the psychrometric chart
\nCalculate the mass flow rate of condensate.
\nCalculate the rate of heat removal at the cooling coil.
\nCalculate the heater electrical power input.
| 94 | 2 | 58 | 58 | 518 | 0 | 30 | 0 | 0 | An air conditioning unit comprises a cooling coil with condensate removal followed by an electrical heater. On test the following temperatures in $^{\circ}\mathrm{C}$ were recorded: table Draw the processes on the psychrometric chart Calculate the mass flow rate of condensate. Calculate the rate of heat removal at the cooling coil. Calculate the heater electrical power input. | 4 |
675f9153-b04f-4db0-a59d-3c8ac43e4430 | 0 | 1 | 0 | 11 | 4 | 2 | 7 | 2 | A fuel is burnt first with the stoichiometric amount of air and then with the stoichiometric amount of pure oxygen. 
| For which case will the adiabatic flame temperature be higher?
| 1 | 0.333333 | 0 | null | null | A fuel is burnt first with the stoichiometric amount of air and then with the stoichiometric amount of pure oxygen. 
For which case will the adiabatic flame temperature be higher?
| 30 | 0 | 0 | 0 | 0 | 0 | 10 | 0 | 0 | For which case will the adiabatic flame temperature be higher? | 1 |
678a7d0c-71db-4140-b767-b5017c34fe00 | 1 | 0 | 3 | 24 | 6 | 1 | 2 | 0 | Consider a surface $S$ defined as the portion of the plane $x + 2y -3z = 0$ with $0<x<1$ and $0<y<1$.
| Show that the surface element $d\vec{S}$ on the plane is given by:
$$
d\vec{S}=\left(\mathbf{\hat{i}}+\frac{\partial z}{\partial x}\mathbf{\hat{k}}\right)\times\left(\mathbf{\hat{j}}+\frac{\partial{z}}{\partial y}\mathbf{\hat{k}}\right)\,dx\,dy
$$
\nHence show that:
$$
d\vec{S} = \left(-\frac{1}{3}\,\mathbf{\hat{i}} - \frac{2}{3}\,\mathbf{\hat{j}}+\, \mathbf{\hat{k}}\right)\,dx\,dy
$$
\nShow that the direction of $d\vec{S}$ agrees (to within a $\pm$ sign) with that extracted directly from the equation of the plane in standard form, i.e., 
$$
\vec{r}\cdot\mathbf{\hat{n}} = x\,n_x + y\,n_y + z\,n_z = d
$$
\nFind $dS \equiv |d\vec{S}|$ and hence calculate the area of the portion of the planar surface with $0<x<1$ and $0<y<1$.
| 4 | 0.333333 | 1 | The position vector on the plane is $\vec{r}=x\mathbf{\hat{i}}+y\mathbf{\hat{j}}+z(x,y)\mathbf{\hat{k}}$.
***
$$
d\vec{S}=\left(\frac{\partial \vec{r}}{\partial x}\times \frac{\partial \vec{r}}{\partial y}\right)\,dx\,dy
$$
***
Leave this in a general form and do not insert the equation of the plane. 
\nWhat is the equation of the plane in the form $z=z(x,y)$?
***
Insert this into the result from part (a) and evaluate the cross product.
\nWhat is $\mathbf{\hat{n}}$ for the plane $x+2y-3z=0$?
***
Show that this is related to the direction of $d\vec{S}$ to within a scalar multiple. 
\nAfter finding $|d\vec{S}|$, recognise that:
$$
\text{Area}= \iint_{S}|d\vec{S}|
$$
***
Evaluate the double integral over the given limits. 
| The position vector on the plane is given by:
$$
\begin{aligned}
\vec{r} &= x\mathbf{\hat{i}}+y\mathbf{\hat{j}}+z(x,y)\mathbf{\hat{k}}\\
\end{aligned}
$$
***
Then, the surface area element is given by:
$$
d\vec{S}=\left(\frac{\partial \vec{r}}{\partial x}\times \frac{\partial \vec{r}}{\partial y}\right)\,dx\,dy
$$
***
$$
\frac{\partial \vec{r}}{\partial x} = \mathbf{\hat{i}} + \frac{\partial z}{\partial x}\mathbf{\hat{k}}
$$
***
$$
\frac{\partial \vec{r}}{\partial y} = \mathbf{\hat{j}} + \frac{\partial z}{\partial y}\mathbf{\hat{k}}
$$
So, as required:
***
$$
d\vec{S}=\left(\mathbf{\hat{i}}+\frac{\partial z}{\partial x}\mathbf{\hat{k}}\right)\times\left(\mathbf{\hat{j}}+\frac{\partial{z}}{\partial y}\mathbf{\hat{k}}\right)\,dx\,dy
$$
\nRe-arranging the equation of the plane into the form $z=z(x,y)$:
***
$$
z = \frac{1}{3}(x+2y)
$$
Then, finding the partial derivatives with respect to $x$ and $y$ so that the result from part (a) can be applied:
***
$$
\begin{aligned}
\frac{\partial z}{\partial x}& = \frac{1}{3}\\\\
\frac{\partial z}{\partial y}&= \frac{2}{3}
\end{aligned}
$$
Therefore, evaluating the cross product:
***
$$
d\vec{S} = \left(\mathbf{\hat{i}} + \frac{1}{3}\mathbf{\hat{k}}\right)\times\left(\mathbf{\hat{j}}+\frac{2}{3}\mathbf{\hat{k}}\right)\,dx\,dy
$$
***
$$
= \begin{vmatrix}\mathbf{\hat{i}}&\mathbf{\hat{j}}&\mathbf{\hat{k}}\\1&0&1/3\\0&1&2/3\end{vmatrix}\,dx\,dy
$$
***
$$
= \left(-\frac{1}{3}\mathbf{\hat{i}}-\frac{2}{3}\mathbf{\hat{j}}+\mathbf{\hat{k}}\right)dx\,dy
$$
\nThe equation of the plane $x+2y-3z=0$, satisfies the form $\vec{r}\cdot\mathbf{\hat{n}} = d$ with $d=0$ and:
$$
\mathbf{\hat{n}} = \frac{1}{\sqrt{13}}(1,2,-3)
$$
***
Multiplying by the scalar multiple $-1/3$, the vector becomes:
$$
(-\frac{1}{3}, -\frac{2}{3}, 1)
$$
This is in the same direction as $d\vec{S}$. 
$$
$$
\n$$
\begin{aligned}
&d\vec{S} = \left(-\frac{1}{3}\,\mathbf{\hat{i}} - \frac{2}{3}\,\mathbf{\hat{j}}+\, \mathbf{\hat{k}}\right)\,dx\,dy\\\\
&\implies |d\vec{S}|=\sqrt{\frac{1}{9}+\frac{4}{9}+1}\,dx\,dy = \frac{\sqrt{14}}{3}
\end{aligned}
$$
***
$$
\text{Area} = \iint_S{|d\vec{S}}| = \frac{\sqrt{14}}{3}\int_{y=0}^{1}\int_{x=0}^{1}dx\,dy
$$
The integral is simply the area of the unit square, so
***
***
$$
\text{Area} = \frac{\sqrt{14}}{3}
$$
| Consider a surface $S$ defined as the portion of the plane $x + 2y -3z = 0$ with $0<x<1$ and $0<y<1$.
Show that the surface element $d\vec{S}$ on the plane is given by:
$$
d\vec{S}=\left(\mathbf{\hat{i}}+\frac{\partial z}{\partial x}\mathbf{\hat{k}}\right)\times\left(\mathbf{\hat{j}}+\frac{\partial{z}}{\partial y}\mathbf{\hat{k}}\right)\,dx\,dy
$$
\nHence show that:
$$
d\vec{S} = \left(-\frac{1}{3}\,\mathbf{\hat{i}} - \frac{2}{3}\,\mathbf{\hat{j}}+\, \mathbf{\hat{k}}\right)\,dx\,dy
$$
\nShow that the direction of $d\vec{S}$ agrees (to within a $\pm$ sign) with that extracted directly from the equation of the plane in standard form, i.e., 
$$
\vec{r}\cdot\mathbf{\hat{n}} = x\,n_x + y\,n_y + z\,n_z = d
$$
\nFind $dS \equiv |d\vec{S}|$ and hence calculate the area of the portion of the planar surface with $0<x<1$ and $0<y<1$.
| 80 | 13 | 24 | 24 | 133 | 8 | 63 | 9 | 0 | Consider a surface $S$ defined as the portion of the plane $x + 2y -3z = 0$ with $0<x<1$ and $0<y<1$. Show that the surface element $d\vec{S}$ on the plane is given by: $ d\vec{S}=\left(\mathbf{\hat{i}}+\frac{\partial z}{\partial x}\mathbf{\hat{k}}\right)\times\left(\mathbf{\hat{j}}+\frac{\partial{z}}{\partial y}\mathbf{\hat{k}}\right)\,dx\,dy $ Hence show that: $ d\vec{S} = \left(-\frac{1}{3}\,\mathbf{\hat{i}} - \frac{2}{3}\,\mathbf{\hat{j}}+\, \mathbf{\hat{k}}\right)\,dx\,dy $ Show that the direction of $d\vec{S}$ agrees to within a $\pm$ sign with that extracted directly from the equation of the plane in standard form, i.e., $ \vec{r}\cdot\mathbf{\hat{n}} = x\,n_x + y\,n_y + z\,n_z = d $ Find $x + 2y -3z = 0$0 and hence calculate the area of the portion of the planar surface with $0<x<1$ and $0<y<1$. | 2 |
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