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classical mechanics
|
Dynamics of controlled overdamped inverted pendulum
|
https://physics.stackexchange.com/questions/69438/dynamics-of-controlled-overdamped-inverted-pendulum
|
<p>I wonder how to properly write the motion equations for the inverted pendulum on a cart in case of overdamped dynamics. Imagine the system <a href="http://en.wikipedia.org/wiki/Inverted_pendulum#Inverted_pendulum_on_a_cart" rel="nofollow noreferrer">illustrated in Wikipedia</a> placed in a liquid with high viscosity $\beta$. I completely understand how the system with the fixed pendulum base behaves. Let's start from the equation for the classic inverted pendulum: $$ml^2 \ddot \theta - \beta \dot \theta + mgl\sin\theta=0.$$ High values of $\beta$ <a href="http://www.physicsforums.com/showthread.php?t=337989" rel="nofollow noreferrer">allow one to neglect the inertia term</a>, so that $$- \beta \dot \theta + mgl\sin\theta = 0.$$</p>
<p>However, as a person who is intrinsically bad at physics, I am greatly confused about what effect the motion of the cart has on the dynamics. The derivations <a href="https://physics.stackexchange.com/questions/35000/elementary-derivation-of-the-motion-equations-for-an-inverted-pendulum-on-a-cart">here</a> are for the classic, non-overdamped case, I guess that here everything should be much simpler. Any help is appreciated.</p>
<p>Update: Probably the equations I ask about should take the simplest possible form in the frame of reference centered at the moving base of the pendulum.</p>
|
<p>Using $\sum F=m a$ and $\sum M = I \ddot{\theta}$ I arrive at</p>
<p>$$ F = \left( \frac{M+m}{m} \frac{I+m L^2}{L \cos\theta} - m L \cos \theta \right) \ddot{\theta} + \left( \frac{M+m}{m} \frac{\beta \dot\theta}{L \cos\theta} + m L \sin\theta \dot{\theta}^2 \right) $$</p>
<p>now you can play with the values to get what you need.</p>
| 0
|
classical mechanics
|
The mighty man and the bridge
|
https://physics.stackexchange.com/questions/69984/the-mighty-man-and-the-bridge
|
<p>Let us say we have a mighty man crossing a bridge, carrying 4 bags of concrete, each of which weighs 50 pounds. </p>
<p>Let us say, for the sake of the argument, that the mighty man himself weighs 300 pounds.</p>
<p>The support capacity of the bridge $x$ is less than the 500 pounds they weigh together - therefore, the mighty man cannot carry the concrete in a stack across the bridge - it will fall.</p>
<p>However, the mighty man does have the ability to juggle all the concrete bags so that he is only in contact with one of them at a time. </p>
<p>Is there such a value of $x$ possible that he can cross the bridge juggling all four bags of concrete at once, or will the bridge necessarily exceed its support capacity? </p>
| 1
|
|
classical mechanics
|
Why doesn't Newton's Second Law include higher-order mass?
|
https://physics.stackexchange.com/questions/93659/why-doesnt-newtons-second-law-include-higher-order-mass
|
<p>I suspect this has been asked here before, but I didn't find anything using Search.</p>
<p>Why is Newton's second law only second-order in position? For instance, could there exist higher-order masses $m_i$ with</p>
<p>$$F(x) = m\ddot{x} + \sum_{i=3}^{\infty} m_i x^{(i)}?$$</p>
<p>Are there theoretical reasons why $m_i$ must be exactly zero for $i>2$? If not, if these masses existed but were extremely small, would we be able to tell experimentally (e.g. by observing galactic motion)?</p>
|
<p>I think Newton's first law, linearity of motion, and the time-reversal symmetry strongly constrain the form of Newton's second law.</p>
<p>Generically, the linear relation between the position and the force is given by</p>
<p>$$
F = \beta_0 x+ \beta_1 \frac{dx}{dt} + \beta_2 \frac{d^2x}{dt^2} + \beta_3 \frac{d^3x}{dt^3}+\cdots.
$$</p>
<p>The time-reversal symmetry demands that all odd derivatives vanish, $\beta_{2k+1} = 0$ for $k = 0,1,2\cdots$. The spatial homogeneity demands $\beta_0 = 0$. Therefore,</p>
<p>$$
F = \beta_2 \frac{d^2x}{dt^2} + \beta_4 \frac{d^4x}{dt^4} + \cdots = \beta_2 a + \beta_4 \frac{d^2a}{dt^2}+\cdots.
$$</p>
<p>Here $a$ is the acceleration. The first law demands that, when $F=0$, the only solution is the trivial solution $a=0$. We see that, if all higher coefficients $\beta_4, \beta_6, \cdots$ are not zero, there will be other solutions. Thus, the only possible form is</p>
<p>$$
F = \beta_2 \frac{d^2x}{dt^2}.
$$</p>
| 2
|
classical mechanics
|
Is it possible to control a treadmill's tread speed such that a plane on the treadmill will be prevented from moving?
|
https://physics.stackexchange.com/questions/32325/is-it-possible-to-control-a-treadmills-tread-speed-such-that-a-plane-on-the-tre
|
<p>I've posed the question in this particular way to avoid the ambiguity usually found in the posing of the "<a href="http://pogue.blogs.nytimes.com/2006/12/11/the-airplane-treadmill-conundrum/" rel="nofollow noreferrer">airplane on a treadmill</a>" puzzle, <a href="https://physics.stackexchange.com/questions/32269/what-will-happen-if-a-plane-trys-to-take-off-whilst-on-a-treadmill">e.g.</a></p>
<p>I'm not specifying how the treadmill is controlled but asking <em>if</em> it can be controlled in such a way that the thrust of the plane's engine is countered with an equal and opposite force. Assume the wheel bearings are frictionless and the wheels rotate freely. Please justify your answer.</p>
<p>[EDIT] Idealize the problem such that we can ignore rolling resistance.</p>
|
<p>yes it is possible. You have to account for the <strong>Tire Friction</strong> and <strong>Rolling resistance</strong>. It is a bit complicated math but you can resort to experiment. Attach a spring balance to the nose and measure the resistance offered by the tires while the treadmill is spinning at desired speed. Now make sure that engine produces just enough thrust to cancel the resistance and there it is, your model airplane is stationary on a running treadmill. In reality, when the engine starts, it produces surplus thrust even while it is idling that it is almost impossible to hold it down without brakes. So a full scale aircraft with engines turned on, without brakes, will find its way out of the treadmill.</p>
| 3
|
classical mechanics
|
Why are bricks typically constructed to have six faces at, or near right-angles to each the other?
|
https://physics.stackexchange.com/questions/32911/why-are-bricks-typically-constructed-to-have-six-faces-at-or-near-right-angles
|
<p>Looking around it appears that bricks, through history, have been constructed in cuboid form i.e. with six faces at right-angles to each other. This is also apparently the case with stone construction too - six-sided volumes.</p>
<p>Is the cuboid form the most efficient form to bear load? </p>
<p>How do other form stack up? </p>
<p>EDIT:
I would argue in favour of a shape such as a pyramid. It would take more effort to fabricate & work. But it may also be better able to bear load - perhaps because of the way the edges dovetail.</p>
<p>Does the cuboid form continue merely because of the psychological inertia that the term 'brick' induces the impression of a cuboid or near cuboid (as in ingots) form, and because the cuboid form is the easiest to grip/position by the mason/robot?</p>
<p>Why are bricks typically constructed to have six faces at, or near right-angles to each the other?</p>
|
<p>Since I argued to re-open this question I should put the effort in to answer it :)</p>
<p>As other answers have pointed out, there <em>are</em> important cultural reasons why we use cubical bricks, and different architectural traditions may predominantly use bricks of different shapes. Not all bricks are cuboid - if you look at a brick building that contains arches (such as an old bridge or railway station) you will find that it contains bricks that are not quite cuboid in shape. However, I would say that cuboid bricks are the optimal shape for the purpose they're designed for, which is to build relatively thin straight-sided vertical walls that are usually at right-angles to one another.</p>
<p>The first consideration is that the wall needs to be straight. Having the top and bottom surfaces of the brick parallel to one another is a good way to do this. You could do it in other ways. For example, by using bricks that are trapezoidal in cross-section. The picture below illustrates this, along with the problem it would bring:</p>
<p><img src="https://i.sstatic.net/Fm5Ct.png" alt="trapezoidal bricks"></p>
<p>Although the wall would be straight-sided, each brick is sitting on an inclined plane, and thus wants to slide out sideways. Although the mortar would prevent this, such a wall would be less stable, and also less practical to build.</p>
<p>Of course, if you want to build an arch or an igloo then this is exactly the sort of brick you need, and if you look at an arch you will see this type of brick:</p>
<p><img src="https://i.sstatic.net/J0hme.png" alt="enter image description here"></p>
<p>but this takes us away from our design goal of a vertical wall.</p>
<p>So if you want to build a straight-sided vertical wall then the bricks need to be rectangular in cross-section. This only leaves the "end" surfaces, which in theory could be at any angle as long as they tesselate. However, making them be at right-angles to the other two sides makes it easy to build two walls that join at right-angles without needing to break any bricks, like this:</p>
<p><img src="https://i.sstatic.net/Jt4ui.png" alt="enter image description here"></p>
<p>This also contrains the length of the sides a bit - the bricks must be twice as wide as they are deep in order for that to work. The only free parameter is the height of the brick, which can vary depending on the type of brick. (Think of the shape of a breeze block compared to a red brick.)</p>
<p>So although there are other types of brick suitable for other purposes, the cuboid form is very good for building rectangular-sided buildings with straight walls, and I would say that the design has survived throughout history because of this.</p>
| 4
|
classical mechanics
|
On the Discretization of Energy Levels
|
https://physics.stackexchange.com/questions/34447/on-the-discretization-of-energy-levels
|
<p>We consider a system of "n" particles whose total energy E and net momentum $\vec{P}$ are fixed are fixed.There no net force on the system(assumed)</p>
<p>$$\Sigma \epsilon_i= E$$
$$\Sigma\vec{p_i}=\vec{P}$$
For an individual particle its momentnum and energy remain constant for the time $\tau$,the relaxation time(average time between successive collisions----a constant). That's an extra constraint for each particle
[Radiational energy density at some point is assumed to be constant for some physically small time interval]</p>
<p>All that seem to indicate that classical theories favor the discretization of energy levels in the equilibrium state.
It is important to note that for each particle to satisfy on the mass shell
condition we should have:
$$\epsilon_i^2-p_i^2=m_0^2c^4$$
The above equation is frame invariant.Any theoretical speculation considering energy and momentum independent of each other should correspond to what we understand by "off the mass shell" situation
An orbiting electron in an atom ,so far as the classical theories are concerned , should radiate energy. But it can also receive energy form other particles.The net radiation from a block of iron at constant temperature is zero.This should favor the discretization of the electronic orbits in the atoms in the expected manner.</p>
<p>[For a cluster of charged particles one should take into account the electromagnetic potential energy of the system( a closed system ) due to the presence of charges and currents.This energy for the system should be considered constant for our model]</p>
<p>Should we use intuition(commonsense classical theories ) to interpret QM with the understanding that the results of observation should not change in view of the system interacting with the measuring instrument and that QM will be no less useful to human activity?</p>
<p>[Some points to examine have been placed on the following link:</p>
<p><a href="http://independent.academia.edu/AnamitraPalit/Papers/1892195/Fourier_Transforms_in_QM_Gravitons_and_Otherons_" rel="nofollow">http://independent.academia.edu/AnamitraPalit/Papers/1892195/Fourier_Transforms_in_QM_Gravitons_and_Otherons_</a> </p>
<p>The document may be downloaded though you get a message that conversion is going on]]</p>
|
<p>This is nonsense--- the "quantization of energy" you are referring to in classical theories is not a quantization at all, it is equipartition. It is only true that independent degrees of freedom have an average energy which is roughly quantized in classical mechanics.</p>
<p>You can't make a classical equilibrium between a field and an atom, because the field always has infinitely many degrees of freedom, and the atom finitely many. So the atom goes to the classical ground state, the electrons sit on top of the nucleus, and the field jitters thermally infinitesimally (because the finite energy is all sucked to the smallest wavelength modes).</p>
<p>This is called the ultraviolet catastrophe, and it dooms attempt to explain quantum behavior from a classical theory.</p>
<p>The simplest argument against a modified classical mechanics reproducing quantum mechanics is that quantum mechanics introduces a dimensional constant $\hbar$ which is not there in the classical limit. So if you start with the classical limit, you have to explain what sets the scale $\hbar$. This was debated in the early quarter of the twentieth century, and these ideas do not work.</p>
| 5
|
classical mechanics
|
How does a treadmill incline mechanism work?
|
https://physics.stackexchange.com/questions/35313/how-does-a-treadmill-incline-mechanism-work
|
<blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="https://physics.stackexchange.com/questions/1639/whats-the-difference-between-running-up-a-hill-and-running-up-an-inclined-tread">What's the difference between running up a hill and running up an inclined treadmill?</a> </p>
</blockquote>
<p>I want to know how a system can give so many small steps of inclination and so strong to against impact from people running. Thanks</p>
| 6
|
|
classical mechanics
|
Complex part of the solution for physical values
|
https://physics.stackexchange.com/questions/35805/complex-part-of-the-solution-for-physical-values
|
<p>What's a physical meaning of, for example, complex part of the solution for coordinate change of the anharmonic oscillator?
Why after substitute (for diff. equation solve) for real x we can earn $x = Re(z) + iIm(z)$? It's because of substitute?</p>
|
<p>When we wish to solve a differential equation like your example, a good guess for when $f(t)=0$ is something of the form $x(t) = e^{i \alpha t}$. This just gives us an algebraic equation for $\alpha$, which we can just solve. I interpret your question to mean: but what is this solution? Why is it complex? Indeed, position should certainly be just a real number. The resolution is that a <em>general</em> solution to the differential equation will not be of the form $x(t)=e^{i\alpha t}$ for $\alpha$ solving our algebraic equation. In general there will be a couple solutions to the algebraic equation for $\alpha$--just two in your example--say $\alpha_1$ and $\alpha_2$. Then a general solution to your diffeq with $f(t)=0$ is of the form $x(t) = A e^{i \alpha_1 t} + B e^{i \alpha_2 t}$ for some complex constants $A$ and $B$. We then need boundary conditions to choose $A$ and $B$ properly. The crux of the biscuit is we can choose $A$ and $B$ so that an $x(t)$ of this form is real for all times $t$ (this assumes $a$, $b$, and $c$ are real in the original equation).</p>
<p>So using the complex guess $x(t) = e^{i\alpha t}$ is just a computational trick which makes things easy because the complex numbers have nice properties with algebraic equations (they are algebraically closed). We could have done the whole thing without $e$, instead just using sines and cosines and it would have worked out too.</p>
| 7
|
classical mechanics
|
Formulas for compressibility of solids (physics)
|
https://physics.stackexchange.com/questions/38246/formulas-for-compressibility-of-solids-physics
|
<blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="https://physics.stackexchange.com/questions/37571/formulas-for-compressibility-of-solids">Formulas for compressibility of solids</a> </p>
</blockquote>
<p>I have a question which is technically a physics question, but since I have yet to find a good physics forum which accepts Latex, I am posting this question here. I hope that's OK. </p>
<p>I am taking a course in mechanics this semester, as well as a course in reservoir physics. Both courses have sections devoted to pressure/compressibility of solids, but the formulas look slightly different, so I wondered if they really mean the same or not.</p>
<p>In my mechanics class I am told that:</p>
<p>$$\Delta P = B \left(\frac{- \Delta V}{V_0}\right)$$</p>
<p>Where $B$ is a constant known as the <strong>bulk modulus</strong> of a given material.</p>
<p>In my reservoir physics class I am presented with the formula:</p>
<p>$$c = -\frac{1}{V} \left(\frac{\partial V}{\partial p}\right)_{T}$$</p>
<p>Where $c$ is referred to as <strong>isothermal compressibility</strong>.</p>
<p>So my question is - are these formulas basically the same, where $c = \frac{1}{B}$? And if they are not the same, can someone please explain the difference to me? I would really appreciate if someone could help me with this!</p>
| 8
|
|
classical mechanics
|
all the 1-dimensional problems in newtonian mechanics are solvable?
|
https://physics.stackexchange.com/questions/51774/all-the-1-dimensional-problems-in-newtonian-mechanics-are-solvable
|
<p>i mean given a system with a conserved Energy in one dimension</p>
<p>$$ E= \frac{p^{2}}{2m}+V(x) $$</p>
<p>then the 'solution' to this problem is implicitly given by</p>
<p>$$ t(x)= \frac{1}{2m} \int_{0}^{x}\frac{du}{\sqrt{E-V(u)}} $$</p>
<p>so apparently from this equation we could know all the quantities $ p(x) $ and $ x(t) $ so for a one dimension all the mechanical problems are solvable isn't it ??</p>
|
<p>Yes, and generally it is solvable numerically just like a differential equation of motion is generally solvable numerically.</p>
| 9
|
classical mechanics
|
Total Mechanical Advantage
|
https://physics.stackexchange.com/questions/51615/total-mechanical-advantage
|
<p>How do you find the net <a href="http://en.wikipedia.org/wiki/Mechanical_advantage" rel="nofollow">Mechanical Advantage</a> (MA) of two joint machines. Do you add or multiply the individual MA?</p>
<p>Suppose I have two sets of wheel and axle connected by a fixed pulley. Each of the wheel has a radius of 100cm and each of the axle has a radius of 10cm. What will be their combined MA? I am quite sure that both of them has a MA of 10. But what will be their total MA? Is it 20 or 100? Should I add them or multiply them?</p>
|
<p>It is multiplication.</p>
<p>When you turn the wheel at the beginning by $x$ degrees, the axle at the end will turn by $x/100$ degrees. This comparison of lengths/angles of movement directly translates to the mechanical advantage. The work is a constant, so because of $W=x_1 F_1=x_2 F_2=x_1/100 F_2=x_1/100\cdot 100 F_1$, it follows that $F_2=100 F_1$, which is exactly the mechanical advantage.</p>
| 10
|
classical mechanics
|
What does net mechanical efficiency mean?
|
https://physics.stackexchange.com/questions/52697/what-does-net-mechanical-efficiency-mean
|
<p>I often see the term "net mechanical efficiency" used in literature, but I am not quite sure what it means, and what the difference is between it and "normal" efficiency. Take this sentence for example: <code>... increased the effectiveness, while reducing net mechanical efficiency.</code> What does exactly does this mean?</p>
|
<p>Mechanical efficiency is usually used as a metric to account for frictional losses in systems. For example, the transmission of a car transmits mechanical work from the engine to the wheels so the mechanical efficiency of the system will be $W_{transmitted}/W_{received/ideal}$. Even within an engine, there is friction between the piston and cylinder walls, bearings of crankshafts etc. Mechanical efficiency is: $W_{output}/W_{obtained\;from\; gas}$. The denominator is the work obtained from the work-fluid, the thermodynamic cycle. Therefor it is a way to quantify the frictional loses in the system.</p>
<p>What I have mentioned above is one of the common use of this term, but there could be other definitions too. You need to look into the context of usage</p>
| 11
|
classical mechanics
|
Lego Blender and gear ratios
|
https://physics.stackexchange.com/questions/56557/lego-blender-and-gear-ratios
|
<p>I bought the Lego Kit LEGO Crazy Contraptions. It allows the learner to build a blender. My son, the engineer, said something to our grandson, his son, about a gear ratio. Can someone translate?</p>
|
<p>A blender spins very fast, presumably faster than the motor supplied in that kit. A gear is a pair of cogs which mesh together to change the rotation speed. If a cog with 20 teeth meshes with one with 10 teeth, then for every 1 turn of the large cog, the small cog will turn twice. This is known as a 1:2 gear ratio, or just a ratio of '1/2'.</p>
<p>So to make the blender spin faster, you need a smaller gear ratio. The ratio is input:output, so you can also take a pair of cogs with a larger gear ratio and turn it around.</p>
<p>There's a good <a href="http://en.wikipedia.org/wiki/Gear_ratio" rel="nofollow">article on wikipedia</a>.</p>
| 12
|
classical mechanics
|
what's the center of mass for triatomic-molecule system
|
https://physics.stackexchange.com/questions/60963/whats-the-center-of-mass-for-triatomic-molecule-system
|
<p>My text use the following example to explain the center of mass. There are three balls (mass $m$) sitting in the origin, at $x=l$ and $x=2l$, each two mass are connected with a spring of constant $k$. The system can only move along $x$ direction. To find the center of mass, I setup the coordinate system with first ball placed at $x=0$, the second ball placed at $x=l$ and the third ball placed at $x=2l$. Set $x_1$, $x_2$ and $x_3$ to be the offset from the corresponding equilibrium positions. To find the center of mass, I do the following</p>
<p>$$
x_{com} = \frac{mx_1 + m(x_2+l) + m(x_3+2l)}{m+m+m} = l + \frac{x_1+x_2+x_3}{3}
$$</p>
<p>The text said since all the ball have the same mass and they separated equally, so the center of the mass will be at the geometrical center of the system, that is, </p>
<p>$$
x_{com} = l
$$</p>
<p>But from the math, we have the last term, I know the conclusion of the text is correct but what's the physical point that we have $x_1+x_2+x_3 = 0$?</p>
|
<p>The center of mas is not what you have defined, the center of mass is:</p>
<p>$$
x_{com}= \frac{mx_1 + mx_2 + mx_3 }{m+m+m} = \frac{x_1 + x_2 + x_3 }{3}
$$ </p>
<p>and if we use $x_1 = 0$, $x_2=l$, $x_3=2l$:</p>
<p>$$
x_{com}= \frac{0 + l + 2l}{3} = l
$$ </p>
<p>acording with your geometrical aproach. </p>
| 13
|
classical mechanics
|
if a simple pendulum is dropped in a elevator with a acceleration greater than acceleration due to gravity then what will be its frequency
|
https://physics.stackexchange.com/questions/65107/if-a-simple-pendulum-is-dropped-in-a-elevator-with-a-acceleration-greater-than-a
|
<p>if a simple pendulum is dropped in a elevator with a acceleration greater than acceleration due to gravity then what will be its frequency ? We know time period depends on frequency.</p>
|
<p>Same formula as usual but with the difference that. $$ g\rightarrow g\pm g'$$ where $g'$ is the acceleration of the elevator. The plus sign is used if the two accelerations are in different directions... Thus if elevator free fall then $g-g'=0 $ and the period goes to infinity. </p>
| 14
|
classical mechanics
|
Convert running speed uphill to equivilent speed on flat
|
https://physics.stackexchange.com/questions/5882/convert-running-speed-uphill-to-equivilent-speed-on-flat
|
<p>Given a certain running pace uphill, I want to be able to determine an equivalent pace running with no elevation change. Assumptions: similar effort in both cases (say for example running at 90% max heart rate), ignore wind, slope is constant for simplicity, ignore physiological and bio-mechanical factors, weight of the runner is 135 lbs if that matters. </p>
<p>Example: Elevation change +236 feet, distance traveled 1 mile, elapsed time 6 minutes 55 seconds. What could I theoretically run for 1 mile with no elevation change given the same effort?</p>
|
<p>The easiest (and roughest) way to to do it would be to convert your running "work" into a VO2 score. </p>
<p>The American College of Sports Med's equation is </p>
<p>VO2= Resting Component + Horizontal Component + Vertical Component
or VO2= 3.5 + (0.2 x Speed) + (0.9 x Speed x Elevation Gain)</p>
<p>So, using your example of 8.67 mph (speed in the equation is in meters per min)</p>
<p>3.5 + (.2*232.67) + (0.9*232.67*.045) = 59.5</p>
<p>Thus running on flat ground should give you a speed of 280 m/min or 10.44 mph 5 min 44 sec per mile</p>
<p>(I'm an exercise scientist, not a physicist)</p>
| 15
|
classical mechanics
|
Name of the guy that Feynman mentioned during a lecture: the diagram is of a chain hanging over a triangle
|
https://physics.stackexchange.com/questions/12792/name-of-the-guy-that-feynman-mentioned-during-a-lecture-the-diagram-is-of-a-cha
|
<p>In a Feynman book, he talks about a man (I believe he lived 400-500 years ago) that discovered something about the dimensions of triangles (I think)by hanging a chain around the triangle. I've searched in "Six Not So Easy Pieces" but haven't been able to find it.</p>
<p>I'm trying to find the name of this man; it would be great if someone remembers this (or know whom I'm talking about off the top of their head).</p>
|
<p>I think it may be Simon Stevin. He lived in years 1548–1620 and did put chain around triangle, which helped him study equilibrium of forces on inclined plane. </p>
<p>This is Wikipedia article about him: <a href="http://en.wikipedia.org/wiki/Simon_Stevin">http://en.wikipedia.org/wiki/Simon_Stevin</a></p>
| 16
|
classical mechanics
|
kinetic energy and conservative force field
|
https://physics.stackexchange.com/questions/13878/kinetic-energy-and-conservative-force-field
|
<p>The kinetic energy of a particle is a periodic function in time. Does it imply that the particle is in a conservative force field and there are no dissipative forces acting on it at any instance of time ?</p>
<p><strong>EDIT</strong> (in view of comment by Willie)</p>
<p>Please consider the question as "Is the force acting on the particle conservative ?" Also consider the kinetic energy to be periodic with non zero period.</p>
|
<p>No. I can take a ball and swing it back and forth periodically with my hand. The motion is periodic, but the situation is not conservative - my body generates a lot of heat.</p>
<p>A simple mathematical example is <a href="http://farside.ph.utexas.edu/teaching/315/Waves/node12.html" rel="nofollow">a forced, damped harmonic oscillator</a>. It has a steady-state periodic solution that dissipates energy.</p>
<p>If you want to know whether a force field is conservative, take its curl. Time-independent force fields (force is a function of position but not time) are conservative iff their curl is zero.</p>
| 17
|
classical mechanics
|
Confusion with the torque
|
https://physics.stackexchange.com/questions/14080/confusion-with-the-torque
|
<p>Consider an imaginary vertical plane. Now say, a body is falling freely (under earth's attraction). If you consider any axis that is perpendicular to that plane. We get a non-zero value for torque. Then why is it that body is not moving in a circular path about that axis?</p>
|
<p>When you exert a torque on something, its angular momentum changes. This doesn't necessarily mean things are moving in a circle, or even rotating. You can have torque and changing angular momentum even when things are going in a straight line. </p>
<p>Most of the time, when we discuss torque and angular momentum in basic physics, we are talking about things like spinning bicycle wheels, precessing tops, or ice skaters doing a fancy maneuver, but that's only because torque and angular momentum are easiest to exemplify and visualize in those situations. The simple facts about torque and angular momentum, such as $\vec{\tau} = \dot{\vec{L}}$, hold generally.</p>
<p>To make your example precise, suppose a particle's trajectory is</p>
<p>$$\vec{r} = x_0\hat{x} - (y_0 - \frac{1}{2}gt^2)\hat{y}$$</p>
<p>Then the force on it is</p>
<p>$$\vec{F} = m\ddot{\vec{r}} = -mg\hat{y}$$</p>
<p>The torque about the origin is</p>
<p>$$\vec{\tau} = \vec{r}\times\vec{F} = -x_0mg\hat{z}$$</p>
<p>The angular momentum about the origin is</p>
<p>$$\vec{L} = \vec{r}\times\vec{p} = \vec{r} \times (-mgt\hat{y}) = -x_0mgt\hat{z}$$</p>
<p>Finally, we see that the time derivative of the angular momentum is the torque.</p>
<p>$$\dot{\vec{L}} = -x_0mg\hat{z} = \vec{\tau}$$</p>
<p>So we can indeed analyze the torque and angular momentum of an object falling in a straight line and find a consistent and accurate description.</p>
| 18
|
classical mechanics
|
What does this infinitesimal Eulerian change describe?
|
https://physics.stackexchange.com/questions/16279/what-does-this-infinitesimal-eulerian-change-describe
|
<p>This is a question I originally <a href="https://math.stackexchange.com/questions/70746/what-does-this-infinitesimal-eulerian-change-describe">posted</a> in math.se which received an answer that was far too mathematically sophisticated for what I wanted; given that basic multivariable calculus was used through out the paper.</p>
<p>I'm reading a paper where the orientation of a coordinate system is specified by <a href="http://mathworld.wolfram.com/EulerAngles.html" rel="nofollow noreferrer">y-convention Euler angles (eqns 30-47)</a>$(\phi_0, \theta_0, \psi_0)$ and rotation matrix $\xi$. It then goes on to say let $$d\xi = (d\psi_0\sin\theta_0\cos\phi_0 - d\theta_0\sin\phi_0,~d\psi_0\sin\theta_0\sin\phi_0 + d\theta_0\cos\phi_0,~d\phi_0 + d\psi_0\cos\theta_0)$$
What sort of infinitesimal change is it describing?</p>
<p>This is equation 3.22 of <a href="http://148.216.10.84/archivoshistoricosMQ/ModernaHist/Thomas1927.pdf" rel="nofollow noreferrer">http://148.216.10.84/archivoshistoricosMQ/ModernaHist/Thomas1927.pdf</a></p>
|
<p>This is a hard question, because Thomas is using terrible language. Most of the same calculation is done easily in my answer to this question: <a href="https://physics.stackexchange.com/questions/14751/how-did-l-h-thomas-derive-his-1927-expressions-for-an-electron-with-an-axis">How did L.H. Thomas derive his 1927 expressions for an electron with an axis?</a></p>
<p>The right way to do it is to just choose the x axis to be in the direction of the velocity at one time, and the x-y plane to include the acceleration and velocity, and the origin to be the position of the electron. Then all of Thomas's manipulations can be quickly done with no vector notation. His results are needlessly obfuscated, especially in this follow-up paper, where the introduction of Eulerian angles so deeply irrelevant to the underlying essentially trivial calculation, that I suspect that Thomas was on a mission to make his paper unreadable.</p>
<h3>If you insist on an answer</h3>
<p>The Eulerian angles introduced by Thomas are not used later, they just serve to parametrize the rotation matrix $\zeta_0$. The quantity $d\zeta_0$ is confusing, because it is not the infinitesimal change in the $\zeta_0$ matrix, which would be a tensor of second rank, but a vector, specifying an axis and a magnitude of a rotation. This vector is given by a well known procedure for specifying an infinitesimal rotation.</p>
<p>When you have two rotation matrices which differ infinitesimally, $R$ and $R+dR$, then $dR$ is a bad quantity, because adding rotation matrices does not produce a rotation matrix. The correct way to measure the infinitesimal difference is to consider $R^{-1}(R+dR) = I + R^{-1}dR$, and this tells you that $R^{-1}dR$ is an element of the Lie algebra (by definition--- the product of two rotations is a rotation, and a rotation which is infinitesimally different from the identity is the identity plus an element of the Lie algebra). There is also $dR R^{-1}$ which works too, they two differ by a rotation.</p>
<p>The Lie algebra in this case consists of all infinitesimal rotations, which are rotations about some axis with an infinitesimal angle. The axis unit vector times the infinitesimal angle is a vector, not a tensor, and this vector tells you what the infinitesimal difference between two nearby rotations is. The interpretation is that R+dR is R followed by a rotation specified by the vector of $dR R^{-1}$. The interpretation of the other order $R^{-1} dR$ is that a rotation by this, followed by a rotation by R, gives R+dR.</p>
<p>The quantity $d\xi_0$ is this infinitesimal rotation vector for the rotation between the two nearby frames he is considering. Rotation vectors are well known from undergraduate mechanics courses--- they are the angular velocities--- the things you cross with the position to find the rate of change of the position under rotation.</p>
<p>It is difficult to see this, because he did not say what he was doing. The only purpose of the Euler angles is to specify that he is talking about the infinitesimal rotation vector, in the most obscure possible way.</p>
<p>Once you know that there is an explicit angular velocity vector which is doing a rotation in addition to the Thomas precession rotation calculated in the linked answer, you just add the rotation vector of this additional rotation to the rotation vector of the Thomas precession to get the full rotation of the frame. This gives his final formula. Nowhere do you need to use his intermediate steps.</p>
<h3>If you want his Euler angle conventions...</h3>
<p>His convention for Euler angles is as follows:</p>
<p>$$ \zeta_0 = A(\phi) B(\theta) C(\psi) $$</p>
<p>Where $A$ is a rotation about the z axis by angle $\phi$, $B$ is a rotation about the y axis by angle $\theta$, and $C$ is a rotation about the z axis by angle $\psi$. From this Euler angle convention, you can find the formula for the infinitesimal rotation vector by the procedure above. Done abstractly, you get:</p>
<p>$$ d\zeta_0 \,\, \zeta_0^{-1} = dA\,A^{-1} + A\,\, dB\,B^{-1} \,\, A^{-1} + AB \,\,dC C^{-1}\,\, B^{-1}A^{-1} $$</p>
<p>The infinitesimal rotation M(\omega) corresponding to the vector $\omega$ has the property that</p>
<p>$$ R M(\omega) R^{-1} = M(R\omega) $$</p>
<p>This is easy to prove--- it is transforming between the the two different orders $R^{-1}dR$ and $dR R^{-1}$, which just changes the axis of rotation by R. Mathematicians say that the Lie algebra is the adjoint representation, which transforms by matrix-on-one-side, inverse-matrix-on-the-other. For 3d rotations, the adjoint representation is the vector representation, and the equivalence between vectors and adjoints is the formula above. But you can ignore all this jawboning--- the above is something you can understand and check directly.</p>
<p>When you turn it into vectors, the first quantity is a rotation purely along the z axis by an amount d\phi, so it's omega vector is purely along the z axis:</p>
<p>$$ \omega(dA A^{-1}) = (0,0,1) d\phi$$</p>
<p>the second starts off as a pure rotation about the y-axis by an amount $d\theta$, but it gets rotated about the z axis by $\phi$ (because of the $A$ and $A^{-1}$), so its omega vector is the y-axis rotated by $\phi$ around the z axis.</p>
<p>$$ \omega(A\,\,dB\,B^{-1}\,\,A^{-1}) = (-\sin(\phi),\cos(\phi),0) d\theta $$</p>
<p>The third would be a rotation about the z axis, or (0,0,1) but it gets rotated around the y axis by $\theta$, so into $(\sin\theta, 0, \cos\theta)$, then it gets rotated around the z axis by $\phi$, so into $(\sin\theta \cos\phi, \sin\theta \sin\phi , \cos\theta)$. This is the rotation vector of the third term.</p>
<p>$$ \omega(AB\,\,dC\,C^{-1}\,\,A^{-1}B^{-1} ) = (\sin\theta \cos\phi , \sin\theta \sin\phi, \cos\theta) d\psi$$</p>
<p>Adding the three angular velocities together is Thomas's formula. Again, it is never used in the equations that follow.</p>
| 19
|
classical mechanics
|
How should a closed-ended terrestrial trajectory be corrected for the Coriolis effect?
|
https://physics.stackexchange.com/questions/17342/how-should-a-closed-ended-terrestrial-trajectory-be-corrected-for-the-coriolis-e
|
<p>I have tried verifying the numerical integration of the Coriolis effect for 1000 to 2000-yard rifle fire by switching ON/OFF the Coriolis correction of a good ballistic simulator program (PRODAS). The program integrates an instantaneously evaluated Coriolis acceleration along with the aerodynamic and gravitational accelerations. My calculations yield about 10 percent larger Coriolis effects than the delta after the same number of timeslice intervals are computed in two otherwise identical simulation runs. I theorize that the integrated Coriolis effect should be <em>independent of the path</em> between the two end-points of any segment of a terrestrial trajectory. If so, the velocity used in the instantaneous Coriolis acceleration can be replaced with the displacement vector of the projectile divided by the time-of-flight over the flight segment. The displacement is just the <em>vector difference</em> between the projectile positions at the ends of the flight segment. And the cumulative Coriolis effect over a segment becomes just the displacement vector crossed with the earth rotation rate vector multiplied by the time-of-flight. I speculate that the directly integrated "instantaneous Coriolis force" produces some along-track component that should not be allowed. An additional "normalizing constraint" is needed. The constraint should be <em>perpendicularity to the displacement vector</em> as above. I can e-mail a 3200-word tech note about this question as an attachment in Word or PDF file format to anyone who would like it. </p>
|
<p>The cumulative Coriolis effect must be <em>independent of the path</em> between the two end-points of the flight segment to which it applies. The Coriolis effect is entirely due to the motion of the observer and not to that of the object. Direct integration of the instantaneous Coriolis acceleration over a curved trajectory introduces error in the resulting Coriolis effect to the same extent that it introduces path-dependence. That commonly used approach only works for straight-line trajectories. My path-independent approach to calculating the Coriolis effect (described in the question), a direct integration of Coriolis acceleration approach, and, indeed, a "conservation of angular momentum" approach to calculating the eastward deflection of an object that is carefully released to fall within a vacuum tower 50 meters in height located at sea-level on the mean equator all produce the same answer: 11.643 mm east of plumb. I used 6,378,137 m as the earth-centric radius to the base of the tower, 9.8066 m/s/s as the effective acceleration of gravity (including the "centrifugal acceleration" effect), and 7.2921 times ten to the minus five rad/s as the rotation rate of the earth. All calculations are carried to five significant figures. It might help in seeing all of this to consider the instantaneous Coriolis acceleration (a function of the instantaneous apparent velocity of the body) as the limiting case of a time-averaged Coriolis acceleration (a function of the object's time-averaged apparent velocity) over shorter time and distance intervals. The time-averaged apparent velocity of the object is just the apparent spatial displacement vector of the object divided by its temporal displacement between the two state vectors bounding the flight segment being considered. My tech note on this has grown to 5500 words and is available upon request.
<strong>Addendum:</strong>
I just worked out the example "tower drop" problem much more carefully. The eastward deflection is proportional to the 3/2 power of the height of the drop or to the cube of the fall time. I used a hybrid height-times-fall-time, but it is trickier. Both a "conservation of angular momentum" approach and a "direct integration of Coriolis acceleration" approach yield 2/3 of my previous result, or 7.7620 mm. The difference would vary with other types of problems. This counter-example blows up my theorem of "independence of path" for the Coriolis effect. The integration of the Coriolis acceleration is still the only way to go in the general case. The tech note mentioned has been withdrawn. "Nevermind..." Jim</p>
| 20
|
classical mechanics
|
How fast will the projectile go the second time?
|
https://physics.stackexchange.com/questions/19395/how-fast-will-the-projectile-go-the-second-time
|
<p>Say I have a linear motor [aka rail-gun] and use a x amount of electrical power. I fire the gun and the object exits at velocity v. I then reuse the same object as my projectile and fire the rail-gun a second time this time with 2x the electrical power. My lessons on momentum suggest it will go 2v. The Work-Energy Theorem says 1.414v. Which is right and why?</p>
|
<p>Your railgun has some fixed length. If you apply some voltage to generate some force, then the energy gained by the projectile will simply be force times distance (i.e. the length of the railgun).</p>
<p>If you now double the voltage, to get twice the force, the work done will be twice as great so the projectile will have twice the kinetic energy, and this is $\sqrt{2}$ times the velocity.</p>
<p>I think what's puzzling you is that you say "My lessons on momentum suggest it will go 2v". Remember that the change of momentum is force times time (called "impulse"). When you double the force the projectile goes faster so it spends less time in the railgun, so even though the force doubles the product $F\times t$ does not.</p>
| 21
|
classical mechanics
|
Does mass affects accelleration of an object in a sloapy movement?
|
https://physics.stackexchange.com/questions/19548/does-mass-affects-accelleration-of-an-object-in-a-sloapy-movement
|
<blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="https://physics.stackexchange.com/questions/19552/would-a-light-or-a-heavy-ball-roll-fastest-down-a-slope">Would a light or a heavy ball roll fastest down a slope?</a> </p>
</blockquote>
<p>Is there a change of the accelleration of an object moving on a sloapy plane if the mass is changed?</p>
| 22
|
|
classical mechanics
|
Does a straight water hose issue water at a greater pressure than a Coiled water hose of same diameter and length?
|
https://physics.stackexchange.com/questions/20656/does-a-straight-water-hose-issue-water-at-a-greater-pressure-than-a-coiled-water
|
<p>I have a one BHP water pump, the water pressure of a coiled hose connected to the water pump output side was not that great. Would an unwound water hose produce greater water pressure? [Friction Losses?]</p>
<p>Thanks,
Alan</p>
|
<p>Yes, but under normal circumstances (e.g. garden hose scale) you wouldn't notice.</p>
<p>Friction losses are higher in bends, especially sharp bends, but the coils in a regular hose aren't very sharp. This is by design: hoses are made fairly sturdy so you don't get unintentional sharp bends. So it's really down to engineering more than physics.</p>
| 23
|
classical mechanics
|
Numerical torque calculation
|
https://physics.stackexchange.com/questions/29022/numerical-torque-calculation
|
<p>Suppose I can compute interaction energy of two rigid bodies as a function of their coordinates of centers of masses and Euler rotation angles (total 6 + 6 degrees of freedom). Now I can numerically compute force acting on the center of mass of the body by calculating numerical derivatives e.g. $F_x = (E(x + dx) - E(x - dx)) / (2 * dx)$.
But if you do the same for Euler angles this doesn't give you torques. So how do I convert numerical derivatives of energy by Euler angles to the resulting torque on a body?</p>
|
<p>OK. I found the answer:</p>
<p>$$
\partial V/\partial \theta = N_x \cos \psi - N_y \sin \psi
$$
$$
\partial V/\partial \phi = N_x \sin \theta \sin \psi + N_y \sin \theta \cos \psi + N_z \cos \theta$$
$$
\partial V/\partial \psi = N_z
$$</p>
<p>Where $\theta, \psi, \phi$ are Euler angles and $N_x, N_y, N_z$ are Torque components.</p>
| 24
|
classical mechanics
|
Path traced out by a point
|
https://physics.stackexchange.com/questions/30614/path-traced-out-by-a-point
|
<p>While studying uniform circular motion at school, one of my friends asked a question:</p>
<p>"How do I prove that the path traced out by a particle such that an applied force of constant magnitude acts on it perpendicular to its velocity is a circle?" Our physics teacher said it was not exactly a very simple thing to prove.</p>
<p>I really wish to know how one can prove it.Thank you!</p>
|
<p>One can prove it in a more-or-less elementary way by solving a pair of simultaneous differential equations. In two dimensions, a vector that is perpendicular to a velocity $$\left(\begin{matrix}u(t)\cr v(t)\end{matrix}\right)\quad\mathrm{is}\quad\left(\begin{matrix}-v(t)\cr u(t)\end{matrix}\right).$$ The acceleration, the time derivative of the velocity, is proportional to this vector, so we have the two differential equations
$$\left(\begin{matrix}\dot u(t)\cr \dot v(t)\end{matrix}\right)=\lambda\left(\begin{matrix}-v(t)\cr u(t)\end{matrix}\right).$$
If $\lambda$ is negative, the circle goes "the opposite way". If $\lambda$ is zero, the circle is a straight line.
If you know differentiation and how to solve differential equations, you should be able to solve this pair of equations, and then integrate it to obtain the way in which the position changes over time. If you don't, then it may be better to be patient and wait until you come across it in the course of your studies. Learning calculus on your own to the level needed to solve this differential equation is possible, however.</p>
| 25
|
classical mechanics
|
Ball bouncing elastically off a wall with no slipping
|
https://physics.stackexchange.com/questions/440212/ball-bouncing-elastically-off-a-wall-with-no-slipping
|
<p>I am a bit confused about what seems like it should be a perfectly straightforward problem in mechanics with a well defined solution.</p>
<p>A sphere approaches a wall from an angle and rebounds elastically. The coefficient of static friction between sphere and wall is large—no slipping occurs during the brief period of contact. Sphere and wall are both made from hard materials so that the contact patch is small. The initial velocity and rotation rate of the sphere are given (rotation is about the <span class="math-container">$z$</span> axis and motion takes place in the <span class="math-container">$xy$</span> plane). What are the final velocity (speed and direction) and rotation rate?</p>
<p>The source of my confusion is that at first glance, the solution seems under constrained: there are three unknowns (<span class="math-container">$[v_x, v_y, \omega]$</span>, say), and only two obvious conserved properties (energy and angular momentum about the point of contact). </p>
<p>I suspect that the third constraint is that the magnitude of the velocity component perpendicular to the wall is the same before and after the collision—that <span class="math-container">$v_{y2}=-v_{y1}$</span> for a wall oriented in the <span class="math-container">$x$</span> direction—but I can’t immediately think of an argument to justify that suspicion with confidence.</p>
|
<p>I'll give here the 2D version of my answer to <a href="https://physics.stackexchange.com/a/429834/197851">this question</a> since there's no guarantee that question (having been closed) won't disappear at some point.</p>
<p>If the wall is in the <span class="math-container">$x$</span> direction, as you suggested, let's assume that the sphere is approaching from the positive-<span class="math-container">$y$</span> side, so initially it has a velocity component <span class="math-container">$v_y<0$</span>. The angular velocity <span class="math-container">$\omega$</span> is defined so that a positive value <span class="math-container">$\omega>0$</span> corresponds to anticlockwise rotation in the <span class="math-container">$xy$</span> plane.</p>
<p>Then the effects of the collision are completely determined by the two components of the impulse <span class="math-container">$(C_x,C_y)$</span> at the point of contact
<span class="math-container">\begin{align*}
mv_x' &= mv_x + C_x \\
mv_y' &= mv_y + C_y \\
I\omega' &= I\omega + RC_x
\end{align*}</span>
where the primed quantities are those after the collision,
<span class="math-container">$R$</span> is the radius of the sphere, <span class="math-container">$m$</span> is the mass, and <span class="math-container">$I$</span> is the moment of inertia.
The third equation here represents the effect of an impulsive torque about the centre of the sphere. So there are two unknowns: <span class="math-container">$C_x$</span> and <span class="math-container">$C_y$</span>.</p>
<p>If the collision is to conserve energy, there is one more condition to apply, but there are a couple of options that will satisfy this condition:
a perfectly smooth wall and a perfectly rough wall. This is a bit subtle. The argument can be applied to a collision between two bodies of <em>finite</em> mass (as mentioned in the answer of @j-murray) and then (if it is appropriate) we can take the mass of one of them to infinity afterwards. But the argument applies equally well to two colliding bodies of finite mass. In the general case, the physical assumption is that the interaction must depend on the <em>relative</em> velocities of the two points that come into contact. It must also respect the symmetry of the collision, which is defined by the direction of that relative velocity vector and the normal to the surfaces. And of course it must be time reversible: if we reverse the post-collisional velocities, and apply the same collision rules, we should get the pre-collisional velocities again. Once those assumptions are made, there are really only these two possibilities which will conserve energy: smooth and rough collisions.</p>
<p>The smooth wall will correspond to <span class="math-container">$C_x=0$</span>. This results in no change to the angular velocity, or to the component <span class="math-container">$v_x$</span>; the component <span class="math-container">$v_y$</span> simply gets reversed <span class="math-container">$v_y'=-v_y$</span> so as to conserve energy.</p>
<p>In the case of the rough wall, the contact condition of roughness means that there is no slip, which in turn implies that the velocity of the point on the surface of the sphere, in contact with the wall, gets reversed. In the <span class="math-container">$x$</span> direction this means <span class="math-container">$(v_x'+\omega' R)=-(v_x+\omega R)$</span>. In the <span class="math-container">$y$</span> direction we get <span class="math-container">$v_y'=-v_y$</span> again.
A little bit of algebra shows that
<span class="math-container">\begin{align*}
C_x &= -2m\left(\frac{I}{I+mR^2}\right)(v_x+\omega R) \\
C_y &= -2m v_y
\end{align*}</span>
Inserting these into the equations for <span class="math-container">$v_x'$</span>, <span class="math-container">$v_y'$</span> and <span class="math-container">$\omega'$</span>,
it is possible to show that the total kinetic energy (translational plus rotational) is the same after the collision as before.</p>
<hr>
<p>[EDIT following OP comment]</p>
<p>I'm assuming the rigid elastic rough hard sphere model which is described by S Chapman and TG Cowling, <em>Mathematical Theory of Nonuniform Gases</em> (3rd edition, Cambridge University Press, 1970).
In this model, there are no internal degrees of freedom of the spheres: each sphere just has its linear and angular momenta, that's all. Also, I've assumed (as is common) that they behave kinematically as spherical tops, with a scalar moment of inertia.
I can only give a simplified argument in the 2D case in favour of associating the energy conservation with the "perfectly rough" condition, i.e. reversal of the relative velocity of the point of impact. For the 3D case I recommend that book, or papers such as JW Lyklema <a href="https://doi.org/10.1016/0378-4371(79)90014-1" rel="nofollow noreferrer"><em>Physica A,</em> <strong>96,</strong> 573 (1979)</a>.</p>
<p>We can write an equation for the change of kinetic energy <span class="math-container">$\Delta K$</span>
<span class="math-container">$$
2\Delta K = m({v_x'}^2-v_x^2) + m({v_y'}^2-v_y^2) + I({\omega'}^2-\omega^2)
$$</span>
Now, I'm just going to assert (as being physically reasonable) that <span class="math-container">$v_y'=-v_y$</span>, as in a standard 1D elastic collision, unaffected by the details of the impulse in the <span class="math-container">$x$</span> direction. This means that the change in <span class="math-container">$v_y$</span> has no effect on <span class="math-container">$\Delta K$</span>, and we can write the remaining terms as
<span class="math-container">\begin{align*}
2\Delta K &= \underbrace{m(v_x'-v_x)}_{C_x}(v_x'+v_x)
+ \underbrace{I(\omega'-\omega)}_{RC_x}(\omega'+\omega) \\
&= C_x [(v_x'+v_x)+R(\omega'+\omega)] \\
&= C_x [g_x' + g_x]
\end{align*}</span>
where I have inserted the equations for the change in <span class="math-container">$v_x$</span> and <span class="math-container">$\omega$</span> in terms of <span class="math-container">$C_x$</span>, as well as defining <span class="math-container">$g_x=v_x+R\omega$</span> as the <span class="math-container">$x$</span>-velocity of the point
on the sphere which is in contact with the wall.</p>
<p>In order for <span class="math-container">$\Delta K$</span> to be zero, there are only two possibilities:
either <span class="math-container">$C_x=0$</span>, in which case we have a smooth hard collision,
or <span class="math-container">$g_x'=-g_x$</span>, in which case the velocity at the point of impact is reversed (and we can derive the equation for <span class="math-container">$C_x$</span> given earlier).</p>
<p>In Lyklema's paper (if you can get hold of it) you'll see that the analogous equation in 3D is <span class="math-container">$|\mathbf{g}_\perp'|^2=|\mathbf{g}_\perp|^2$</span>,
where <span class="math-container">$\mathbf{g}_\perp$</span> is that part of the vector <span class="math-container">$\mathbf{g}$</span> which is perpendicular to the normal to the surfaces in contact. (The derivation is for two colliding spheres, but it can easily be adapted to one sphere colliding with another massive, huge, sphere, which acts like a wall). This leaves open the possibility of conserving energy by <em>rotating</em> <span class="math-container">$\mathbf{g}_\perp$</span> through some arbitrary angle in the plane of contact to give <span class="math-container">$\mathbf{g}_\perp'$</span>. However, it is hard to argue that this makes physical sense, without ascribing some exotic properties to the surfaces (chirality). The non-chiral option is to make the angle <span class="math-container">$180^\circ$</span> giving <span class="math-container">$\mathbf{g}_\perp'=-\mathbf{g}_\perp$</span>.</p>
<p>To be honest, in many places this reversal of the velocities at contact is often taken to be the defining property of the perfectly rough sphere model; the detailed justification involves a fair bit of work.</p>
<p>[2nd, hopefully last, EDIT following chat]</p>
<p>I'm going to stick with my previous point: the hard, elastic, rough sphere model is fully specified by this collision rule, that the relative velocity of the points of impact on the two bodies is reversed. The quotations below (going back to 1894!) explain the <em>physical thinking</em> of those who proposed the model, which (in my opinion) largely matches the description of the collision given in the question. </p>
<p>I accept that it is possible to extend the model so as to have different collision rules, while still satisfying the conservation laws: Lyklema's paper (cited above), I think, goes into this in some detail. So one can argue that the mechanical problem is <em>not</em> fully specified (although I am still happy with the assumption made in my answer, that the reversal of velocity perpendicular to the wall is not affected by the impulse tangential to the wall). There are plenty of other papers around which attempt similar extensions of the model, but beware, it's easy to inadvertently make a mistake. </p>
<p>Finally, having established the collision rule for two spheres of different radius, mass, and moment of inertia, respecting all the conservation laws, one can carefully take the limit that one of them becomes very large and heavy, and initially motionless, hence acting as a wall.</p>
<p>Quoting S Chapman and TG Cowling 3rd ed (full ref above) p217 <span class="math-container">$\S$</span>11.6,</p>
<blockquote>
<p>These results are first applied to the rough elastic spherical
molecules of Bryan and Pidduck, these being by definition such that at
a collision the relative velocity of the spheres at their point of
impact is exactly reversed.</p>
</blockquote>
<p>Quoting GH Bryan <a href="https://www.biodiversitylibrary.org/item/95243#page/7/mode/1up" rel="nofollow noreferrer"><em>Rep Brit Assoc Advan Sci</em> (1894)</a>, p83:</p>
<blockquote>
<p>The cases of <em>perfectly rough</em> spheres or circular discs having both
their normal and tangential coefficients of restitution unity furnish
interesting examples for solution. We may imagine the spheres and
discs covered over with perfectly elastic fine teeth or minute
projections by whose action the tangential components of the relative
velocity are reversed at impact ....</p>
</blockquote>
<p>Quoting FB Pidduck <a href="https://dx.doi.org/10.1098/rspa.1922.0028" rel="nofollow noreferrer"><em>Proc Roy Soc A,</em> <strong>101,</strong> 101 (1922)</a></p>
<blockquote>
<p>Imagine two spheres to collide and grip each other, so as to bring the
points of contact to relative rest. A small elastic deformation is
produced, which we suppose to be released immediately afterwards, the
force during release being equal to that at the corresponding stage of
compression. Thus the relative velocity of the points of contact is
reversed by collision.</p>
</blockquote>
<p>That's pretty much all I can contribute here!</p>
| 26
|
classical mechanics
|
Invariance of Space and Time in Different Reference Frames
|
https://physics.stackexchange.com/questions/451280/invariance-of-space-and-time-in-different-reference-frames
|
<p>I'm studying dynamics by Anil Rao's book. He said something that made me really confused. The citations below construct the problem.</p>
<blockquote>
<p>[...]</p>
<p>An assumption of Newtonian mechanics is that space is invariant with
respect to changes in reference frames, i.e., observations made of
space are the same in all reference frames. As a result of the
assumption of the invariance of space, observations of an arbitrary
vector <strong>b</strong> are the same regardless of the reference frame. Consequently,
for any two reference frames A and B we have</p>
<p><span class="math-container">$ {}^{A} {\bf b} = {}^{B} {\bf b} = {\bf b}$</span></p>
<p>[...]</p>
</blockquote>
<p>Then he talks about rate of change of a vector function:</p>
<blockquote>
<p>[...] given an arbitrary vector function of time, at every instant of time
observers in all reference frames see the same vector. In other words,
a vector is independent of the reference frame in which it is
observed. However, unlike a scalar function of time, observers in
different reference frames do not see the same rate of change of a
vector function of time. Contrariwise, observers fixed in different
reference frames will, in general, make different observations about
the rate of change of a vector function of time. As a result, the
phrase “rate of change of a vector” must always be qualified by
stating explicitly the reference frame in which the observation is
being made.</p>
<p>[...]</p>
</blockquote>
<p>Further, about velocity:</p>
<blockquote>
<p>It is important to understand that the notation <span class="math-container">$ {}^{A} {\bf v} $</span> is
merely a shorthand for <span class="math-container">$ {}^{A} d{\bf r}/dt $</span>. In other words, when we
write <span class="math-container">$ {}^{A} {\bf v} $</span>, we mean that the vector <span class="math-container">$ {}^{A} {\bf v} $</span>
arises from taking the rate of change of another vector (in this case
position) in reference frame A. Furthermore, it is important to
understand that, although <span class="math-container">$ {}^{A} {\bf v} $</span> arises from taking the
rate of change of r in reference frame A, the quantity <span class="math-container">$ {}^{A} {\bf v} $</span>
is itself a vector. Consequently, by assumption observations of
the vector <span class="math-container">$ {}^{A} {\bf v} $</span> are the same in all reference frames.</p>
</blockquote>
<ol>
<li>The rate of change of the position is dependent of the frame of reference.</li>
<li>Velocity is the rate of change of the position.</li>
</ol>
<p>So why velocity is independent of reference frame?</p>
<p>Or more generally: i can't understand why vectors are independent of reference frame, but their rate of change are.</p>
| 27
|
|
classical mechanics
|
Does the axis of rotation of a rigid body depend on the frame of reference?
|
https://physics.stackexchange.com/questions/452360/does-the-axis-of-rotation-of-a-rigid-body-depend-on-the-frame-of-reference
|
<p>When studying the kinematic motion of a rigid body, angular velocity <span class="math-container">$\omega$</span> is a vector that not seem to specify a unique axis of rotation... When looking at the free rigid body motion of a wheel rolling without sliding, we can talk about the wheel's rotation from the point of view of a fixed frame of reference and in that case talk about rotation about the instantaneous center of rotation (which is the contact point) or we can talk about rotation from the center of mass frame of reference and in that acase the center of rotation is the center of mass itself. From a frame of reference that is fixed with the wheel (body axes), the wheel does not rotate at all because the frame of reference rotates and every point looks stationary. Chasles theorem states that a rigid body can pass from one configuration to the next one via one of the infinite combinations of translation/rotation about any <strong>arbitrary</strong> point which becomes the point of rotation for that transformation. All transformations share the same <span class="math-container">$\omega$</span>... Does that mean that rotation is a relative concept and there is no unique, physical, axis of rotation for a rigid body? I have read about the instantaneous screw axis where the points of the rigid body with the same velocity parallel to the axis reside...
Certainly, when a free rigid body rotates while translating, maybe tumbling in some random fashion, the initial conditions (how the body starts, the forces in action) should uniquely determine the rigid body's configuration at every instant <span class="math-container">$t$</span> and how it kinematically moves from one configuration to the next: even if Chasles theorem states that there are infinite possible combinations (translation+rotation), the body certainly moves from its current configuration to the next configuration in a very specific way...Can anyone shine some clarity on this topic?</p>
|
<p>I don't understand the question.</p>
<p>At any instance, in any <em>one</em> coordinate frame if the position <span class="math-container">$\boldsymbol{r}_A$</span>, velocity <span class="math-container">$\boldsymbol{v}_A$</span> of a point <strong>A</strong> in a rigid body rotating with <span class="math-container">$\boldsymbol{\omega}$</span> is known or measured then the location of the instantaneous rotation axis is given by the following calculation</p>
<ul>
<li><p><strong>Direction</strong> (Vector) - The direction of rotation is <span class="math-container">$$ \boldsymbol{e} = \frac{ \boldsymbol{\omega} }{ \| \boldsymbol{\omega} \| }$$</span></p>
</li>
<li><p><strong>Position</strong> (Vector) - The point on the rotation axis closest to the coordinate frame is <span class="math-container">$$\boldsymbol{r}_{\rm COR} = \boldsymbol{r}_A + \frac{ \boldsymbol{\omega} \times \boldsymbol{v}_A} { \| \boldsymbol{\omega} \|^2 }$$</span></p>
</li>
<li><p><strong>Magnitude</strong> (Scalar) - The magnitude of rotation <span class="math-container">$$ \omega = \| \boldsymbol{\omega} \| $$</span></p>
</li>
<li><p><strong>Pitch</strong> (Scalar) - The ratio of the parallel motion (translation) along the rotation axis to the rotation magnitude <span class="math-container">$$ h = \frac{ \boldsymbol{\omega} \cdot \boldsymbol{v}_A }{ \| \boldsymbol{\omega} \|^2 } $$</span></p>
<p><sub>Here <span class="math-container">$\cdot$</span> is the vector inner product and <span class="math-container">$\times$</span> the vector cross product. Vector quantities are shown in <strong>boldface</strong>.</sub></p>
</li>
</ul>
<p>I can provide proof of the above if needed. So from the basis of the statements above can you comment below and rephrase./summarise your question.</p>
<hr />
<h2>Equations of Motion</h2>
<p>Equations of motion for a rigid body are derived from the time derivative of momentum. When expressed at the center of mass (Point <strong>C</strong>) the momentum equations are</p>
<p><span class="math-container">$$ \begin{aligned}
\boldsymbol{p} & = m \boldsymbol{v}_C \\
\boldsymbol{L}_C & = \mathrm{I}_C \,\boldsymbol{\omega}
\end{aligned} $$</span></p>
<p><sub>Where <span class="math-container">$\mathbf{I}_C$</span> is the mass moment of inertia 3×3 tensor about the center of mass, and <span class="math-container">$\boldsymbol{v}_C$</span> the velocity of the center of mass.</sub></p>
<p>As you can see, as expressed at the center of mass the equations are rather simple.</p>
<p>Now transform the above equations to an arbitrary point <strong>A</strong> and the above become</p>
<p><span class="math-container">$$ \begin{aligned}
\boldsymbol{p} & = m ( \boldsymbol{v}_A - \boldsymbol{c} \times \boldsymbol{\omega}) \\
\boldsymbol{L}_A & = \mathrm{I}_A \boldsymbol{\omega} + \boldsymbol{c} \times m \boldsymbol{v}_A
\end{aligned} $$</span></p>
<p><sub>Where <span class="math-container">$\boldsymbol{c}$</span> is the position vector of the center of mass from the reference point <strong>A</strong>. Also <span class="math-container">$\mathbf{I}_A$</span> is the mass moment of inertia tensor at <strong>A</strong>.</sub></p>
<p>As you can see, the level of complexity has increased significantly since now there are cross terms (linear momentum depends on <span class="math-container">$\boldsymbol{\omega}$</span> and angular momentum on <span class="math-container">$\boldsymbol{v}_A$</span></p>
<p>Now ask the question if resolving the equations of motion onto the <em>center of rotation</em> makes things any simpler. The answer is just marginally yes. If <strong>A</strong> is the center of rotation, then <span class="math-container">$\boldsymbol{v}_A = h\, \boldsymbol{\omega}$</span> where the scalar <span class="math-container">$h$</span> is the pitch. That is it. That is all the simplifies at the center of rotation.</p>
<p><span class="math-container">$$ \begin{aligned}
\boldsymbol{p} & = m ( h\,\boldsymbol{\omega} - \boldsymbol{c} \times \boldsymbol{\omega}) \\
\boldsymbol{L}_{\rm COR} & = \mathrm{I}_{\rm COR} \boldsymbol{\omega} + m\,h\,(\boldsymbol{c} \times \boldsymbol{\omega})
\end{aligned} $$</span></p>
| 28
|
classical mechanics
|
Linear acceleration on a spinning satellite with an unbalanced force
|
https://physics.stackexchange.com/questions/510025/linear-acceleration-on-a-spinning-satellite-with-an-unbalanced-force
|
<p>If there is a satellite in orbit in space, with an off centre booster providing an unbalanced force, it will experience rotational acceleration. However, I was wondering if it also undergoes linear acceleration while spinning, or only for a brief moment at the start...</p>
|
<p><img src="https://i.sstatic.net/2mea2.png" alt="Force divided into components"></p>
<p>Here I have drawn a simple diagram of what might answer your question. <span class="math-container">$\overrightarrow{a}$</span> is the acceleration provided by the off-centred booster. <span class="math-container">$\overrightarrow{r}$</span> is a vector from centre-of-mass of the rocket to the point at which booster is applying force.</p>
<p>We divide, for our convenience, <span class="math-container">$\overrightarrow{a}$</span> into it's components along <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> axes, where <span class="math-container">$X$</span>-axis is perpendicular to <span class="math-container">$\overrightarrow{r}$</span>.</p>
<p>Let the <span class="math-container">$x$</span>-component of acceleration be <span class="math-container">$\overrightarrow{a_x}$</span> and <span class="math-container">$y$</span>-component be <span class="math-container">$\overrightarrow{a_y}$</span>.</p>
<p>Now you can easily spot which component provides linear acceleration (<span class="math-container">$\overrightarrow{a_y}$</span>) and which one provides rotational acceleration (<span class="math-container">$\overrightarrow{a_x}$</span>).</p>
<p>So, answering your original question - It undergoes <strong>both linear and rotational acceleration</strong> (throughout the burnout of the booster). The satellite would be linearly accelerated the whole time, although not along the direction of it's nose.</p>
<p>Edit: Please note that <span class="math-container">$\overrightarrow{r}$</span> and <span class="math-container">$\overrightarrow{a_y}$</span> are overlapping each other since <span class="math-container">$\overrightarrow{r}$</span> <span class="math-container">$\perp$</span> <span class="math-container">$X$</span>-axis and <span class="math-container">$Y$</span>-axis <span class="math-container">$\perp$</span> <span class="math-container">$X$</span>-axis. They have been drawn a bit off in the diagram by mistake. </p>
| 29
|
classical mechanics
|
How does a drop of water evaporate and still maintain its drop shape?
|
https://physics.stackexchange.com/questions/534574/how-does-a-drop-of-water-evaporate-and-still-maintain-its-drop-shape
|
<p>Let's say we have a drop of water on a surface. Its surface tension maintains the shape. But was the water evaporates, the molecules leaving it, leave from the surface. Shouldn't the evaporation disturb the surface tension causing the water droplet to fall apart?</p>
|
<p>As the drop gets smaller the ratio of the number of molecules on the boundary compared to those in the main volume gets higher. A smaller drop creates a stronger sphere. </p>
<p>As an exiting molecule leaves the boundary the molecules that were holding it in place lose the outward pull of that molecule and so become pulled only inwards, renewing the contraction of the surface.</p>
| 30
|
classical mechanics
|
Box Sliding Down an Inclined Plane - Underlying Assumption
|
https://physics.stackexchange.com/questions/546228/box-sliding-down-an-inclined-plane-underlying-assumption
|
<p>The classic introductory mechanics problem considers the motion of a box sliding down an inclined plane. As I'm reviewing the early chapters in Taylor's <em>Classical Mechanics</em>, I was struck by a question: in such a problem, we always <em>say</em> that the box has no motion in the y-direction (as is standard for such problems our coordinate system has axes parallel and perpendicular to the plane, the x- and y-axis, respectively) and therefore we can equate the normal force with the perpendicular component of gravity so that we can proceed with the analysis. For instance, in solving the problem, Taylor waves his hand, saying</p>
<blockquote>
<p>Since the block does not jump off the incline, we know there is no motion in the y-direction[...]</p>
</blockquote>
<p>What allows us, <em>a priori</em>, to say that the box will stay on the plane. Is it simply common sense, or is there some other way of showing it that is more rigorous?</p>
|
<p>The forces on the box in such questions are gravity, friction, and the reaction of the plane. The reaction of the plane is the key. Gravity has a component in the y direction, and would cause the object to fall through the plane if not for the reaction. </p>
<p>The plane is a rigid object. That means when a force tries to deform it, it pushes back with just enough force to prevent the deformation. An object sitting on it will not penetrate the surface. On the other hand, the plane will not push back strongly enough to lift the object off the surface. </p>
<p>This doesn't explain the reaction force. It is more or less a restatement of the fact that the plane is rigid. And that is as far as this kind of problem usually takes it. </p>
<p>To explain the normal force you need to explain atomic bonds. They have a more or less fixed length. They can stretch, but less than the separation between atoms. So you can model them as rigid links, or perhaps very stiff springs, and leave it at that. </p>
<p>Or you can look at the orbitals of molecules and calculate the configuration that results in the lowest energy. In spirit, this is like calculating orbitals of an atom. But instead of one nucleus, there are two for a diatomic molecule. Or a periodic array of them for a crystal. </p>
<p>There is a lowest energy at a particular distance. If the nuclei get too close, potential energy increases because they repel each other. If they are too far apart, energy increases because the electrons are not close enough to both nuclei. </p>
| 31
|
classical mechanics
|
Conservation of Energy Question (train example)
|
https://physics.stackexchange.com/questions/581827/conservation-of-energy-question-train-example
|
<p>So we have a train going 5 m/s that gets loaded with 20,000 kg of coal while going over a 10m platform for 2 s. There is a 50,000 N horizontal force applied to the train during that period in order to keep the train going a constant 5 m/s.</p>
<p>That 50,000 N force does 500,000 J of work. But the KE of the coal at the end of the platform is only 250,000 J and the KE of the train itself (minus the coal) is unchanged. Where does the other 250,000 J go?</p>
|
<p>The addition of coal essentially constitutes as a series of inelastic collisions. Therefore, kinetic energy will be lost. The energy loss comes from work done by internal forces between the coal and the train.</p>
| 32
|
classical mechanics
|
Coordinate transformation to non-inertial frame: do we assume displacement is invariant?
|
https://physics.stackexchange.com/questions/658470/coordinate-transformation-to-non-inertial-frame-do-we-assume-displacement-is-in
|
<p>I'm a math student who's studying classical mechanics for the first time, so forgive me if this question sounds pedantic, but I find it hard to think about such problems without stating assumptions clearly.</p>
<p>If a frame K' is accelerating away from an inertial frame K, we wish to transform from the inertial to the non-inertial coordinates. Now assuming we know the displacement between the origins <span class="math-container">$\vec{R}$</span>, we can write:</p>
<p><span class="math-container">$$\vec{r} = \vec{r}{}' + \vec{R} \qquad \Longrightarrow \qquad
\vec{r}{}' = \vec{r} - \vec{R}$$</span></p>
<p>This is usually accompanied by a drawing illustrating the vector algebra. However, don't we assume implicitly here that both frames will observe the same displacement? In other words, that displacement is invariant in all frames of reference (not just inertial ones)?</p>
<p>All references I am able to find giving an axiomatic approach to classical mechanics take great pains to emphasize that, in the Galilean relativity of classical mechanics, one can only assume quantities to be preserved under transformations of the Galilean group, which is then shown to exclude accelerating frames (translation velocity must be constant), e.g. in Arnold:</p>
<p>'The galilean group is the group of all transformations of a galilean space which preserve its structure (...) every Galilean transformation of the space R×R<span class="math-container">$^3$</span> can be written in a unique way as the composition of a rotation, a translation, and a uniform motion.'</p>
<p>So, does a non-inertial observer measure the same displacement? And if not, does it make sense to transform into that frame from our non-inertial displacement measurement?</p>
|
<p>The primary assumption that you are talking about: I like to call that the 'inertial dead reckoning' assumption.</p>
<p>So the metaphor is 'dead reckoning'.<br />
Quick recapitulation of 'dead reckoning'.<br />
Two ships are on a motionless body of water. The two ships depart from each other, both keep an exact log of their velocity and direction of velocity. The two ships can plan to rejoin at some future point in time. As long as a ship keeps an accurate log it can plot a course to the point of departure.</p>
<p>In theory of motion we have the following three tiers:<br />
-position<br />
-velocity<br />
-acceleration</p>
<p>'Inertia dead reckoning' is analogous to dead reckoning, but the measurement is one tier higher. (The usual name for such navigation is: 'inertial guidance system')</p>
<p>Two spaceships in space, far enough from sources of gravity that gravitational effect is negligable. The two spaceships can depart from each other and plan to rejoin at some point in the future. By maintaining an accurate log of acceleration and direction of acceleration the two ships can at al times plot a course back to the rendez-vous point.</p>
<br>
<p>So there is this assumption that all of the space of the solar system is a uniform space. More specifically, we assume that the space presents itself as a perfectly uniform space because <em>inertia is uniform throughout space</em>.</p>
<br>
<p>We have the following corroborating <em>evidence</em> for the uniformity of inertia:</p>
<p>If inertia would be in any way non-uniform then planets of the Solar system will not move according to a simple motion law.</p>
<p>Wat we observe is that the planets of the solar system do move along highly regular orbits. The orbits of the planets of the Solar system are described to a high level of accuracy by the inverse square law of gravity.</p>
<p>Once planet-planet interactions are accounted for only an anomalous precession of Mercury's orbit remains (which is in the order of seconds of arc per <em>century</em>).</p>
<br>
<p>So:<br />
For the time being we can count the Solar system as 'our system' (Because it is for the extent of the Solar system that we have <em>observation of motion</em> that can serve as corroborating evidence.)</p>
<p>We have strong corroborating evidence that inertia is uniform throughout the Solar system.</p>
<p>That evidence for the uniformity of inertia is the justification for relying on the Principle of Galilean relativity.</p>
| 33
|
classical mechanics
|
Particle collides with hard wall
|
https://physics.stackexchange.com/questions/716422/particle-collides-with-hard-wall
|
<p>When a particle elastically collides with a hard wall at some velocity, can the point at which the particle instantenously jumps be assumed to have zero velocity for potential to be continuous?</p>
|
<p>Assuming a head-on collision (or only considering the normal component of the velocity), we could apply the <a href="https://en.wikipedia.org/wiki/Mean_value_theorem" rel="nofollow noreferrer">mean value theorem</a> to argue as follows: If velocity is given as <span class="math-container">$v(t) = dx(t)/dt$</span> where <span class="math-container">$x(t)$</span> is the distance from the wall at time <span class="math-container">$t$</span>, and we assume that <span class="math-container">$x(t)$</span> is continuous and differentiable (i.e. changes smoothly), then according to the mean value theorem there will be a time <span class="math-container">$t_0$</span> where <span class="math-container">$v(t_0)=0$</span> as long as a collision takes place (i.e. there are times such that <span class="math-container">$v>0$</span> at one point and <span class="math-container">$v < 0$</span> at another).</p>
<p>If instead the velocity changes instantaneously upon collision, <span class="math-container">$v(t)$</span> is discontinuous and there will be no derivative of <span class="math-container">$x(t)$</span> defined at the moment of collision. This would mean that there is no point where the velocity is zero, but this would generally be considered a less accurate model.</p>
| 34
|
classical mechanics
|
Can a physical system have initial velocity zero, but non-zero constant everywhere else?
|
https://physics.stackexchange.com/questions/716442/can-a-physical-system-have-initial-velocity-zero-but-non-zero-constant-everywhe
|
<p>Does there exist a particular physical system where the initial velocity can be zero, but non-zero constant everywhere else?</p>
|
<p>If the velocity is constant the force is zero. So we want a system which produces a very high force in an infinitesimally short time at <span class="math-container">$t=0$</span> and then drops back to zero. The simplest system that I can come up with right now is an ideal<span class="math-container">$^*$</span> billiard ball at rest which is hit by another billiard ball at <span class="math-container">$t=0$</span>. It will have velocity zero before <span class="math-container">$t=0$</span> and then a constant velocity afterwards</p>
<p><span class="math-container">$*$</span> with ideal I mean in the limit of an infinitely hard ball. If the ball is not infinitely hard the force will exist for a finite amount of time (the ball will act like a spring).</p>
| 35
|
classical mechanics
|
Isn’t $(c^2 - v^2/c^2)^(1/2)$ the mathematical equivalent of $( 1 - v^2/c^2)^(1/2)$ ? Why do we never see it?
|
https://physics.stackexchange.com/questions/730910/isn-t-c2-v2-c21-2-the-mathematical-equivalent-of-1-v2-c21
|
<p>Isn’t <span class="math-container">$\sqrt{c^2 - v^2/c^2} $</span> the mathematical equivalent of
<span class="math-container">$\sqrt{1 - v^2/c^2}$</span>? Why do we never see it presented this way?</p>
|
<p>Are you asking why we never see <span class="math-container">$\sqrt{\frac{c^2-v^2}{c^2}}$</span> as opposed to <span class="math-container">$\sqrt{1-\frac{v^2}{c^2}}$</span>?</p>
<p>If so, what advantage does the former have? The latter expression makes it clear that the relevant quantity is <span class="math-container">$v/c$</span>, which varies between 0 and 1. It's easier to expand in a Taylor series, it's easier to actually <em>compute</em> for any given value of <span class="math-container">$v$</span>, it is <em>aesthetically</em> nicer to look at (at least to my eyes) ... It's hard for me to imagine why you'd prefer the former expression.</p>
| 36
|
classical mechanics
|
Why may tensions calculated in different ways be slightly different?
|
https://physics.stackexchange.com/questions/213262/why-may-tensions-calculated-in-different-ways-be-slightly-different
|
<p>An question I am doing gives this situation:
<img src="https://i.sstatic.net/EMo57.jpg" alt="enter image description here"></p>
<p>Mass1 is 4kg. And mass2 is 7kg. </p>
<p>I expect the tension to be the same, no matter which mass I use in the calculation. But...</p>
<p>a=5.17</p>
<p>T1=7(9.81-5.17)=32.48N (which is not the same as T2=0.3*9.81*4+5.17*4=32.452N</p>
<p>I use the same number for gravitational acceleration, why other reason may there be for which the tensions turns out not to be the same?</p>
|
<p>The two objects are connected by a string, and should therefore have the same acceleration. Based on some of the equations you are showing, I believe that object 1 is subject to friction with a coefficient of 0.3. We can then write the equations for the forces:</p>
<p>$$F_1 = T - m_1 g \mu\\
F_2 = m_2 g - T\\
a_1 = \frac{F_1}{m_1}\\
a_2=\frac{F_2}{m_2}\\
a_1=a_2$$</p>
<p>These should all be consistent, as you said. We have five equations; we can work out the acceleration as follows:</p>
<p>$$\frac{F_2}{m_2} = \frac{F_1}{m_1}\\
g - \frac{T}{m_2} = \frac{T}{m_1} - g\mu\\
T = \frac{gm_1 m_2(1+\mu)}{m_1+m_2}$$</p>
<p>Substituting the values, we get $T = \frac{9.81*4*7*1.3}{11}=32.46 \rm{N}$</p>
<p>You can use the same equations to solve for the acceleration, which becomes</p>
<p>$$a = \frac{T-m_2 g}{m_2} = \frac{g m_1 (1+\mu)}{m_1+m_2}-g = -5.173 \rm{m/s^2}$$</p>
<p>Since these come from the same set of equations, these give consistent answers. The only reason you are getting slightly different answers is because you are trying to compare numbers to 4 significant figures when the intermediate result is given to only 3.</p>
<p>Rounding errors can creep in once you go from symbols to numbers. Keep your calculation symbolic as long as possible, and you will see that the error goes away. Or you discover the problem with your math...</p>
| 37
|
classical mechanics
|
Friction of a rolling cylinder
|
https://physics.stackexchange.com/questions/214643/friction-of-a-rolling-cylinder
|
<p>I was wondering why friction vectors are drawn differently regarding a cylinder rolling on a surface and a cylinder rolling down an inclined surface. Since friction is responsible for the rotational motion shouldn't it be always pointing in the same direction (given that the cylinder is rotating to the right)?</p>
<p>Here are two pics I googled so you can see what I mean:</p>
<p><a href="https://i.sstatic.net/fLcXM.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/fLcXM.png" alt="enter image description here"></a></p>
<p><a href="https://i.sstatic.net/56488.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/56488.gif" alt="enter image description here"></a></p>
|
<p>This is because static friction, in the second case, tries to oppose the force which would otherwise result in the movement of the cylinder, which is the component of gravitational force along the inclined plane.</p>
<p>In the second case, I will assume that the rolling friction has been referred to. Then, the shown direction of friction at that point actually does oppose the direction of rolling at that point, which is the point of contact. </p>
<p>Edited diagrams to help you understand better: (Sorry for the small text)</p>
<p><img src="https://i.sstatic.net/K34C0.gif" alt="Case 2"></p>
<p><img src="https://i.sstatic.net/Pa62j.png" alt="Case 1"></p>
| 38
|
classical mechanics
|
Again, why is kinetic energy and velocity independent of position coordinates in Cartesian coordinates
|
https://physics.stackexchange.com/questions/233073/again-why-is-kinetic-energy-and-velocity-independent-of-position-coordinates-in
|
<p>This might be a very simple question. I read one previous post
<a href="https://physics.stackexchange.com/questions/231185/can-the-kinetic-energy-be-a-function-of-the-position-vector">Can the kinetic energy be a function of the position vector?</a></p>
<p>I know that in Cartesian coordinates, the kinetic energy $T=\frac{1}{2}mv^2$. And $T$ is not an explicit function of position. So $\frac{\partial T}{\partial x}=0$, where we suppose $x$ is a coordinate.</p>
<p>But I got confused by one example, we have a ball move vertically from the origin O, with a velocity of $\vec V_0$.</p>
<p><a href="https://i.sstatic.net/zkAU5.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/zkAU5.jpg" alt="enter image description here"></a></p>
<p>then when the ball reach $y_1$ in the positive $y$ axis, we have
$$mgy=\frac{1}{2}mv_0^2-\frac{1}{2}mv_1^2$$
so the velocity at y is
$$v_1=\sqrt {\frac{1}{2}mv_0^2-mgy}$$
Does it means that velocity and kinetic energy both are explicit function of position y in this case? I know this is a special case, but the statement that $\frac{\partial T}{\partial x}=0$ in Cartesian coordinates seems to be quite general. So where have I missed so far? Thanks guys!</p>
|
<p>In classical mechanics calculate the evolution of a particle means to know its position and its velocity for any time.</p>
<p>In general if I say that a particle is in position $x$ and has velocity $v$ and ask you about the kinetic energy. That is $\frac{1}{2}mv^2$ and this is independent of the system you have. The answer is always correct.</p>
<p>Solving the equations of the system means finding, for each moment of time, what is the position and the velocity. Once you solve the equations you will have a position and velocity for each time $t$. Therefore it means you can for any system, not only the one you described, write the velocity as a funtion of the position.</p>
<p>The kinetic energy, however, will never depend explicity on the position but will depend implicity. Informally this means that it does not depend on the position directly but if you can associate to each position a speed then it will depend indirectly.
In fact taking the partial derivative as you mentioned $\frac{∂T}{∂x}$ means "how much the kinetic energy change if I vary the position while I keep the velocity constant" and this is always zero. If you do however $\frac{dT}{dx}$ you are saying "how much the kinetic energy change if I vary the position while I varying the velocity accordingly" and that is non-zero.</p>
<p>I think the mixing of the two concepts is the source of your question. </p>
| 39
|
classical mechanics
|
Moving a Car the quickest path
|
https://physics.stackexchange.com/questions/240156/moving-a-car-the-quickest-path
|
<p>I have a car, that has a current angle and location, and a destination angle and location. The car has a maximum linear and angular acceleration. Assume the car will always travel in the direction it is facing. I'm trying to figure out how to determine the path that this car should travel that will be the quickest path. An example setup might be as follows:</p>
<p><a href="https://i.sstatic.net/5Dihh.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/5Dihh.png" alt="enter image description here"></a></p>
<p>The car has to end up in the direction and location of the rightmost arrow, and is facing the direction of the leftmost.</p>
<p>What I'm trying to do is to figure out the optimal path, or at least a reasonable approximation of the optimal path, that will get between the two points. Assume an empty field, the car can travel over anything without restriction. The car might be moving beforehand as well. </p>
<p>Ultimately, I'm trying to figure out the formula that will take the current state and the destination state and figure out what the next step should be.</p>
| 40
|
|
classical mechanics
|
Sliding sphere wear shape
|
https://physics.stackexchange.com/questions/267665/sliding-sphere-wear-shape
|
<p>Please refer to the figure attached. Consider a normal force is acting on the top of sphere. A constant coefficient of friction causes frictional force throughout the sliding. I want to know after this sphere slides (pure sliding no rolling) for sometime and assuming that it wears as it slides, what should be the shape of sphere after sliding? The one shown in (a) or (b)? In short, I want to know whether the worn side of sphere will be a straight line or a curved one? What will happen if a sphere slides against a sphere? Also, I will highly appreciate if someone can refer to some good papers / books about this.</p>
<p>Edit: Additional assumptions</p>
<p>Lets assume that hardness of both surfaces is the same. Also assume that material is removed but is not attached to any of the surfaces (no adhesive wear). The phenomenon under consideration is abrasive wear (but no accumulation of wear debris). If a lubricant is in circulation, it is easy to realize this kind of wear</p>
<p><a href="https://i.sstatic.net/498fE.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/498fE.png" alt="enter image description here"></a></p>
|
<p>Given the diagrams you present, I believe a straight wear(A) on the sphere is most likely. If the hardness of the surface is less than that of the sphere, there should be no, or almost no, wear on the sphere. </p>
<p>(b) Would primarily occur if your surface was curved, such as if it were U shaped or in the case of sphere on sphere wearing. As you move along the curved surface, the moving sphere will be worn away along its "sides" which come into contact with the curved surface. </p>
<p>Hope this helps!</p>
| 41
|
classical mechanics
|
Is this a metorite captured on film or a bird/plane etc?
|
https://physics.stackexchange.com/questions/304770/is-this-a-metorite-captured-on-film-or-a-bird-plane-etc
|
<p>It is a shiny object in broad daylight, the distance seems quite far away, is there sufficient information in the footage to ascertain if it is a meteorite captured on film or something else?</p>
<p><a href="https://i.sstatic.net/PC3gE.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/PC3gE.gif" alt="enter image description here"></a></p>
<p>PS: Maybe somebody can grab a high def picture from somewhere?
I originally <a href="https://movies.stackexchange.com/questions/66417/meteorite-flash-above-congress-building-around-3502-mark-in-sleepy-hollow">posted this question on movie SE</a></p>
<p>But I think it might be more suited for Phsysics SE.</p>
| 42
|
|
classical mechanics
|
Transport Theorem in analytical dynamics: basis of the vectors
|
https://physics.stackexchange.com/questions/399424/transport-theorem-in-analytical-dynamics-basis-of-the-vectors
|
<p>If we have two reference systems, <span class="math-container">$N$</span> and <span class="math-container">$B$</span>, with common origins <span class="math-container">$O_N=O_B$</span> and <span class="math-container">$B$</span> being rotating around <span class="math-container">$N$</span> with angular velocity <span class="math-container">$\omega_{B|N}$</span>, the time derivates of any vector <span class="math-container">$\vec{u}$</span> in both systems are related by the next theorem:</p>
<p><span class="math-container">$$\frac{d \vec{u} }{dt}_N=\frac{d \vec{u} }{dt} _{B}+ \vec{\omega}_{B|N} \wedge \vec{u} $$</span></p>
<p>In which reference frame are expressed the vectors <span class="math-container">$\vec{\omega}$</span> and <span class="math-container">$\vec{u} $</span> from the second term of the RHS: in the basis of <span class="math-container">$N$</span> or in the basis of <span class="math-container">$B$</span>?</p>
|
<p>We could rewrite the equation in the following way:</p>
<p><span class="math-container">$$\frac{d ({}^{B}\vec{u}) }{dt}\Biggr\rvert_N=\frac{d ({}^{B}\vec{u}) }{dt} \Biggr\rvert_{B}+ ({}^{B}\vec{\omega}_{B|N}) \wedge ({}^{B}\vec{u}) $$</span></p>
<p>The angular velocity vector <span class="math-container">$\vec{\omega}_{B|N}$</span> is typically written in the B frame.</p>
<p>However, it is not necessary for the vector <span class="math-container">$\vec{u}$</span> to be written in the B coordinate frame, because <span class="math-container">$\vec{u}$</span> is simply one of the infinity of possible components of the unique vector <span class="math-container">$\vec{u}$</span>. Rather, components can be written in any arbitrary coordinate frame.</p>
| 43
|
classical mechanics
|
How does the actual elimination of dependent coordinates takes place?
|
https://physics.stackexchange.com/questions/601508/how-does-the-actual-elimination-of-dependent-coordinates-takes-place
|
<p>In the textbooks of classical mechanics I have been through, it is often quoted that, given an <span class="math-container">$N$</span> particle system having <span class="math-container">$3N$</span> Cartesian coordinates <span class="math-container">$(x_i,y_i,z_i)$</span> connected by <span class="math-container">$r$</span> holonomic constraints <span class="math-container">$\phi(x_i,y_i,z_i,t)=0$</span>, any <span class="math-container">$r$</span> of the <span class="math-container">$3N$</span> coordinates <em>can be eliminated</em>........ . How? For instance, I began with <span class="math-container">$3N$</span> equations of motion,
<span class="math-container">$$
\ddot {x_i}=X_i,
$$</span>
<span class="math-container">$$
\ddot {y_i}=Y_i,
$$</span>
<span class="math-container">$$
\ddot {z_i}=Z_i,
$$</span>
Then I differentiated the constraint equations twice to get <span class="math-container">$r$</span> equations of the form;
<span class="math-container">$$
\sum_i \phi_{x_i} \ddot {x_i} + \sum_i \phi_{y_i} \ddot {y_i} + \sum_i \phi_{z_i} \ddot {z_i} + c =0,
$$</span>
where <span class="math-container">$c$</span> depends on the quantities <span class="math-container">$\dot {x_i}$</span>,...,<span class="math-container">$\phi_{x_iy_j}$</span>., etc. So now I have <span class="math-container">$r$</span> linear equations in <span class="math-container">$3N$</span> variables <span class="math-container">$\ddot {x_i},\ddot {y_i},\ddot {z_i}$</span> as well as the original <span class="math-container">$3N$</span> equations in these variables. My question is, how do I eliminate any <span class="math-container">$r$</span> of <em>these</em> <span class="math-container">$3N$</span> variables (i.e. the actual acceleration components)? By some substitution? Or by setting up the <span class="math-container">$3N \times r$</span> matrix equation? Thanks!</p>
|
<p>you forgot in your equations the constraint forces:</p>
<p>for the free body diagram (1 particle) with constraint equations you obtain:</p>
<p><span class="math-container">$$m\ddot x=F_x+F_{cx}$$</span>
<span class="math-container">$$m\ddot y=F_y+F_{cy}$$</span>
<span class="math-container">$$m\ddot z=F_z+F_{cz}$$</span></p>
<p>in vector notation</p>
<p><span class="math-container">$$m\,\vec{\ddot{w}}=\vec{{F}}_a+\vec{{F}}_c\tag 1$$</span>
where <span class="math-container">$\vec F_a~$</span> is the applied force and <span class="math-container">$\vec F_c~$</span> is the constraint force</p>
<p>and the constraint equations</p>
<p><span class="math-container">$$\vec\phi(\vec w)=\vec 0$$</span></p>
<p>with :
<span class="math-container">$$\vec{{F}}_c=\left[\frac{\partial \vec \phi}{\partial \vec w}\right]^T\,\vec{\lambda}=\boldsymbol C_b^T\,\vec \lambda$$</span></p>
<p>where <span class="math-container">$\vec \lambda~$</span> is the generalized constraint force</p>
<p>thus Eq. (1):</p>
<p><span class="math-container">$$m\,\vec{\ddot{w}}=\vec{{F}}_a+\boldsymbol C_b^T\,\vec \lambda\tag 2$$</span></p>
<p>to solve Eq. (2) you differentiate twice the constraint equation and obtain</p>
<p><span class="math-container">$$\boldsymbol C_b\,\vec{\ddot{w}}+\frac {d}{dt}\boldsymbol C_b=\vec 0\tag 3$$</span></p>
<p>with Eq (2) and (3) you obtain <span class="math-container">$~\vec{\ddot{w}}~$</span> and <span class="math-container">$~\vec \lambda$</span></p>
<p>you can also write Eq (2) and (3) in matrix notation</p>
<p><span class="math-container">$$\boldsymbol A\,\boldsymbol z=\boldsymbol b$$</span></p>
<p>where :</p>
<p><span class="math-container">$$\boldsymbol z=\begin{bmatrix}
\vec{\ddot{w}} \\
\vec \lambda\\
\end{bmatrix}$$</span>
<span class="math-container">$$\boldsymbol b=\begin{bmatrix}
\vec{F}_A \\
-\frac {d}{dt}\boldsymbol C_b\\
\end{bmatrix}~,\frac {d}{dt}\boldsymbol C_b=\frac{ \partial \left[\boldsymbol C_b\,\vec{\dot{w}}\right]}{\partial \vec w}\,\vec{\dot{w}}$$</span></p>
<p>and</p>
<p><span class="math-container">$$\boldsymbol A=\begin{bmatrix}
m\,E_3 & -\boldsymbol C_b^T \\
\boldsymbol C_b & \boldsymbol 0 \\
\end{bmatrix}$$</span></p>
<p><strong>Remark</strong></p>
<p>to solve the differential equations you need the initial conditions those must full filled the constraint equations <span class="math-container">$~\vec \phi(\vec w)=\vec 0$</span></p>
<p><strong>Example: 3D Pendulum</strong></p>
<p>the applied force is:</p>
<p><span class="math-container">$$\vec F_a=\begin{bmatrix}
0 \\
0 \\
-m\,g \\
\end{bmatrix}$$</span></p>
<p>the constraint equation is:</p>
<p><span class="math-container">$$\phi=x^2+y^2+z^2-L^2=0$$</span></p>
<p><span class="math-container">$\rightarrow$</span></p>
<p><span class="math-container">$$\boldsymbol C_b=\left[ \begin {array}{ccc} 2\,x&2\,y&2\,z\end {array} \right] $$</span></p>
<p><span class="math-container">$$\frac{d}{dt}\boldsymbol C_b=
\left[ \begin {array}{ccc} 2\,\dot x^2&2\,\dot y^2&2\,\dot z^2\end {array} \right] $$</span></p>
<p><span class="math-container">$$\boldsymbol A=\left[ \begin {array}{cccc} m&0&0&-2\,x\\ 0&m&0&-2
\,y\\ 0&0&m&-2\,z\\-2\,x&-2\,y&-2
\,z&0\end {array} \right]
$$</span></p>
<p><span class="math-container">$$\boldsymbol b= \left[ \begin {array}{c} 0\\0\\
-mg\\ -2\,{{\dot x}}^{2}-2\,{{\dot y}}^{2}-2\,{{\dot z}}^{2}\end {array} \right]$$</span></p>
<p><span class="math-container">$\Rightarrow$</span></p>
<p><span class="math-container">$$\boldsymbol z=\left[ \begin {array}{c} {\frac {x \left( z\,g+{{\dot{x}}}^{2}+{{\dot{y}}
}^{2}+{{\dot{z}}}^{2} \right) }{{L}^{2}}}\\ {\frac {y
\left( z\,g+{{\dot{x}}}^{2}+{{\dot{y}}}^{2}+{{\dot{z}}}^{2} \right) }{{L}^
{2}}}\\ -{\frac {g\,{x}^{2}+g\,{y}^{2}-z{{\dot{x}}}^{2}-z
{{\dot{y}}}^{2}-z{{\dot{z}}}^{2}}{{L}^{2}}}\\ \frac 12\,{
\frac {m \left( z\,g+{{\dot{x}}}^{2}+{{\dot{y}}}^{2}+{{\dot{z}}}^{2}
\right) }{{L}^{2}}}\end {array} \right]
$$</span></p>
| 44
|
classical mechanics
|
The angular momentum and angular velocity roles in precession
|
https://physics.stackexchange.com/questions/602576/the-angular-momentum-and-angular-velocity-roles-in-precession
|
<p>I'm not sure I understand this topic so I would be happy if someone could clarify it for me:</p>
<p>Is precession (say, of a spinning top) generally speaking a change in the <strong>direction</strong> of the angular velocity vector over time? Is that why it is sometimes said that whenever <span class="math-container">$\vec{\omega}||\vec{L}$</span> (angular velocity parallel to the angular momentum) then there is no precession considering there is no torque on the system? In the classical mechanics Landau and Lifshitz book they say that <span class="math-container">$\omega_p$</span> (the "precessing" component of the angular velocity) is (when there is no torque), the projection of <span class="math-container">$\vec{\omega}$</span> onto <span class="math-container">$\vec{L}$</span>, why is that? In that case I see that precession is described as the axis of symmetry of a body tracing a circle around the angular momentum vector. <strong>In other words, I fail to see how and why <span class="math-container">$\vec{L}$</span> and <span class="math-container">$\vec{\omega}$</span> both define whenever there is precession and whenever there isn't.</strong></p>
<p>Thank you!</p>
|
<p>If a symmetrical object is spinning with an angular velocity, ω, then the angular momentum, L = I ω (with both vectors along the axis of rotation). If an external torque (often from gravity) is applied, then the angular momentum vector (and the axis of rotation) swing in the direction of the torque vector: τ = dL/dt.</p>
| 45
|
classical mechanics
|
Why don't we consider pressure as the driving agency for motion, instead of force, because the force applied is on an area/part of the body only?
|
https://physics.stackexchange.com/questions/603949/why-dont-we-consider-pressure-as-the-driving-agency-for-motion-instead-of-forc
|
<p>Because the force applied on a body obviously would act on a particular area, and not the whole body.</p>
|
<p>Because it's a lot easier to work with. You could use pressure to calculate how an object will move/behave but you would have to work out the average pressure acting on the body and multiply it by the surface area which would give you the force anyway.</p>
<p>It is similar to how we model gravity to be acting on an object's center of mass even though in reality it is acting on every atom in the object separately</p>
| 46
|
classical mechanics
|
Is the total energy conserved in a moving reference frame?
|
https://physics.stackexchange.com/questions/4827/is-the-total-energy-conserved-in-a-moving-reference-frame
|
<p>I have a ball attached to a spring and the spring is attached to a wall. There is no gravity for simplicity. In the rest RF the oscillating ball energy is conserved: T + U = const. In a moving RF it is not conserved. I would like to see the shortest answer to the question "Why?".</p>
<p>| <-- --><br>
|/\/\/\/\/\/\/<strong>O</strong><br>
|</p>
<p>=====> $V_{RF}-->$ </p>
<p><strong>Edit:</strong> For those who has doubt - I choose the moving RF velocity equal to the maximum ball velocity, for example. Then the ball with its maximum velocity in the rest RF looks as still and does not have any potential energy in the moving RF (the maximum velocity is attained at the equilibrium position where no force acts). So T = 0 and U = 0. </p>
<p>At the farthest right position the ball looks as moving with $-V_{RF}$ and having a potential energy $kx^2/2$, both energy terms are positive ($v^2 > 0, x^2 > 0$). So now the total energy of the ball is positive and thus is not conserved. It oscillates from zero to some maximum value. Why?</p>
|
<p>Energy is <strong>not</strong> a Lorentz scalar (or Galilean scalar). In different reference frames, the values of energies wil also be different, but this does not mean that there is no energy conservation. The energy is still conserved in <strong>each</strong> reference frame.</p>
| 47
|
classical mechanics
|
Why does the net lift of an airship remain constant when the Kg of air it displaces should fall as altitude increases and pressure decreases?
|
https://physics.stackexchange.com/questions/583498/why-does-the-net-lift-of-an-airship-remain-constant-when-the-kg-of-air-it-displa
|
<p>In the paper, <a href="https://www.researchgate.net/publication/228609262_Development_of_an_Aerodynamic_Model_and_Control_Law_Design_for_a_High_Altitude_Airship_presented_at" rel="nofollow noreferrer">Development of an Aerodynamic Model and Control Law Design for a High Altitude Airship</a> I found a passage that makes no sense to me.</p>
<p>This in regards to the lift of airships.</p>
<p><strong>Note:</strong> pressure altitude = max safe altitude.</p>
<blockquote>
<p>It can be shown that the net lift is constant over all altitudes, up
to the pressure altitude. This is based upon the assumption that the
density of the lifting gas changes at the same rate as the atmospheric
density.</p>
</blockquote>
<p>An airship's lift is based on the mass of air it displaces as per <a href="https://en.wikipedia.org/wiki/Archimedes%27_principle" rel="nofollow noreferrer">Archimedes Principle</a>. How can net lift remain constant when air density falls at increasing altitudes? Although the density of the helium in the airship may also decrease the mass of helium stays constant because it is contained inside the airship.</p>
<p><strong>For example:</strong></p>
<p>0 meters altitude and 15 degrees Celsius.</p>
<ul>
<li>Air pressure in Pascals is 101,325</li>
</ul>
<p>6,000 meters altitude and -24 degrees Celsius.</p>
<ul>
<li>Air pressure in Pascals falls to 41,413</li>
</ul>
<p>If air pressure falls doesn't the mass of the displaced air also fall? And if the mass of displaced air falls wouldn't the airship's lift also fall?</p>
|
<p>Bouyant force is given by</p>
<p><span class="math-container">$F_B = \rho Vg$</span></p>
<p>As the airship rises, the outside air pressure does decrease as well, meaning the helium in the airship expands (which means you are correct in that its density decreases for the same mass). Therefore, the amount of displaced volume of air will also increase. This means the buoyancy force increases.</p>
<p>There must be a mechanism where the pilots balance the air pressure with the helium pressure inside the airship, to maintain a neutral net buoyancy.</p>
| 48
|
classical mechanics
|
Matching torque SI units from two different formula
|
https://physics.stackexchange.com/questions/587464/matching-torque-si-units-from-two-different-formula
|
<p>Torque = r x F = r*F*sin(theta) => So units will be kg*m^2/s^2</p>
<p>Torque = I*alpha => SI units is kg*m^2*rad/s^2</p>
<p>There is a rad unit extra in the second formula. They both should match right because they are calculating torque.</p>
|
<p>Rad or radians has no physical dimensions and so both these equations are correct.</p>
| 49
|
classical mechanics
|
Non Classical mechanic answer to : length of time a thrown object spends in rest before falling down?
|
https://physics.stackexchange.com/questions/3372/non-classical-mechanic-answer-to-length-of-time-a-thrown-object-spends-in-rest
|
<p>When an object is thrown upwards, when it eventually comes to rest and starts falling, for how long is it stationary? What about an particle in electric field having an initial velocity towards it's same charge? That too would come to rest and reverse velocity, the question is for how long is it at full stop with rest? </p>
<p>Classical physics gives a time of 0, but is that correct? Is it really at rest for 0 seconds? That answer seems a bit counter intuitive.</p>
|
<p>Another result is <a href="http://en.wikipedia.org/wiki/Earnshaw%27s_theorem" rel="nofollow" title="Earnshaw's_theorem">Earnshaw's_theorem,</a> which tells us that no particle can ever be held in a stable equilibrium in the presense of any other collection of interacting particles.</p>
| 50
|
classical mechanics
|
max velocity of SHM
|
https://physics.stackexchange.com/questions/4189/max-velocity-of-shm
|
<p>So I think my algebra is wrong somewhere.
Lets say you have an object that goes under SHM with some initial conditions (amplitude is 5m. The piston is at 5m at t = 0 and period is 20 seconds).</p>
<p>Okay so $x(t) = A sin(\omega_0 + \phi)$. $\phi$ for us is 0. The frequency = $0.05 Hz$, angular frequency = $\frac{\pi}{10}$ and period of the piston = $20s$.
The maximum velocity is is when the first derivative $x'(t) = \omega_0 A cos(\omega_0 t)$</p>
<p>The maximum velocity is when the cosine function is 1 at x = 0. So the maximum velocity is $v(t) = A\omega_0$ so that means $cos(\omega_0 t)= 1$ but $\omega_0 = \frac{\pi}{10}$ so the missing value is t. But thats trivial to find, since cosine function is 1. So $t = 20s$</p>
<p>So here is my issue. The maximum velocity can only be at x = 0 (when kinetic energy is max, and potential energy is zero). But the value t = 20s represents the object at x = Amplitude. </p>
<p>So what am I doing wrong?</p>
<p>edit: I just realized that if $x(t) = A sin(\omega_0 + \phi)$ then the initial conditions are not satisfied. When t = 0, sin function goes to 0 and hence everything is 0 but that dosnt satisfy initial conditions. When $t = 0$, displacement should be $5m$ However, if i replace $sin$ with $cos$ then everything works. Is that okay?</p>
|
<blockquote>
<p>But the value t = 20s represents the object at x = Amplitude.</p>
</blockquote>
<p>Check this calculation again. Remember your formula is $x(t) = A sin(\omega_0 t)$, and you should get full credit on your homework.</p>
<p>(or at least mostly full credit. There are many values of $t$ that satisfy $cos(\omega_0 t)=1$, right? you've found only one, which isn't the most general answer you could give.)</p>
| 51
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classical mechanics
|
Force on suspension when hitting a step
|
https://physics.stackexchange.com/questions/574523/force-on-suspension-when-hitting-a-step
|
<p>I'm trying to calculate the magnitude and direction of the force exerted on a wheel when encountering a step. The wheel and the steps are non-deformable.</p>
<p>I found two different approaches:</p>
<ul>
<li><p>energy approach, from which I can calculate the difference in speeds before and after, but I have no idea how to continue from there.
<a href="https://engineering.stackexchange.com/questions/23524/vertical-component-of-force-for-rigid-wheel-hitting-a-step">https://engineering.stackexchange.com/questions/23524/vertical-component-of-force-for-rigid-wheel-hitting-a-step</a></p>
</li>
<li><p>impulse and angular momentum, the procedure described by the user QiLinXue on the bottom of the post bellow roughly makes sense to me, but I have no idea what is the direction of the average force he calculates. I also don't really understand what the velocity change represents.
<a href="https://physics.stackexchange.com/questions/515895/what-is-the-term-for-shock-absorbed-by-car-suspension">What is the term for “shock” absorbed by car suspension?</a></p>
</li>
</ul>
<p>If anyone has any pointers how to continue from here, it would be much appreciated. I'm trying to calculate the vertical force to select the appropriate spring for the given speed.</p>
|
<p>If you had a lone wheel of known size and mass which experiences no friction at the point of contact, you might be able to simulate this situation. The problem is that the force of contact changes in both magnitude and direction as the wheel rises up over the step.</p>
| 52
|
classical mechanics
|
Constaints in Rotation matrix
|
https://physics.stackexchange.com/questions/575225/constaints-in-rotation-matrix
|
<p>In Rigid body rotation, we need only 3 parameters to make a body rotate in any orientation. So to define a rotation matrix in 3d space we only need 3 parameters and we must have 6 constraint equation (6+3=9 no of elements in rotation matrix)</p>
<p>My doubt is if orthogonality conditions R.Transpose(R)=I must be satisfied for rotation matrix (no of constraints=6) and the determinant should be equal to +1 which makes it total 7 constraints, shouldn't be there only two parameters left( instead of 3)</p>
| 53
|
|
classical mechanics
|
Torque when jumping from AT-REST Merry Go Round, Conservation of Angular momentum
|
https://physics.stackexchange.com/questions/577125/torque-when-jumping-from-at-rest-merry-go-round-conservation-of-angular-momentu
|
<p>Say there a child, that stands on the edge of an AT REST merry go round. When they jump off, the child has a linear velocity, and the merry go round begins to turns.</p>
<p>They say there is no net torque on the merry go round/child system and angular momentum is conserved.</p>
<p>I understand the torque applied to the merry go round from the child(force of child x radius of merry go round), but how does the merry go round provide a torque to the child. Doesn't it just apply a force?</p>
<p>This is the same issues as a child jumping on a merry go round that is coming in straight/tangentially to the edge.</p>
|
<blockquote>
<p>I understand the torque applied to the merry go round from the child(force of child x radius of merry go round), but how does the merry go round provide a torque to the child. Doesn't it just apply a force?</p>
</blockquote>
<p>Torques are "just forces", but interpreted about a specific point or axis. If we have an axis of consideration (probably the platform axis is convenient), then a force applied off-axis is also a torque.</p>
<p>So any torque the child applies to the platform, the opposite torque is applied to the child (even though the child doesn't start rotating). The sum of torques and the sum of angular momentum remains zero.</p>
<blockquote>
<p>Why doesn't the child rotate though if they are experiencing a torque?</p>
</blockquote>
<p>Unbalanced torques create a change in angular momentum. Rotations are only one form of angular momentum. Motion that is not collinear with the axis is also a form of angular momentum.</p>
<blockquote>
<p>Does this change the work done by the child also, since now they are accelerating linear and yet gaining angular momentum.</p>
</blockquote>
<p>No. The work done is the same. Just because we account for the motion as having angular momentum in this case doesn't change the energy transfer.</p>
| 54
|
classical mechanics
|
Angular velocity and rotation matrices
|
https://physics.stackexchange.com/questions/585208/angular-velocity-and-rotation-matrices
|
<p>I'm running into some confusion understanding rotations as being instantaneous cross products of angular velocity vectors. I would like to highlight two approaches which seem like they contradict each other, in the hopes that somebody can explain the conceptual error being made.</p>
<p><strong>Approach 1</strong></p>
<p>Don't think about frames. Just imagine I have some vector rotating according to some rotation matrix <span class="math-container">$A(\omega t).$</span>
<span class="math-container">$$\vec{r}(t) = A \vec{r_0}$$</span>
<span class="math-container">$$ A^T \vec{r}(t) = \vec{r_0}$$</span>
Now, I want to find how this changes with time.
<span class="math-container">$$\frac{d \vec{r}(t)}{dt} = \frac{dA}{dt} \vec{r_0}$$</span>
<span class="math-container">$$\dot{\vec{r}(t)} = \dot{A} \vec{r_0}$$</span>
<span class="math-container">$$\dot{\vec{r}(t)} = \dot{A} A^T \vec{r}(t)$$</span>
Now, it so happens that <span class="math-container">$\dot{A} A^T $</span> is an antisymmetric matrix. So I can associate it with the cross product of some <span class="math-container">$\vec{\omega.}$</span>
<span class="math-container">$$\dot{\vec{r}(t)} =\vec{ \omega} \times \vec{r}(t).$$</span>
<strong>Approach 2</strong></p>
<p>Now think about frames.
We have the same rotation matrix which transforms a basis set <span class="math-container">$\vec{e}_j'=A_{ij} \vec{e}_i.$</span></p>
<p>Then for any vector <span class="math-container">$\vec{r}$</span> you have</p>
<p><span class="math-container">$$\vec{r}=r_j' \vec{e}_j' = A_{ij} r_j' \vec{e}_i.$$</span></p>
<p>Now you look at how this changes with time</p>
<p><span class="math-container">$$\dot{\vec{r}}=(A_{ij} \dot{r}_j' + \dot{A}_{ij} r_j') \vec{e}_i.$$</span></p>
<p>We put this in the components of the body-fixed frame</p>
<p><span class="math-container">$$\dot{\vec{r}} = (A_{ij} \dot{r}_j' + \dot{A}_{ij} r_j') A_{ik} \vec{e}_k' = (\dot{r}_k' + \Omega_{kj} r_j') \vec{e}_k'.$$</span>
In the case of a vector that is just rotating, this yields
<span class="math-container">$$\dot{r_i}\vec{e}_i= \dot{A}_{ij} A_{ik} r_j' \vec{e}_k' $$</span>
But when we transform this back to the "space" system we get
<span class="math-container">$$\dot{\vec{r}(t)} = A^T \dot{A} \vec{r}(t).$$</span>
So why do I get that <span class="math-container">$\vec{\omega} = \dot{A} A^T$</span> in the first approach, but <span class="math-container">$\vec{\omega} = A^T \dot{A} $</span> in the second approach?</p>
| 55
|
|
classical mechanics
|
Is it possible that a satellite rotates planet infinitely without air resistance?
|
https://physics.stackexchange.com/questions/593615/is-it-possible-that-a-satellite-rotates-planet-infinitely-without-air-resistance
|
<p>I am ignorant of physics besides college, and I would like to know if it is possible that a moon rotates its mother planet for an infinite time assuming there is no resistance of any form.</p>
<p>I came up with this question since once a moon starts rotating, it keeps rotating according to law of gravity. Isolate the system to the planet and its satellite. If I recall correctly however, second law of thermodynamics says that entropy is always non-decreasing and it is impossible to create an eternal machine. I was thinking about drawing an equivalence between eternal motion of the planet and an eternal motion of a machine, but this is clearly a contradiction; hence, this eternal motion of the planet cannot viewed as an eternal machine.</p>
<p>So what's the difference? What makes the planet to be moving forever given that the initial speed is not 0? I forgot everything in physics, but I am fortunately quite familiar with elementary math, so any mathematical explanation suffices. Thank you very much in advance.</p>
|
<p>In an isolated Kepler system, the orbits will continue indefinitely.</p>
<p>This is not a problem as regards the second law of thermodynamics, as no work is being extracted from the system. The entropy of the system is constant, consistently with the non-decreasing constraint.</p>
<p>(In the real world, of course, there are multiple mechanisms that can cause the orbit to decay or change, including air resistance, <a href="https://en.wikipedia.org/wiki/Mass_concentration_(astronomy)#Effect_of_lunar_mascons_on_satellite_orbits" rel="nofollow noreferrer">irregularities in the gravitational field</a>, interference from other planets, and gravitational-wave emission, among others.)</p>
| 56
|
classical mechanics
|
Are all kinds on Forces between particles only dependent on the distance between the particles?
|
https://physics.stackexchange.com/questions/535953/are-all-kinds-on-forces-between-particles-only-dependent-on-the-distance-between
|
<p>The 'total energy' of a particle in an isolated system is conserved only if all the forces on it are conservative. For a conservative force, we need a force dependent only on the current position of the particle, right? Is every force in the universe like that? Coloumb and Gravitation Forces are like that, and I think the forces which we use to push/pull in daily lives are also coloumbic forces at their core. What about weak forces and nuclear forces? Are they also of this type?</p>
|
<p>No, not all of them, actualy most of them do not. Coulomb and Gravitational Newton force have such simple form because they describe interaction between particles at rest (unmoving) and with no other qualities that scalar electric charge and mass. The situation they are supposed to describe is so simple there isn't much that then can depend on.</p>
<p>For example the magnetic force (which is an aspect of electromagnetic force) also depends on the velocities of the particles. What's more, because of the limited speed of propagation of the electromagnetic field, it actualy depends on how they were moving in the past.</p>
<p>Same with gravitation: if you consider the full equations of General Relativity, you can see that the force depends on the velocity of the object, it's just that for low velocities it can be approximated with the force desbribed by Newton's equation of gravity.</p>
<p>Weak interaction depends on a certain quantum property of particles called chirality. It doesn't affect right-chiral matter particles and left-chiral antimatter particles
at all - it only affects left-chiral matter and right-chiral antimatter.</p>
<p>The field that describes strong interaction (gluon field) follows so complicated equations, that it's difficult to describe this interaction as forces between particles. we have a number of particles affecting the gluon field, and gluon field affecting the particles back, but often in a way than cannot even be calculated, only simulated.</p>
| 57
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classical mechanics
|
What happens if a particle moving in a hollow sphere is given a horizontal velocity
|
https://physics.stackexchange.com/questions/536095/what-happens-if-a-particle-moving-in-a-hollow-sphere-is-given-a-horizontal-veloc
|
<p>Consider a heavy particle moves inside a frictionless hollow sphere with radius a.<br>
The particle is
held on the inner surface of the sphere at z = −a/2, where z is measured upwards from the centre of the sphere. Given a horizontal velocity U. If U squared = 4ag,
find the maximum and minimum height that the particle reaches during the ensuing motion.</p>
<p>There is actually a first question about finding the energy equation of the particle moving in the sphere depends on h(angular momentum) z(vertical displacement) and vertical velocity. I managed to solve it. But I have no idea how to solve the one I give. </p>
<p>z = -a/2 means the position vector makes a angle pi/3 with downward vertical. But what happens after the particle is released with a horizontal velocity U? If the particle is given a horizontal velocity, can I still use the former energy equation? Or will it leave the surface and conduct free fall? Can anyone tell me how to do it?</p>
| 58
|
|
classical mechanics
|
Free falling bodies: why air resistance is not considered?
|
https://physics.stackexchange.com/questions/536270/free-falling-bodies-why-air-resistance-is-not-considered
|
<p>We affirm that free falling bodies are by definition bodies that are subject to the sole force of gravity. We state that free falling bodies fall through equal distances in equal times regardless their different weight. Most of the experiments that show this evidence are executed in normal conditions here on earth; for example, we see that dropping a bowling ball and a ping-pong ball, the two objects touch the ground at the same moment despite their different weight.
What I don’t understand is how we can state something about free falling objects from experiments that are not about free falling objects, because the ping-pong ball and the bowling ball, for example, are also under the influence of air inertia.</p>
|
<blockquote>
<p>we see that dropping a bowling ball and a ping-pong ball, the two objects touch the ground at the same moment despite their different weight.</p>
</blockquote>
<p>Not when I do the experiment. In fact Galileo had a hard time convincing his colleagues that, without air resistance, they would fall at the same rate. His genius was to design experiments which reduced the effect of air resistance to a negligible amount.</p>
| 59
|
classical mechanics
|
Shooting a moving target from a moving target?
|
https://physics.stackexchange.com/questions/536716/shooting-a-moving-target-from-a-moving-target
|
<p>What is the physics of shooting a moving target from a moving target? Bullet turns, target moves, and one have to pre-aim to compensate the difference such as World War II fighter plane. Does it even have a name?</p>
|
<p>Shooting at a moving target, whether or not you are moving, is known as deflection shooting. There is no special name just because the gun platform is also moving.</p>
<p>Besides the obvious things like relative motion, distance and the time of flight of the projectile, you need to know many other things. The attitude of the gun platform and its heave (in a rough sea or a jinking bomber) affect your aim. The airspeed of the plane or wind speed and direction if you are on board a ship or a tank will deflect the path of the projectile. The air pressure, especially the altitude of a plane, will affect the force of the wind, the time of flight and the downward curve due to gravity.</p>
<p>Until computing gunsights came along during WWII, mobile gunnery was a pretty hit-amd-miss affair. Standard Naval and Army practice was to "bracket" your target (fire shots too near and too far) with the first few rounds to get its range, before really letting rip.</p>
| 60
|
classical mechanics
|
Is a free particle one on which there's no NET force or one on which there's no force at all?
|
https://physics.stackexchange.com/questions/537933/is-a-free-particle-one-on-which-theres-no-net-force-or-one-on-which-theres-no
|
<p>I'm getting different definitions from different sources. Some claim that free particles have no forces acting on them at all (i.e. even if a particle has forces acting on it such that they cancel, it's not free). Other sources explicitly state that there is no net force (like <a href="https://www.tau.ac.il/~tsirel/dump/Static/knowino.org/wiki/Free_particle.html" rel="nofollow noreferrer">this</a>). Other sources, when I google about this, state that free particles are "free from external influence" - which is kind of vague and can be interpreted either way.</p>
<p>Can someone resolve the confusion?</p>
|
<p>Forces often come from force fields like gravitational or electromagnetic. These differ in different parts of space, so they may cancel in one point but not in another.</p>
<p>Now, suppose a particle is at a point where all forces cancel (and there's a non-empty set of forces being in superposition). Let's now perturb its position. In general, the forces now won't cancel, and the particle will experience acceleration. Suppose that the forces are all directed into the initial point where no net force acted on the particle. Then this initial point would be the point of <a href="https://en.wikipedia.org/wiki/Mechanical_equilibrium#Stability" rel="nofollow noreferrer">stable equilibrium</a>. Now, if the particle has too small kinetic energy, it will be bound in the potential well, unable to escape it.</p>
<p>I think it's fair to say that in the case described above a particle, even at the point of equilibrium, where all forces cancel, is not free (it can't escape arbitrarily far given arbitrarily long time). So, I think the case where there's no <em>net</em> force, but the set of forces is non-empty shouldn't be included in the definition of a free particle.</p>
| 61
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classical mechanics
|
What are the fields in this problem?
|
https://physics.stackexchange.com/questions/41925/what-are-the-fields-in-this-problem
|
<p>In problem 3 of chapter 2 of Landau Lifshitz "Mechanics," I don't understand the meaning of the fields as defined in the following statement:</p>
<blockquote>
<p>Which components of momentum and angular momentum are conserved in motion in the following fields?</p>
<p>(a) the field of an infinite homogeneous plane,<br />
(b) that of an infinite homogeneous cylinder,<br />
(c) that of an infinite homogeneous prism,<br />
(d) that of two points<br />
etc.</p>
</blockquote>
<p>I just don't get what he's trying to say. Does he mean an electric field produced by a uniformly charged infinite plane, or an uniformly charged cylinder, etc.?</p>
|
<p>Yes, that is what Landau and Lify are getting at. I don't really see another interpretation. I mean, are momentum and angular momentum conserved under the following applied potential fields? I think that is a reasonable interpretation. Haven't looked at Landau and co. Mechanics in a bit, but this seems reasonable. </p>
| 62
|
classical mechanics
|
Spiral path of particle moving under circular force
|
https://physics.stackexchange.com/questions/565672/spiral-path-of-particle-moving-under-circular-force
|
<p>Imagine a situation where a long masless stick is inserted through a ring, and fix one of the ends of the stick. Then, the system is given an impulsive force such that it starts rotating with an initial velocity <span class="math-container">$\omega_0$</span>, the ring will start moving outwards, and thus, the system will experience a negative torque. I want to find the conditions when the system finally comes to (angular) rest: i.e: distance from centre, time elapsed, final velocity of the ring. I have gotten the following two equations using angular momentum change, and simple free body diagrams, and I'm now stuck.
<span class="math-container">$$\ddot r= \dot \theta^2 r$$</span>
<span class="math-container">$$r \ddot \theta = -2 \dot \theta \dot r$$</span>
How do I further reduce these?</p>
<p>Note: Initial distance from axis can be taken as <span class="math-container">$r_0$</span></p>
|
<p><strong>Hint:</strong>
You can write <span class="math-container">$\omega= \dot{\theta}$</span>, and cancel the <span class="math-container">$dt$</span> in the second equation, so it becomes seperable in <span class="math-container">$r$</span> and <span class="math-container">$\omega$</span>.</p>
| 63
|
classical mechanics
|
Are forces from different directions applied to a hanging mass the same?
|
https://physics.stackexchange.com/questions/565901/are-forces-from-different-directions-applied-to-a-hanging-mass-the-same
|
<p>Let's assume a mass that is hanging somewhere. My question is basically:</p>
<ol>
<li>We are applying a force from the side to move the mass sidewards</li>
<li>We are applying a force from below to move the mass upwards</li>
</ol>
<p>Are these forces the same?</p>
<p>In this context, a bit more concise and realistic:
A box, weighing 100 kg, is hanging from a crane. An architect wants to figure out which way it is easier to move that weight.</p>
|
<p>Yes, because the mass <span class="math-container">$m$</span> is a scalar quantity and when you describe Newton's 2nd law in vector form you have</p>
<p><span class="math-container">$$ \vec{F} = m\, \vec{a}$$</span></p>
<p><span class="math-container">$$ \pmatrix{F_x \\ F_y} = m \pmatrix{a_x \\ a_y} = \pmatrix{m\, a_x \\ m\, a_y} $$</span></p>
<p>So if the resulting acceleration is to be the same in both directions (<span class="math-container">$a_x = a_y$</span>) then the forces applied are the same (<span class="math-container">$F_x = F_y$</span>)</p>
<p><strong><sub>The above is in theory only, as it ignores gravity.</sub></strong></p>
<p>With gravity, and with <span class="math-container">$F_x$</span> and <span class="math-container">$F_y$</span> defined as the <strong>applied forces</strong> on the body, you have <span class="math-container">$$ \pmatrix{F_x \\ F_y} + \pmatrix{0 \\ -m g} = \pmatrix{m\,a_x \\ m\,a_y} $$</span></p>
<p>Remember Newton's 2nd law involves the <em>sum</em> of forces applied to a body. In this case, I am adding the weight (pointing downwards).</p>
<p>So the applied force to achieve acceleration is different by the weight of the body</p>
<p><span class="math-container">$$ F_x = m (a_x) $$</span>
<span class="math-container">$$ F_y = m ( a_y + g) $$</span></p>
| 64
|
classical mechanics
|
Disintegration of many particles Landau (Mechanics 3rd Ed page $43$)
|
https://physics.stackexchange.com/questions/571565/disintegration-of-many-particles-landau-mechanics-3rd-ed-page-43
|
<p>On page <span class="math-container">$41$</span> Landau states that the total momentum in the C system is <span class="math-container">$0$</span>.</p>
<p>On page <span class="math-container">$43$</span> for the disintegration of many particles, Landau states: In the C system... every resulting particle (of a given kind) has the same energy...</p>
<p>Why is this?</p>
<p>I understand why the statement is true if the momentum of a primary particle is <span class="math-container">$0$</span>.</p>
| 65
|
|
classical mechanics
|
horizontal motion inside a cone (cylindrical polars)
|
https://physics.stackexchange.com/questions/104836/horizontal-motion-inside-a-cone-cylindrical-polars
|
<p>I have a question from an example we done in lecture</p>
<p>Suppose we have a particle moving inside the surface of a cone given by $r = wz$ where $w$ is a constant,</p>
<p>and also suppose initially the particle is moving horizontally at height $b$ and speed $U$,the lecturer then stated these "facts"</p>
<p>$\underline{\dot r} = U \underline{e_\theta}$ so $\dot r = 0$, $\dot z = 0$</p>
<p>I understand where he got $\underline{\dot r} = U \underline{e_\theta}$, however, I do not understand why $\dot r = 0$ and $ \dot z = 0$, all we know is that $z(0) = b$, and $\dot r(0) = U$, but that does not mean general $\dot z, \dot r$ are $0$, </p>
<p>if someone could clear this up please</p>
<p>edit: <img src="https://i.gyazo.com/9c5d292390687ecc4adf13bbcfc680c5.png" alt="a"> !</p>
|
<p>It says "<em>initially</em>, $\dot r = \dot z = 0$". This is synonymous with $\dot r(0) = \dot z(0) = 0$. Vague wording like that is usually to trick students.</p>
| 66
|
classical mechanics
|
How to reconcile angular momentum measurements?
|
https://physics.stackexchange.com/questions/109211/how-to-reconcile-angular-momentum-measurements
|
<p>Imagine a particle tracing a counter-clockwise circular path on a flat table with a certain speed. The particle is tied with a massless string of length $R$ to a point $P$ at the center of the circular path. Will the particle rotate about $P$ forever at constant speed in the absence of any external force? Consider using different origins to measure the physical quantities.</p>
<p><strong>Measurement Case 1 (origin $O$ at $P$):</strong></p>
<p>If I choose the origin $O$ of a Cartesian coordinate system to be at $P$, the flat table is the xy-plane and the particle rotates about the point $O$ as described. Specifically, the angular velocity $\vec{\omega}$ of the particle points in the direction of the positive z-axis, the position of the particle at any time is specified by the position vector $\vec{r}$, the angle between $\vec{\omega}$ and $\vec{r}$ is always $\frac{\pi}{2}$, and the angular momentum $\vec{L}$ is as follows:</p>
<p>$\vec{L} = \vec{r}\times\vec{p}\;\ldots\text{ definition} \\
\hphantom{\vec{L}} = m\,(\vec{r}\times\vec{v})\;\ldots\text{ definition of linear momentum }\vec{p} \\
\hphantom{\vec{L}} = m\,(\vec{r}\times(\vec{\omega}\times\vec{r}))\;\ldots\text{ definition of linear velocity in terms of angular velocity} \\
\hphantom{\vec{L}} = m\,(\vec{\omega}\,(\vec{r}\cdot\vec{r}) + \vec{r}\,(\vec{r}\cdot\vec{\omega}))\;\ldots\text{ scalar triple product}\\
\hphantom{\vec{L}} = m\,(\vec{\omega}\,(r^2) + \vec{r}\,(0))\;\ldots\;\vec{r} \perp\vec{\omega}\\
\hphantom{\vec{L}} = (m\,r^2)\,\vec{\omega}$</p>
<p>That is, $\vec{L}$ and $\vec{\omega}$ have the same direction, and $\vec{L}$ does not change direction and magnitude while the particle is rotation about $P$. Therefore, the particle will rotate about $P$ forever at constant speed in the absence of any external force because the angular momentum $\vec{L}$ is conserved (the presence of a centripetal force does not affect the angular momentum $\vec{L}$ in any way because a centripetal force is always parallel to the moment arm, and therefore, does not give rise to a torque $\vec{\tau}$).</p>
<p><strong>Measurement Case 2 (origin $O'$ vertically directly beneath $P$):</strong></p>
<p>Now if I choose to describe the same problem by choosing an origin $O'$ of a Cartesian coordinate system to be vertically directly beneath $P$, the particle's position vector $\vec{r}'$ makes an angle $\phi$ with the z-axis and the particle's angular momentum $\vec{L}'$ also makes a $\phi$ angle with the z-axis. As the particle rotates, however, $\vec{r}'$ also rotates about the z-axis, and therefore, the angular momentum $\vec{L}'$ keeps changing direction as shown below:</p>
<p>$\vec{L}' = \vec{r}'\times\vec{p}'\;\ldots\text{ definition} \\
\hphantom{\vec{L}'} = m\,(\vec{r}'\times\vec{v}')\;\ldots\text{ definition of linear momentum }\vec{p}' \\
\hphantom{\vec{L}'} = m\,(\vec{r}'\times(\vec{\omega}\times\vec{r}'))\;\ldots\text{ definition of linear velocity in terms of angular velocity} \\
\hphantom{\vec{L}'} = m\,(\vec{\omega}\,(\vec{r}'\cdot\vec{r}') + \vec{r}'\,(\vec{r}'\cdot\vec{\omega}))\;\ldots\text{ scalar triple product}\\
\hphantom{\vec{L}'} = m\,(\vec{\omega}\,(r'\,^2) + \vec{r}'\,(r'\,\omega\,\cos(\angle(\vec{r}', \vec{\omega})))) \\
\hphantom{\vec{L}'} = (m\,r'\,^2)\,\vec{\omega} + (r'\,\omega\,\cos\phi)\,\vec{r}'$</p>
<p>Because the angular momentum $\vec{L}'$ is not conserved due to changing direction, and a changing $\vec{L}'$ requires the presence of an external force to give rise to a net torque $\vec{\tau}'$ about the point that is used to measure $\vec{L}'$, which is the origin $O'$, the particle will not rotate about $P$ forever in the absence of any external force.</p>
<p>But then, a contradiction arises: <strong>the same phenomenon has a different outcome depending on the choice of origin $O$ or $O'$!</strong> That is <em>unacceptable</em> in Physics!</p>
<p>How to resolve this contradiction?</p>
|
<p>Your error lies in your last paragraph:</p>
<blockquote>
<p>Because the angular momentum $\vec{L}'$ is not conserved due to changing direction, and a changing $\vec{L}'$ requires the presence of an external force to give rise to a net torque $\vec{\tau}'$ about the point that is used to measure $\vec{L}'$, which is the origin $O'$, the particle will not rotate about $P$ forever in the absence of any external force.</p>
</blockquote>
<p>Everything in that sentence is factually correct, but the conclusion, "the particle will not rotate about $P$ forever in the absence of any external force", does not follow from it.</p>
<p>Let's step back and think about what the system actually looks like. In the first case, a particle rotates in a circle with the origin $O$ at its center, presumably due to some force (like gravity towards $O$) which keeps the motion circular. </p>
<p>In the second case, the origin is moved to $O'$ but the system itself is not changed, ie, the particle now rotates in a circle <em>above</em> the new origin $O'$ (again, due to gravity towards $O$).</p>
<p>In the first case, the centrifugal force is parallel to the moment arm, since the particle is attracted to $O$, which is the chosen coordinate origin. Thus $\vec{L}$ is conserved, which makes sense.</p>
<p>However, in the second case, <strong>the centrifugal (gravity) force is no longer parallel to the moment arm</strong>, since the particle is attracted to $O$, not $O'$. Thus there is no reason to expect that $\vec{L}'$ is conserved. Naturally this means that there is a nonzero torque $\vec{\tau}'$ with this choice of coordinates.</p>
<p>What is the physical significance of this nonzero torque? In the first case, the torque $\vec{\tau}$ is zero, because the particle orbits the origin $O$. In the second case, the particle no longer orbits the origin $O'$, but instead orbits a location $O$ above it, and thus <em>from this coordinate viewpoint</em>, there must be a nonzero torque $\vec{\tau}'$, as otherwise if $\vec{\tau}'=0$ the particle would orbit $O'$, rather than $O$!</p>
| 67
|
classical mechanics
|
Two-Body with external force - energy confusion
|
https://physics.stackexchange.com/questions/109239/two-body-with-external-force-energy-confusion
|
<p><strong>Setup</strong></p>
<p>Imagine a two-body system of masses under a classical mechanics model. The separation and mass-ratio doesn't matter for this example. Presume they are initially stationary.</p>
<p>Now suppose that we hold one of the masses, m1, permanently stationary with an external force opposing the gravitational force due to the second mass. This external force changes over time such that it always balances the gravitational force on m1.</p>
<p>Take the frame-of-reference centred on m1 - the stationary mass.</p>
<p><strong>Problem</strong></p>
<ol>
<li>If we think of the two bodies as as system, then we would say that
the centre-of-mass will accelerate from stationary, and will be
displaced in the same direction as the external force. Therefore,
positive work is being done on the system by the external force
(energy is added to the system).</li>
<li>However, consider the bodies individually. For m1, it experiences no
net force. Nor is it displaced. Therefore, we must conclude that no
work is done on m1. m2 only experiences a gravitational force, and
so is accelerated from stationary in the direction of the force.
Therefore work is done on m2.</li>
</ol>
<p>But, if energy is transferred to the system from an external source (as suggested from 1), which of the two objects received this energy?</p>
<p>It can't have been m1, since it was stationary. But how m2 gain this external energy when the external force acted on m1?</p>
|
<p>The system of the two masses does not experience work due to the external force. You are using the wrong displacement. You should use the displacement of the "point of application" of the external force to calculate work for a given force, even if that force is applied to some system. In particular, don't use the center-of-mass displacement.</p>
<p>For your problem, if you do choose to view the situation as a single system composed of two bodies, the displacement of the point of application of the force is zero, and hence no work. The external force does not add energy into this system. The kinetic energy of m2 comes from the stored gravitational potential energy of the system. </p>
<p>Or, if you want to forgo the system treatment, then most people would consider the kinetic energy of m2 coming from the (external) gravitational work done on it by m1 rather than from potential energy.</p>
<hr>
<p>In the treatment above, I'm using the work-energy theorem written in the form
$$W_\text{net,ext} = \Delta K + \Delta U,$$
where the left-hand side is the sum of works done by external forces and I'm being careful to use the displacement of the point of application. The change in potential energy $\Delta U$ arises only from <em>internal</em> interactions, and thus exists only when you consider the two objects as a system. This is not the only way to treat such systems, but it is the one that is most consistent with other areas of physics, like thermodynamics.</p>
| 68
|
classical mechanics
|
upwind vehicle exceeding the wind speed
|
https://physics.stackexchange.com/questions/110395/upwind-vehicle-exceeding-the-wind-speed
|
<p>Has anyone done any research about the upwind vehicle <a href="http://www.popsci.com/cars/article/2012-07/wind-powered-car-travels-upwind-twice-speed-wind" rel="nofollow">http://www.popsci.com/cars/article/2012-07/wind-powered-car-travels-upwind-twice-speed-wind</a>?</p>
<p>I think it is impossible but get a surprising number of disputes from some pretty educated people. It seems to me that if it were possible for a wind powered vehicle to go into the wind at greater than the speed of the wind X, then it would follow that on a windless day, a Ford F-150 towing the same vehicle at speed X (equaling the relative wind), would allow the vehicle to accelerate and pass the Ford, which is absurd, especially considering that the towed vehicle only needs to overcome the rolling coefficient of friction while the self starting vehicle must overcome the static coefficient of friction. </p>
<p>Am I overlooking something?</p>
|
<p>There no hard theoretical upper limit on the true wind multiple a wind-powered vehicle can achieve. It's just a matter of practically achievable efficiency. This applies to all directions, including directly upwind and directly downwind.</p>
<p>The mechanical principle can be demonstrated with a simple gear-toy, where the air mass is replaced with a strip of paper, and the rotor with another set of wheels. Note that by changing the gear ratios you can achieve speeds greater than the paper-speed in both directions:</p>
<p><a href="http://www.youtube.com/watch?v=pw_B2MnMqZs" rel="nofollow">http://www.youtube.com/watch?v=pw_B2MnMqZs</a></p>
<p>Dealing with a fluid like air adds slippage making it less efficient and more difficult to achieve. But that is just an engineering challenge. Nothing in physics prevents it from working. Here some theory on the subject:</p>
<p>Theory and Design of Flow Driven Vehicles Using Rotors for Energy Conversion
Mac Gaunaa, Stig Øye, Robert Mikkelsen</p>
<p><a href="http://orbit.dtu.dk/fedora/objects/orbit:55484/datastreams/file_3748519/content" rel="nofollow">http://orbit.dtu.dk/fedora/objects/orbit:55484/datastreams/file_3748519/content</a></p>
| 69
|
classical mechanics
|
How was it possible to beat a soldier with a full plate armor?
|
https://physics.stackexchange.com/questions/113124/how-was-it-possible-to-beat-a-soldier-with-a-full-plate-armor
|
<p>From my point of view, it seems that a soldier armed with a <strong>full metal plate armor</strong> was almost invulnerable at the time their opponents yielded swords, spears or bows. I understand that it couldn't be the case, but I'm not sure about the physics behind it.</p>
<p>More specifically:</p>
<ol>
<li><p>How could an archer beat this soldier? Would it matter whether the archer aimed at more vulnerable spots at the soldier, or any arrow shooted with full power would do?</p></li>
<li><p>How could a swordsman or a spearman beat this soldier? Would the impact of these weapons suffice?</p></li>
<li><p>Just out of curiosity, a modern gun would penetrate into the plates that easily? One headshot would do?</p></li>
</ol>
<p>I'm thinking on a soldier wearing an armor like this one below, or even more bulkier.</p>
<p><img src="https://upload.wikimedia.org/wikipedia/commons/thumb/3/34/Italian_-_Sallet_-_Walters_51580.jpg/378px-Italian_-_Sallet_-_Walters_51580.jpg" alt="Italian suit of Armor, c.1450">[1]</p>
<p>[1] Courtesy of Wikipedia</p>
|
<p>A crossbow would easily penetrate the armor. The bolt has more kinetic energy than a 45 magnum. </p>
<p>Long bows might penetrate the armor, especially the thinner areas. </p>
<p>Wielding a sword is tiring. Most people would need to rest after 15 minutes, warriors might go 30 minutes before needing to rest. When you are tired, you make mistakes and leave yourself vulnerable to your opponent.</p>
<p>I don't know how a spear would do against the armor.</p>
| 70
|
classical mechanics
|
Does lever needs gravitation to work?
|
https://physics.stackexchange.com/questions/131044/does-lever-needs-gravitation-to-work
|
<p>Simple question - Does lever needs gravitation force to work or it just needs fulcrum and could work in vacuum as well?</p>
|
<p>A lever can work in zero-gravity conditions as long as the fulcrum and lever arm are physically attached. They also work regardless of atmospheric presence.</p>
| 71
|
classical mechanics
|
Which way to lean when driving a gokart?
|
https://physics.stackexchange.com/questions/132476/which-way-to-lean-when-driving-a-gokart
|
<p>Given a car that has two lines of wheels, the center of gravity at constant height above the ground, constant turn angle and given surface and wheel material.</p>
<p>What is the maximum speed the car can drive without the wheels sliding given the position of center of gravity (leaning in, leaning out or central)?</p>
|
<p>I believe the correct approach in a go-kart is leaning outwards.
I think this is useful because go-karts do not have a differential between the left and right wheels. So the wheels are forced to turn at the same rate. Because the turn radius is not the same for the inner and outer wheels there must be some slip. Leaning outwards facilitates this without the go-kart loosing grip.</p>
| 72
|
classical mechanics
|
Concept map/graph
|
https://physics.stackexchange.com/questions/134046/concept-map-graph
|
<p>Where can I find maps of physical concepts illustrating the connection & logical relations between concepts . I think it can be of a great help to summarize what I learnt . </p>
| 73
|
|
classical mechanics
|
How to calculate when an object will fall over
|
https://physics.stackexchange.com/questions/150582/how-to-calculate-when-an-object-will-fall-over
|
<p>TL;DR Given the point of centre of mass, width of base and height, is there a way to calculate the angle where the object will fall over?</p>
<p>The TL;DR of this question pretty much sums it up, however I wanted to give an example.</p>
<p>Let us say that we have a cuboid made of uniform weight wood with the dimensions 10 (height) x 5 (width) x 5(breadth). Due to the fact that it is uniform weight, we can tell the CoM (centre of mass) will be at 5 x 2.5 x 2.5.</p>
<p>If we push the object, it will rotate around a turning point, for this let's say the right side (from our perspective) and it's angle will increase compared to the table. When the CoM reaches over the turning point, it will fall over and collapse.</p>
<p>Is there a mathematical way of working this out? Without experimentation, or are there other variables I need.</p>
<p>Note - First question, and I wasn't sure whether to post it on mathematics or physics. Do you think I posted on the wrong stackexchange site?</p>
|
<p>The answer is actually more simple that you think. The tipping point is reached when the x coordinate of the center of mass is the same as the horizontal position of the last contact point. In the case of your cube there is only a contact point. But for instance, if you had a trapezoid at a table, it would tip if the center of mass is beyond the critical point at the end of any of the two sides of the base. </p>
| 74
|
classical mechanics
|
How do I turn my bicycle?
|
https://physics.stackexchange.com/questions/156197/how-do-i-turn-my-bicycle
|
<p>No doubt a very simple question with an easy answer that's been puzzling me:</p>
<p>If I'm riding my bicycle in the $x$ direction with speed $v$ and turn my handlebars I can end up travelling in the $y$ direction with almost the same speed without having to provide any additional energy. I don't have to come to a halt, so there must be some mechanism coupling my $x$ kinetic energy to my $y$ kinetic energy without the need for a force to decrease one and increase the other separately. But what is this mechanism?</p>
<p>I assume it's the friction of my tires when I turn the wheel but I don't quite see it: this force always seems to act away from my intended direction of travel.</p>
|
<p>Actually unless you lean over you will tip when you turn the wheel. It is the leaning that changes the direction of travel and the handlebars only provide fine tuning of the motion.</p>
<p>The component of gravity perpendicular to the plane of the bike provides for a force that does no work (since its perpendicular to motion also). So the speed is unchanged, but the velocity (vector) rotates until the lean stops.</p>
<p>You can try turning without turning the handle bars by just leaning and it will work just fine.</p>
| 75
|
classical mechanics
|
Taking moments about two different points in a system of forces
|
https://physics.stackexchange.com/questions/165074/taking-moments-about-two-different-points-in-a-system-of-forces
|
<p>If you have a system of forces and you take moments about two different points will the moment be the same?</p>
|
<p>No. You can see now why dealing with torques and angular momentum is so difficult: It depends on what point in space you are taking it with respect to. The torque about one point may be different in both magnitude and direction of another arbitrary point. </p>
<p>If you are dealing with a system in static equilibrium however, it's a requirement that the torque about ANY point you choose is zero. </p>
| 76
|
classical mechanics
|
Are launch angles relative to observers?
|
https://physics.stackexchange.com/questions/171988/are-launch-angles-relative-to-observers
|
<p>Supposed we have someone on a moving platform which is at constant velocity. Lets say the person launches a mass at some speed relative to the platform an some angle with respect to the platform. Does the launch angle differs from say someone on the ground? </p>
<p>I know velocities are relatives, but I really can't see how the launch angle can be relative.</p>
|
<p>The angle between the object being launched and the platform from which it is launched should remain the same to all observers unless they are moving at speeds comparable to the speed of light relative to the platform. At these speeds they will observe length contraction, the length along the axis which the objects are moving relative to each other will appear to be smaller. Because of this, the angle can easily change in different ways depending on the way that the objects are moving relative to each other.</p>
<p>When dealing with small speeds, the angles will look the same for all reference frames as the x component of the velocity of the platform (assuming the platform is moving on the x axis) will add to the velocity of the ball creating the same relative motion forward for the ball. At the same time the y axis stays the same relative to the platform, so the angle measured will be the same.</p>
| 77
|
classical mechanics
|
Definition of kinetic energy without the second Law of Newton
|
https://physics.stackexchange.com/questions/176436/definition-of-kinetic-energy-without-the-second-law-of-newton
|
<p>As I see it, the definition of kinetic energy
$$T= {1\over2} m u^2 \text { where $u<<c$}$$
comes by using the definition of work
$$W= {\int F\cdot\ dx }$$
and we use for the meaning of F(force) the Second Law of Newton:
$$F={dp\over dt}=ma$$</p>
<p>Do I understand correctly that the kinetic energy from this point and on becomes a connection link between Newtonian mechanics, theoretical mechanics(Lagrange, Hamilton) and relativity?</p>
<p>If so, this the question: Can there be a definition of energy without the law of Newton?(it seems to me that to use the law of newton is not wrong but strange to the point of view that comes with theoretical mechanics. We remain somehow bounded to a definition of kinetic energy associated with a law that makes the mass something a little more fundamental from let's say charge-by fundamental I mean that something without mass, Newton says it cannot move or have any type of interaction. But if the charge is another way of interaction-via the electromagnetic fields or potentials- shouldn't we search for the possibility of defining the kinetic energy via the charge. What should become of the potentials then? And why mass in the Newton's law?).</p>
<p>Thank you.</p>
<p>PS:if anyone has a recommendation for further study, he can suggest it.</p>
|
<p>You can define the kinetic energy as you want. As long as they are consistent with each other, definitions are yours to choose.</p>
<p>For example you can take
$$T = \frac{1}{2}m v^2 \, ,$$
as a definition and use it to <strong>prove</strong> Newton's law by requiring energy to be conserved,
$$ \frac{dE}{dt} = \frac{dT}{dt} - \frac{dW}{dt} = m v a - F v = 0\, .$$</p>
<p>If you want a less silly answer, you can use the relativistic definition
$$ T = \sqrt{p^2 c^2 +m^2 c^4} \, ,$$
($c$ is the speed of light.) which stays valid for $m=0$.</p>
<p>About your parenthesis: It should not bother you that mass has a special place in the laws of physics as compared to electric charge for example. It is true that in classical physics the mass is the gravitational charge, but is it also (and more importantly) a measure of the inertia of things. Newton's second law, $F = m a$, tells us that it takes more force to accelerate heavy objects. This has nothing to do with gravitational interactions. See <a href="https://physics.stackexchange.com/questions/173767/what-exactly-is-the-mass-of-a-body-what-determines-it/173780#173780">this</a> post.</p>
| 78
|
classical mechanics
|
name of this bouncing balls separator model
|
https://physics.stackexchange.com/questions/184308/name-of-this-bouncing-balls-separator-model
|
<p><a href="https://www.youtube.com/watch?v=SRGf0Mq2Zwg" rel="nofollow">https://www.youtube.com/watch?v=SRGf0Mq2Zwg</a></p>
<p>I want to read the physical and mathematical model of this "bouncing balls separator " in the above link .</p>
<p>What is name of this experiment so I can search it in literature ?</p>
| 79
|
|
classical mechanics
|
What are possible explanations for the permeability of balloon rubber, PET plastic and other synthetic materials for carbon dioxide?
|
https://physics.stackexchange.com/questions/187868/what-are-possible-explanations-for-the-permeability-of-balloon-rubber-pet-plast
|
<p>Balloons are definitely not gas-tight. Carbon dioxide just leak by the rubber away. A balloon is filled with carbon dioxide. Knot in it. And play. Shrinkage. After an hour or two the carbon dioxide balloon has dramatically shrunk in size. Apparently carbon dioxide creeps much faster through the rubber balloon than "ordinary air".</p>
<p>But why carbon dioxide leaks through balloon rubber, and oxygen and nitrogen barely? Is an oxygen molecule not a whole lot more compact than a carbon dioxide molecule?</p>
<p>Similarly, it is still a hassle to keep bubbles inside plastic bottles, says a friendly product developer of a large beverage manufacturer. PET bottles are light and firmly and cheap, but especially the small disposable bottles of thin plastic, lose their carbon dioxide very quickly. You can get expensive, extra thick 'multilayer' bottles, those thick bottles, but usually is the choice simply filling under higher pressure, combined with short expiration date. The cola in the bottle can be just after the expiry date still safe to drink, but by that time there is not enough carbon dioxide left in order to be tasty.</p>
<p>What are possible explanations for that permeability of the balloon rubber, PET plastic and other synthetic materials for carbon dioxide?
Is one possible explanation that the gas easier 'dissolves' into all kinds of substances - including rubber?
Has the gas leakage through a balloon only to do with the "creep" of the gas molecules through the cracks of the rubber molecules maze?
Or has the gas leakage also to do with the degree of attraction and repulsion between the rubber and the gas molecules?</p>
| 80
|
|
classical mechanics
|
Proof of vertical and horizontal velocity component in projectile motion
|
https://physics.stackexchange.com/questions/191201/proof-of-vertical-and-horizontal-velocity-component-in-projectile-motion
|
<p>Why is it that $v\cdot sin(x)$ gives the vertical component and $v \cdot cos(x)$ gives the horizontal component, where $v$ is the speed? What logic is there behind it, or even better is there a proof to back it up?</p>
<p>I know by drawing a right angled triangle you can find out the components, but I want a deeper explanation than that.</p>
|
<p>Tl;dr: Vector components act like sides of a triangle, where the length of the vector coincides with the triangle's hypotenuse.</p>
<hr>
<p>Mathematically, this follows from the fact that velocity in the plane is an element $v \in \mathbb{R}^2$. Consider a Euclidean coordinate system whose origin coincides with the spatial location of the particle in motion$^1$. Let the vertical $y$-axis be perpendicular to the ground, with increasing values of $y$ corresponding to an increasing altitude. In the standard orthonormal basis,</p>
<p>$$ v = v_x e_x + v_y e_y,$$</p>
<p>where $v_i = v \cdot e_i$. Defining $\theta$ as the angle between $e_x$ and $v$, it follows that the dot product gives</p>
<p>$$v_x = ||v|| ||e_x|| \cos \theta = ||v|| \cos \theta .$$</p>
<p>Determining $v_y$ is fairly simple as well. Because $e_x$ and $e_y$ are orthogonal, it follows that the angle between $v$ and $e_y$ is $\pi / 2 - \theta $. Using the dot product gives</p>
<p>$$v_y = ||v|| ||e_y|| \cos (\pi / 2 - \theta) = ||v|| \sin \theta .$$</p>
<p>It should now be apparent that talking about $v_x$ and $v_y$ is only meaningful when one has defined a coordinate system. They are arbitrary, and and any orthogonal coordinate system is equally valid in defining the velocity's components. In the case of projectile motion relative to flat ground, it is <em>convenient</em> to the coordinate system as defined above.</p>
<ol>
<li>I have disregarded the fact that $v$ is technically a "free" vector with no position, but this is irrelevant for our purposes.</li>
</ol>
| 81
|
classical mechanics
|
Why is a bending rod assumed to be undergoing torsion?
|
https://physics.stackexchange.com/questions/188334/why-is-a-bending-rod-assumed-to-be-undergoing-torsion
|
<p>If I take a rod and bend it at both ends as far as it will go, why is there an assumption that I am also exerting a torsion along with my bending?</p>
<p>Referencee: ccording to the third edition of "Theory of Elasticity": </p>
<blockquote>
<p>"It is important to note that, when a rod undergoes larding
deflections, there is a general twisting of it as well, so that the
resulting deformation is a combination of pure bending and torsion."</p>
</blockquote>
|
<p>Without further context I believe what the book is saying is that when one applies a pure bending moment stress in a beam that the resulting strain is not nessiarily purely in bending, but will also include torsional strain.</p>
<p>For example, if one looks purely at bending moments and assumes that no torsional strain will take place, then a C beam would be just as strong as an I beam. However, if one tried to use a single C beam in place on an I beam it will deflect tortionaly. If this deflection is significant then, the forces will no longer be aligned with the strong direction of the beam and it may cause buckling failure.</p>
| 82
|
classical mechanics
|
How does rotational energy transfer to linear energy?
|
https://physics.stackexchange.com/questions/196946/how-does-rotational-energy-transfer-to-linear-energy
|
<p>So I have recently started looking into moments of inertia, and all that stuff. I have come to a question which has a plane inclined at some angle theta and a sphere at the peak. The G.P.E at the top is mgh. The plane is frictionless.The book then states by conservation of energy $mgh = 1/2(mv^2) + \text{rotational KE from the sphere}$. </p>
<p>Firstly I don't understand how if the ball is initially rotating how it gains rotational energy from the top of the plane to the bottom, since there is no friction to provide torque. (Or how released from rest how it begins to rotate) Similarly is it the case that mass is always treated as if it is all concentrated at the center of mass, because intuitively I would have thought that the individual mass particles making up the sphere might have provided some torque?</p>
<p>Sorry if this is a bit of a basic question but I've only just started the topic.</p>
<p>Any help would be much appreciated!</p>
|
<blockquote>
<p>Firstly I don't understand how if the ball is initially rotating how it gains rotational energy from the top of the plane to the bottom, since there is no friction to provide torque. </p>
</blockquote>
<p>Your understanding is mostly correct. We can choose any point on the object to measure rotation around; let us choose the center of the ball. Both the gravitational force and the normal force point through this center and cannot rotate the ball.</p>
<p>However, "no friction" could <em>also</em> mean "it rolls without slipping and there is negligible rolling friction." In that case, there <em>is</em> static friction providing torque, it just does not do work on the ball. If this is the case, the explanation will involve the moment of inertia $I$ as:$$m~g~h = \frac 12 ~m ~v^2 + \frac 12 ~I~ \omega^2$$and the fact that $\omega = \pm v/R$ will then be substituted to get $$2~g~h = \left(1 + \frac I{m ~R^2}\right) ~ v^2$$For a hollow sphere this ratio $I/mR^2$ is $2/3$; for a solid, uniform ball it is $2/5$ since more of it is closer to the center and therefore easier to rotate.</p>
<blockquote>
<p>Similarly is it the case that mass is always treated as if it is all concentrated at the center of mass, because intuitively I would have thought that the individual mass particles making up the sphere might have provided some torque?</p>
</blockquote>
<p>For any rigid body we can decouple the dynamics into two parts. The first part is the <strong>dynamics of the center of mass</strong>, which can be treated as having a mass equal to the total mass of the rigid body and as having forces equal to the sum of all external forces on the rigid body (internal ones tend to cancel due to Newton's third law, so you can include them too if you're careful to always include them in pairs). If the net force on particle $i$ is Newton's law, $$m_i \frac{d\vec v_i}{dt} = \vec F_{i}, $$ then we can find the center of mass with the formula $\vec R = \sum_i \frac{m_i}{M} ~ \vec r_i$ where $M = \sum_i m_i.$ This in turn moves with a velocity $\vec V = \sum_i \frac{m_i}{M} ~ \vec v_i$, so then $M ~ \frac{d\vec V}{dt} = \sum_i \vec F_i$. If you're paying very close attention you can even see that there's no absolute meaning to $M$, except that it keeps the center of mass $\vec R$ somewhere which is an "average" of all the positions of the little masses, which is nice (but not essential) for understanding the meaning of $\vec V$.</p>
<p>The second part is the <strong>rotation of the body about the center of mass</strong>. This is a little more complicated. Fortunately we have a nice theorem about the rotation group SO(3): every rotation is about some axis, also in 3D space, which it leaves stationary. We can therefore identify rotation with a vector $\vec \omega$, with each particle involved in the rotation having position $\vec r_i$ and velocity $\vec v_i = \vec \omega \times \vec r_i$, which is called the "cross product" of the two vectors. We can sum all of the details up by an "angular momentum" about a fixed point $\vec r_0$, $\vec L = \sum_i m_i~(\vec r_i - \vec r_0)~\times~\vec v_i$. Since $\vec v_i \times \vec v_i = 0$ the sum of Newton's rotational laws can be written, $$\frac {d\vec L}{dt} = M \vec V \times \frac{d\vec r_0}{dt} + \sum_i m_i ~(\vec r_i - \vec r_0) \times \frac{d\vec v_i}{dt} $$ So when $\vec r_0 = \vec R$ and we get a $\vec V \times \vec V = 0$ term, this is simply$$\frac{d\vec L}{dt} = = \sum_i (\vec r_i - \vec R) \times \vec F_i = \sum_i \vec \tau_i.$$ So, we sum together these "torques" about the center of mass and they tell us how this "angular momentum" of the object changes. Then it turns out that the angular momentum always exists in a fixed relationship to $\vec \omega$ given by the <a href="http://farside.ph.utexas.edu/teaching/336k/Newtonhtml/node64.html" rel="nofollow">"moment of inertia tensor"</a> $\mathbf I$ as $\vec L = \mathbf I \cdot \vec \omega$, so any change in the angular momentum creates the corresponding change in the angular velocity vector. </p>
| 83
|
classical mechanics
|
spin-orbit coupling for a rigid body
|
https://physics.stackexchange.com/questions/200574/spin-orbit-coupling-for-a-rigid-body
|
<p>Consider the motion of a coffee cup in the gravitation field of earth. The force acting on the cup apparently depends on the orientation of the cup. Therefore, the internal rotation (with respect to its center of mass) of the cup is coupled with its center-of-mass motion. Is this a classical counterpart of the spin-orbit coupling in quantum mechanics? </p>
<p>To what extent is the center-of-mass motion modified by the internal motion? </p>
| 84
|
|
classical mechanics
|
A metal block in a tub of frictionless ball bearings
|
https://physics.stackexchange.com/questions/197187/a-metal-block-in-a-tub-of-frictionless-ball-bearings
|
<p>If you place a metal block in a tub of small frictionless ball bearings of the same metal, would it stay on top or sink?..</p>
|
<p>The block will have higher density because of the air gaps between the steel balls so it should sink to the bottom.</p>
| 85
|
classical mechanics
|
Limits of integration
|
https://physics.stackexchange.com/questions/200830/limits-of-integration
|
<p>In the following video can someone explain why did he take the limits of integration to be from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$ ?
<a href="https://www.youtube.com/watch?v=bJWFgJTxIFk&index=9&list=PLYVDsiuOZP5pNzoB-e4ugTz96dGpspdDI" rel="nofollow">https://www.youtube.com/watch?v=bJWFgJTxIFk&index=9&list=PLYVDsiuOZP5pNzoB-e4ugTz96dGpspdDI</a></p>
|
<p>The person in the video took $\theta$ to be from the vertical axis. This was stated in the video itself.</p>
<p>You might be more familiar with taking an angle to measure going counter-clockwise from the 3 o'clock position. If you want to measure angles that way then you can go from $0$ to $\pi$ and this was explained by the maker of the video in the comments to the video.</p>
| 86
|
classical mechanics
|
Hausdorff spaces and finite elements
|
https://physics.stackexchange.com/questions/202505/hausdorff-spaces-and-finite-elements
|
<p>Must the shape functions and the interpolation functions (which are the same in an isoparametric element) in a finite element model be elements of a Hausdorff space? If so, is this necessary to ensure convergence to a unique solution when the mesh is refined? If not, what are the requirements for these functions?</p>
| 87
|
|
classical mechanics
|
Why must an inertial navigation system take the Coriolis effect into account?
|
https://physics.stackexchange.com/questions/205126/why-must-an-inertial-navigation-system-take-the-coriolis-effect-into-account
|
<p>I read somewhere that an inertial navigation system, in order to be accurate, must take the Coriolis effect into account. Why is this so? If I go a 500 mph velocity in a given direction, I'm going 500 mph, so in 1 hour, I should be 500 miles from the airport whether I'm traveling N, E, or W. </p>
|
<p>To a stationary observer on Earth watching you travel in the same direction in which the Earth rotates, you might be going 500 mph. Galilean relativity says that you are moving 500 mph relative to Earth, and the Earth is moving 500 mph relative to you.However, since the Earth is rotating, that stationary observer is also rotating. So if you're travelling in a different direction, what would otherwise be 500 mph is now 500 + how quickly the Earth is rotating.</p>
| 88
|
classical mechanics
|
A system of particles interacting via conservative forces, will also respect $U_2+K_2=U_1+K_1$
|
https://physics.stackexchange.com/questions/205145/a-system-of-particles-interacting-via-conservative-forces-will-also-respect-u
|
<p>Studying vector calculus you learn to prove that a particle moving in a gravitational field will, in that field respect that $dU=-dW$. From this you can conclude $U_2+K_2=U_1+K_1$. </p>
<p>This is easy to prove in here but I fail to see how to prove it for, suppose, $n$ charged particles or massive particles.</p>
<p>How can I prove such a thing? Namely, prove that $U_2+K_2=U_1+K_1$ is true if the nature of all the forces in your system are conservative (irrotational). Obviously here $U_j=\sum_i U_i$ and $K_j=\sum_i K_i$, i.e., at snapshot (2) $U_2$ is the sum of all the potential energies in the system, and the same for kinetic energy.</p>
|
<p>A force is conservative iff there exists a potential $\Phi$ such that ${\bf F} = -\nabla\Phi$. The Lagrangian for a system of $n$ particles acting under (any number of) conservative forces can be written on the general form</p>
<p>$$L = \sum_{i=1}^n \frac{1}{2}m_i{\bf \dot{r_i}}^2 - \Phi({\bf r_1},{\bf r_2},\ldots,{\bf r_n})$$</p>
<p>which leads to the equation of motion</p>
<p>$$m_i{\bf \ddot{r_i}} = -\nabla_{{\bf r_i}}\Phi$$</p>
<p>Multiplying with ${\bf \dot{r_i}}$ and summing over all particles gives us the desired result</p>
<p>$$\frac{d}{dt}\sum_{i=1}^n\frac{1}{2}m_i{\bf \dot{r_i}}^2 = -\sum_{i=1}^n{\bf \dot{r_i}}\cdot\nabla_{\bf r_i}\Phi \equiv -\frac{d\Phi}{dt} \implies K+U={\rm const.}$$</p>
<p>where $K=\sum_{i=1}^n\frac{1}{2}m_i{\bf \dot{r_i}}^2$ and $U = \Phi$ are the total kinetic and potential energy respectively. This argument covers the case where we have more than one force in play for which we can write $\Phi=\Phi_{\rm force~1}+\Phi_{\rm force~2}+\ldots$.</p>
| 89
|
classical mechanics
|
Can we change the rotation speed of Earth if all cars, planes, ships, etc travel towards east or west together?
|
https://physics.stackexchange.com/questions/207444/can-we-change-the-rotation-speed-of-earth-if-all-cars-planes-ships-etc-travel
|
<p>I know when cars drive, there is action and reaction pair on the wheel, then is it possible to speed up or slow down the rotation of Earth if all cars and other machines travel towards one direction together?</p>
|
<p>Earth + machines is an isolated system so angular momentum is conserved
\begin{align}
I_{\text{Earth}}\omega_{\text{Earth}}^{\text{before}} &= I_{\text{Machine}}\omega_{\text{Machine}} + I_{\text{Earth}}\omega_{\text{Earth}}^{\text{after}}\\
\Rightarrow
&\frac{\omega_{\text{Earth}}^{\text{after}}}{\omega_{\text{Earth}}^{\text{before}}} =1 - \frac{I_{\text{Machine}}\omega_{\text{Machine}}}{I_{\text{Earth}}\omega_{\text{Earth}}^{\text{before}}}
\end{align}
Therefore the percent change in earths angular rotation speed is
\begin{align}
\frac{\omega^{\text{after}}_{\text{earth}}}{\omega^{\text{before}}_{\text{earth}}}=\frac{I_{\text{Machine}}\omega_{\text{Machine}}}{I_{\text{Earth}}\omega_{\text{Earth}}^{\text{before}}} &=\frac{5}{2}\frac{M_{\text{Machine}}}{M_{\text{Earth}}} \frac{v_{\text{Machine}}}{R\omega_{\text{Earth}}^{\text{after}}} \approx \left(10^{-27} \text{sec}/ \text{kg}\cdot\text{m}\right) \times p_{\text{Machine}}
\end{align}
In other words assuming all machines are moving at the equator, their momentum must be of the order of $10^{27}\text{kg}\cdot\text{m}/\text{sec}$ to get a $1\%$ change in earths angular speed, i.e. to increase/decrease the day by around a quarter of an hour.</p>
<p>In comparison notice that if all the cars on earth where to move at a very illegal speed, their combined momentum wont exceed $10^{14}\text{kg}\cdot\text{m}/\text{sec}$ which could only change the length of the day by less than $10$ nano seconds...</p>
| 90
|
classical mechanics
|
How physics opens a knife with a flick of the wrist?
|
https://physics.stackexchange.com/questions/212150/how-physics-opens-a-knife-with-a-flick-of-the-wrist
|
<p>A person holds the handle of a folding knife in their hand with the pivot point of the blade between their index finger and thumb. The blade points away from the body and the other fingers grip the handle in a way not to impede the opening of the blade. The friction at the pivot point is just enough for the blade to not open by gravity. The wrist rotates 120 degrees horizontally as fast as humanly possible, accelerating throughout the motion. Then the rotation stops as fast as humanly possible. Shortly after the rotation stops the folding blade has swung open 180 degrees at the pivot point and locks in place. </p>
<p>Question what physics force(s) opened the knife?</p>
<p>Observation: The heavier the blade the easier it is to open with a flick of the wrist</p>
<p><a href="http://www.digitalcanoe.ca/flick.jpg" rel="nofollow noreferrer">Folding knife http://www.digitalcanoe.ca/flick.jpg</a></p>
|
<p>Think of the example of a passenger in a car. When the car is moving in uniform motion (acceleration = 0) the passenger is not moving relative to the car. However if the car slows down or speeds up, the passenger begin to resist the motion of the car (i.e. car slows down passenger moves forward relative to the car). [Newton's first law]</p>
<p>Now picture the blade as the passenger and the handle as the car. When you accelerate the handle to the right the blade is resisting that motion and is moving toward the left relative to the blade, however the pivot point forces the blade into the handle.</p>
<p>When you bring the blade to a stop, you accelerate it to the left, making the blade move right relative to the blade. This pivot point forces the blade to rotate upwards.</p>
| 91
|
classical mechanics
|
Rotating disc problem
|
https://physics.stackexchange.com/questions/218166/rotating-disc-problem
|
<p>This question comes into my mind this evening. </p>
<p>Suppose I have a rotating disc whose maximum rotation speed is $500$ rpm (say). On this rotating disc I have placed another small rotating disc with same property and in such a way that both can rotate in this configuration. Now if I started the disc below, it starts to rotate with it maximum speed with the smaller disc on its top. After some time i start the smaller disc and it also starts to rotate with maximum speed. So my question is what kind of effect the larger disc will put on the smaller disc if (a) they both rotate in same direction (b) one disc clock wise and other anti-clock wise. </p>
<p>N.B.- I am not a physics student so don't know how to interpret. </p>
|
<p>If they rotate in the same direction at the same speed, there will be no effect. This is because the disks do not move relative to one another.</p>
<p>If they rotate at different speeds or in different directions and there is friction between the disks, their speeds will gradually come closer to one another's until their rotation rates are the same.</p>
<p>A good way to think about this is to imagine you are standing on one of the disks. Then the disk you are standing on is holding still, from your perspective. If the other disk is moving, friction will slow it down until it eventually stops.</p>
| 92
|
classical mechanics
|
Probability of light being refracted in deep space by H 2(Hydrogen)
|
https://physics.stackexchange.com/questions/228538/probability-of-light-being-refracted-in-deep-space-by-h-2hydrogen
|
<p>Is it a valid question to ask what is the percentage of light that gets refracted by H 2 for a certain distance(Lets say 1 light year) in deep space. Lets assume a few(3) Hydrogen atoms per cubic meter. </p>
| 93
|
|
classical mechanics
|
When an object is moving at a constant velocity,would the normal force and $mg$ be equal?
|
https://physics.stackexchange.com/questions/228701/when-an-object-is-moving-at-a-constant-velocity-would-the-normal-force-and-mg
|
<p>Does the object's normal force and $mg$ cancel out, resulting in the two force becoming equal, or would one force be greater than the other? Thank you!
Edit: Also would the $mg$ be considered weight or would it be more correct to just refer to the force as $mg$? </p>
|
<p>If the object is moving on a horizontal frictionless surface, then the normal force equals $mg$.</p>
<p>Yes, $mg$ is the weight of the object.</p>
| 94
|
classical mechanics
|
Detached wheel from a moving vehicle speed
|
https://physics.stackexchange.com/questions/228301/detached-wheel-from-a-moving-vehicle-speed
|
<p>If a wheel is detached from a moving vehicle, would it speed be higher than the vehicle? Why?</p>
|
<p>Assume the car travels at speed $v$ and the wheel has a radius $R$. Assume also that the wheel was rolling without slipping, then:</p>
<p>$$v=\omega R,$$</p>
<p>where $\omega$ is the <em>angular speed</em> of the wheel.</p>
<p>When the wheel detaches from the car and assuming no torques or forces act on it from then on, then Newton's Laws tell us the <em>state of motion of the wheel will not change</em>: $v$ and $\omega$ will be conserved, <em>'forever'</em>.</p>
<p>In the real world, forces like air drag, bumps in the road and rolling resistance <em>do</em> cause the wheel to decelerate somewhat and eventually come to a halt.</p>
| 95
|
classical mechanics
|
How am I able to keep my footing on an accelerating platform?
|
https://physics.stackexchange.com/questions/226483/how-am-i-able-to-keep-my-footing-on-an-accelerating-platform
|
<p>When I'm standing in a train car and the train starts slowing down relatively quickly, I instinctively flex certain muscles in my legs and that helps me keep my footing. </p>
<p>What muscles am I flexing and how does flexing them help me maintain my balance and avoid falling?</p>
|
<p>I'll answer this as a physics question. To avoid falling over, you need to keep your center of gravity "over" your feet, where "over" means that the vector sum of gravitational and train acceleration points from your $CG$ down to between your feet. To do this, you lean towards the front of the train so as to move your $CG$ forward in order to counteract the vector acceleration's pointing backward.</p>
<p>Once the train has reached its cruising speed and stopped accelerating, that vector will again be just due to gravity, so you'll straighten up to move your $CG$ back to its usual stable position. When the train approaches the station and starts decelerating, you'll lean towards the rear of the train to compensate, and when the train finally stops moving you'll again straighten up.</p>
| 96
|
classical mechanics
|
Kinetics of gas molecule
|
https://physics.stackexchange.com/questions/231889/kinetics-of-gas-molecule
|
<p>In the explanation of the nature of gases we use the kinetic formula $PV=1/3(mnc)$
where $p$ is pressure, $v$ volume, $m$ mass, $n$ number of molecules and $c$ means root mean square velocity. But in its explanation I don't understand why everyone uses time to measure the change of momentum of a particular molecule, and in that case everyone uses the time for the molecule to go from one side to another, but it is true that the molecule has a constant velocity.
In order to prove the equation we firstly think there r some gas in a cubic container.Lets think about a single gas
molecule.it is moving with it's kinetic energy.suppose the leangth of the cubic container is $l$ and the molecule needs t second to reach from one side of the container to other.suppose the molecule doesn't hit other molecule bt the wall.so before hitting the wall its velocity doesn't change.bt within short time its velocity changes its direction after hitting the wall and moves
towards the opposite side of the container.now i want to measure the action and reaction force by using the change of momentum.</p>
<p>When that hits the wall the direction of velocity changes within a short time limit to 0 as it is a spontaneous collision, so the change of momentum must be measured by using this time not the time for the molecule to go from one side of the wall of a container to the other side. Am I right or not?</p>
| 97
|
|
classical mechanics
|
If a bullet penetrates a bag, how come the repulsive force is constant?
|
https://physics.stackexchange.com/questions/232326/if-a-bullet-penetrates-a-bag-how-come-the-repulsive-force-is-constant
|
<p>I was doing a question on energy and forces, and it goes as follows:(Doesn't require knowledge of calculus):</p>
<blockquote>
<p>If a bullet with velocity $v$ penetrates a bag upto a distance $x$,
then find the distance which bullet penetrates if velocity becomes
$2v$.</p>
</blockquote>
<p>I tried to solve it, like this: </p>
<p>$
1/2mv^2 = -F * x$ because energy is conserved and retarding force given by bag would be negative.</p>
<p>So I multiplied both sides by 4, because velocity becomes 2 times, and the square makes it 4. Then, on the R.H.S of the equation, I got the distance to become 4 times, if force remains constant in both cases. And, my answer came out to be correct. However, I was thinking why would force be constant. By Newton's third law of motion, bullet with higher velocity will provide a greater force, but we are taking force to be constant. And, what are the pairs of equal and opposite forces in this case? If second bullet provides a greater force, shouldn't the bag provide the same greater force?</p>
|
<p>The force is most likely not constant but your teacher wanted you to be able to do the problem in not too long a period of time.</p>
| 98
|
classical mechanics
|
Can point mass have vibrational motion?
|
https://physics.stackexchange.com/questions/234561/can-point-mass-have-vibrational-motion
|
<p>Can a point mass have vibrational motion. I have read that reason for point mass is to ensure that we can idealize translational motion and don't have to worry about rotational and vibrational motion. Is that correct?</p>
|
<p>It depends what you mean by vibrational motion. A point mass indeed does not have <em>internal</em> vibrational degrees of freedom, it can just have translational motion, as you pointed out.</p>
<p>However, if you attach this point mass to a spring you create a harmonic oscillator. It has a translational motion, but this oscillating motion can describe <em>vibrations</em> such as vibrations of atoms around their equilibrium state in a solid.</p>
| 99
|
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