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changed too.
Now we pause briefly to note that our foregoing introduction oŸtorque, through
the idea of work, gives us a most important result for an object in equilibrium: 1Ý
all the forces on an object are in balance both for translation and rotation, not
only is the net ƒorce zero, but the total of all the #orgues is also zero, because if an
object is in equilibrium, 0ö 0ork ?s done bụ the ƒorces [or a small đisplacement.
Therefore, since AW = r AØ =0, the sum of all the torques must be zero. So
there are two conditions for equilibrium: that the sum of the forces 1s zero, and
that the sum of the torques is zero. Prove that it suffices to be sure that the sum
of torques about any øne axis (in two dimensions) is zero.
Now let us consider a single force, and try to ñgure out, geometrically, what
this strange thing # — „ amounts to. In Eig. 18-2 we see a force ' acting at
a point r. When the object has rotated through a small angle A0, the work done,
of course, is the component of force in the direction of the displacement times the
Fig. 18-2. The torque produced by a force.
--- Trang 340 ---
displacement. In other words, ¡È is only the tangential component of the force
that counts, and this must be multiplied by the distance r A0. Therefore we see
that the torque is also equal to the tangential component oŸ force (perpendicular
to the radius) times the radius. That makes sense in terms of our ordinary idea
of the torque, because 1f the force were completely radial, ít would not put any
“twist” on the body; it is evident that the twisting efect should involve only the
part of the force which is not pulling out from the center, and that means the
tangential component. Eurthermore, it is clear that a given force is more effective
on a long arm than near the axis. In fact, if we take the case where we push right
ơn the axis, we are not twisting at alll So ¡it makes sense that the amount of
twist, or torque, is proportional both to the radial distance and to the tangential
component of the force.
There is still a third formula for the torque which is very interesting. We
have Jjust seen that the torque is the force times the radius times the sine of the
angle œ, in Fig. 18-2. But if we extend the line of action of the force and draw
the line Ø5, the perpendicular distance to the line of action of the force (the
lcuer œrm of the force) we notice that this lever arm is shorter than r in just the
same proportion as the tangential part of the force is less than the total force.
Therefore the formula for the torque can also be written as the magnitude of the
force times the length of the lever arm.
The torque is also often called the mmomen# oŸ the force. "The origin of this
term is obscure, but it may be related to the fact that “moment” is derived from
the Latin mouữnentum, and that the capability of a force to move an object
(using the force on a lever or crowbar) increases with the length of the lever arm.
In mathematics “moment” means weighted by how far away it is om an axis.
18-3 Angular momentum
Although we have so far considered only the special case of a rigid body,
the properties of torques and their mathematical relationships are interesting
also even when an object is not rigid. In fact, we can prove a very remarkable
theorem: just as external force 1s the rate of change of a quantity ø, which we
call the total momentum of a collection of particles, so the external torque is the
rate of change of a quantity Ù which we call the angular mmormnentum oŸ the group
of particles.
To prove this, we shall suppose that there is a system of particles on which
there are some forces acting and fñnd out what happens to the system as a result
--- Trang 341 ---
O7
Fig. 18-3. A particle moves about an axis Ó.
of the torques due to these forces. First, of course, we should consider just øne
particle. In Fig. 18-3 is one particle of mass ?n, and an axis Ó; the particle is not
necessarily rotating in a cirele about Ó, it may be moving in an ellipse, like a
planet going around the sun, or in some other curve. Ït is moving somehow, and
there are Íorces on it, and it accelerates according to the usual formula that the
#-component of force is the mass times the z-component of acceleration, etc. But
let us see what the #orgue does. The torque equals ø„ — ;„, and the force in
the ø- or -direction is the mass times the acceleration in the z- or -direction:
T—#Èu T— Uy =
= zm(dŠu/dt2) — m(d°+/d12). (18.14)
Now, although this does not appear to be the derivative of any simple quantity,
1E is in fact the derivative of the quantity zrm(dụ/đf) — ym(d+z/dÐ):
d dụ d+z dˆụ + d+z dụ
— |zm| —- | —m| — || —=zm| —> — |Jm| —-
dt dt) ”“Vi di? di dí
(18.15)
d2 dụ dœ d2 d2+z
—m| —c |— | TT |m| —| =zm| —— ]_—-m| —- |:
MAV di di d2) — ”“Ẳp
So ït is true that the torque is the rate of change of something with timel So we
pay attention to the “something,” we give it a name: we call it b, the angular
1mmomentum:
= ~zm(dụ/đĐ) — ym(d+z/dt)
= #Ðụ — Da. (18.16)
Although our present discussion is nonrelativistic, the second form for Ù, given
above is relativistically correct. So we have found that there is also a rotational
--- Trang 342 ---
analog for the momentum, and that this analog, the angular momentum, is given
by an expression in terms of the components of linear momentum that is jus$
like the formula for torque in terms of the force componentsl Thus, iŸ we want
to know the angular momentum of a particle about an axis, we take only the
component of the momentum that is tangential, and multiply it by the radius. In
other words, what counts for angular momentum is not how fast it 1s goïng œa
from the origin, but how much it is going around the origin. Only the tangential
part of the momentum counts for angular momentum. Eurthermore, the farther
out the line of the momentum extends, the greater the angular momentum. And
also, because the geometrical facts are the same whether the quantity ¡s labeled
por F) it is true that there is a lever arm (nø the same as the lever arm of the
force on the particlel) which is obtained by extending the line of the zmomentưm
and fñnding the perpendicular distance to the axis. Thus the angular momentum
is the magnitude of the momentum tỉmes the momentum lever arm. So we have
three formulas for angular momentum, just as we have three formulas for the
torque:
Ù = tDụ — UPz
— Ttang
= p- lever arm. (18.17)