For simplicity, in the beginning we are going to focus on rigid moving with its one point fixed. Thus it will change only by changing its orientation. We will further simplify the problem by considering rotation about an axis fixed in space. In the next step, we will allow the axis to translate but without changing its orientation. Finally we will also let the orientation of the axis change. Thus we will increase the complexity of the problem gradually.


Dynamics of rigid body: The dynamics of a rigid body is best described by considering its angular momentum. You can think of angular momentum as the rotational counter part of linear momentum. This quantity is central to describing rotational motion of a rigid body. So let us first spend some time in understanding this quantity. Although we are introducing angular momentum here in the context of rigid bodies, the treatment below is quite general.

For a single particle moving with linear momentum at a distance from the origin the angular momentum is defined as

You can immediately see that it is an origin-dependent quantity. If we calculate it with respect to some other point, it will come out to be different. If a particle of mass m is moving in a plane then using the polar coordinates for it, it is easily shown that its angular momentum is . Let us now find out what is the rate of change of angular moment? It is calculated below.

With , where is the force on the particle, the equation above is simplified to

Thus rate of change of angular momentum is equal to the torque applied on the body. From the equation above, the law of conservation of angular momentum follows immediately: If the applied torque the angular momentum does not change, i.e. it is a constant. The equation

is the angular momentum equivalent of Newton 's II^nd law. Let us now illustrate the ideas presented so far with the example of a conical pendulum.

Example 1: A conical pendulum is like the regular pendulum with a light (mass m = 0 ) rigid rod carrying a bob of mass m at one of its ends. The other end is fixed and the bob moves in a circle with speed v (see figure 6). We wish to calculate the tension in the rod and the angle θ it makes from the vertical by applying the angular momentum-torque equation.

Let us first calculate angular momentum about point O . We will use cylindrical co-ordinates because of the symmetry of the problem. With respect to O

The vector looks as shown in figure 7, when the bob of the pendulum is in the paper plane.

So the angular momentum is perpendicular to the rod (take the dot product with for mass m and see for yourself) and as the particle rotates the horizontal component of are rotates with it and the vertical component remains a constant. Let us now apply the equation

We have

We now calculate the torque acting on the pendulum. There are two forces, the tension and the weight , acting on the particle as shown in figure 8.

But passes through O and does not give any torque. Thus

Substituting these in the angular momentum-torque equation then gives

The angular momentum-torque equation therefore gives us the angle θ that the pendulum makes with the vertical. How do we find the tension T ? On the other hand, applying Newton 's second Law we get

giving

These equations give us both T and θ, but the equation gives only the angular relationship. Does this mean that the angular-momentum torque equation is not equivalent to Newton 's second law? The answer is that it is. It so happens that in applying the equation about O, when cross products are taken, some components of the force do not contribute to the torque and drop out of the equation. For example in this case becomes zero. To get full solution, therefore, we now apply about point A. Taking A as the origin we have

Since all the quantities in are constants, we have

Let us calculate the torque about A. With A as the origin, the forces are given as

Therefore

which gives

Thus applying about two different points gives exactly the same solution as that obtained from . Thus the two ways of solving the problem are equivalent. Through this example I have shown you (a) the origin dependence of , and (b) equivalence of and .

Let me now illustrate conservation of angular momentum by a well known example: that of Kepler's Law of equal area concept in equal time. Accordingly, when planets are going around the sun, the rate at which their position vector from the sun sweeps the area is a constant. Recall from the lecture on polar coordinates that for a particle moving under a radial force, we had obtained that is a constant. This is nothing but two times the rate of area sweep by the radius vector. We now want to get this law from the conservation of angular momentum.

For a planet, we know that the force is in redial direction. So that the torque

Thus

Since , its constancy means

which is Kepler's second law.