Example 1: A thin massless rod of length 2l has a point mass m at both its ends. It is rotating with angular speed w about a vertical axis passing through its centre and at an angle θ from it, as shown in figure 1. If the axis of rotation is held at its two ends by ball bearings, calculate the force that the ball bearings apply on the axis. The ball bearings are placed symmetrically from the centre of the rod at a distance d each.

Recall from the previous lecture that I had taken the principal axes (1,2,3 ) with (1,2) as shown in figure 1 and axis 3 perpendicular to them. The moments of inertia about the principal axes are

The angular velocity and the angular momentum of the rod-mass system are

and

All the parameters - mass m , length l and angle θ - in the equation above are constant so the magnitude of the angular momentum is also a constant. As such we can apply the formulae given above to get the components of the torque to be applied as

Thus the torque needed to keep the rotating rod in its position is in the direction of principal axis 3 of the body. As was noted above, the torque is indeed perpendicular to . The torque is provided by the forces applied by the bearings. When the rod is in the plane of the paper, as shown in the figure, the force would be to the left at the upper end and to the right at the lower end of the rod (see figure 1). And their magnitudes will be equal since the CM of the rod has zero acceleration. Thus the forces provide a couple equal to . Their magnitude is

There is another method of calculating that we describe now. has one component in the direction of and the other component   perpendicular to (see figure 2).

As the rod rotates L_V remains unchanged but L_H sweeps a circle with angular frequency . The rate of change of is therefore the same as that of L_H . The magnitude of the latter is ωL_H . Since at the position shown, the tip of L_H is moving out of the paper, the direction of the change in L_H   is also the same. This is the direction of principal axis 3. It thus follows that

in the direction of principal axis 3. For completeness I also calculate the kinetic energy of the rod-mass system. It is

I now give you a couple of exercises similar to the problem above.

Exercise 1: In the problem above, if the axis of rotation passes through a different point than the centre of the rod (see figure 3), what will be the forces applied by the bearings with everything else remaining the same? ( Hint: the CM is now moving in a circle )

Exercise 2: For the rotating objects shown below in figure 4, calculate the rate of change of their angular momentum by the two methods employed in the example above.

If you have followed the example above, and have also done the exercises suggested, then you will be in a position to understand the explanation of two of the three observations I started my previous lecture with. The two observations were the precession of a spinning top and only one roller of the three shown being able to go over a curved track entirely.

Example 2: Let me take the case of the precession of a spinning top. In this case we observe that when a spinning top is put on a floor and its lower point is held at one point, it starts precessing about the vertical axis (see figure 5)

I take the mass of the top to be m, its moment of inertia about the spinning axis I , distance of its CM from the pivot point l and its spinning rate to be ω_s. The top's axis is making an angle θ from the vertical. Let us take the rate of precession, i.e. the angular speed at which the top starts to rotate about the vertical to be Ω. It is observed that Ω is usually much smaller than ω_s. So in calculating angular momentum we are going to take it as arising from the spin only and neglect any contribution of Ω to it. The angular momentum is then along the spin axis of the top and its magnitude is , where I is top's moment of inertia about its axis. Further, there is torque acting on the top due to its weight. The magnitude of the torque is mgl sinθ and it is perpendicular to the plane formed by the vertical and the spin axis (the direction of ). At the position shown in figure 5, the torque is going into the plane of the paper. The problem then reduces to the following. A rigid body has an angular momentum and is being acted upon by a torque of magnitude mgl sinθ perpendicular to . What will happen to the body?

Since the angular momentum is being acted upon by a torque perpendicular to it, it changes continuously with time with its magnitude remaining unaffected. Thus it moves on the surface of a cone as shown in figure 6.

Let me now calculate the frequency of rotation of vector . For this I again look at the vertical L_V and horizontal L_H components of the angular momentum, as shown in figure 6. The vertical component remains unchanged and the horizontal component changes at the rate as the vector rotates. This gives , which should be equal to the torque. Substituting , I thus get

This is the rate at which the vector rotates. Since is attached to the top, the top also rotates at the same rate. is then the rate of precession of the cone.

As the top precesses, its CM moves in a circle. You may now wonder where does the centripetal force for this come from? This is provided by the horizontal reaction or the frictional force at the pivot point. Second question you may raise is why is it that the component L_H starts moving in a horizontal circle due to the torque while the vertical component does not move in a vertical circle. In the actual motion, it does. So in addition to the precession, the axis of the top also oscillates up and down with very small amplitude. If you are careful in you observations, you will see this motion. This is known as the nutation of the top. In our present treatment, we have ignored this motion and solved the problem only to get the precession rate.

I now wish to explore if to get this answer, I could equivalently have used the equations

To do this, let me first identify the principal axes of the cone at the pivot point and label them. The principal axes are the spin axis and two other axes perpendicular to it. These are shown and labeled (1,2,3) in figure 7; in this position axes 1 and 2 are in the plane of the paper and axis 3 is coming out of it.

The moments of inertia about the principal axes are . The components of at the instant (I take it to be time t = 0 ) shown in figure 7 are

Substituting the values of moments of inertia and the angular velocity components in the equations for the components of the torque gives

This is not the same answer as obtained earlier. Where have we gone wrong? Is the previous answer correct or is this answer correct? We will see later that in applying the equations above, we have not taken into account the fact that due to the spin of the top, its principal axes also spin about axis 1 and that makes the components of along them time-dependent. For now I move on to explain the observation about only one of the rollers being able to go over all the curves of a track.

Example 3: If you have performed the experiment, you would have seen that only roller 1 (see figure 8) that is tapering down as we move away from its centre is able to go over all the curves. Let me now explain that.

As a roller goes over a curve, its centre of mass moves requires a centripetal force to do so. At the same time, the angular momentum of the roller also changes direction and that requires a torque. Both the centripetal force and the torque are provided by the normal reaction of the track on the rollers. These reaction forces on the three rollers are shown in figure 9.

In analyzing the motion of these rollers, I am taking them to be moving into the paper. Thus the direction of their angular momentum is to the left, as shown in the figure. Now if these rollers have to make a turn, the normal reactions should provide the required centripetal force in the horizontal direction. This rules out the plain cylindrical roller (roller 2) from making any turn because both normal reactions on it are in the vertical direction. This leaves the other two cylinders for further consideration. For those rollers, the torque of the normal reaction forces about the CM should change their angular momentum vector in the appropriate direction. Let us look at roller 1 first.

Roller 1: For a left turn, N₁ < N₂ for centripetal force. Therefore the torque generated by them is in the direction coming out of the page. As the roller makes a left turn, the associated change in its angular momentum also is in the direction coming out of the page, consistent with the torque generate. For a right turn by this roller, the centripetal force is to the right so N₁ > N₂ . This generates a torque about the CM that goes into the page. For the right turn, the change in the angular momentum is also into the page, consistent with the torque generated. Thus for roller 1 , the centripetal force and the torque generated are consistent with the centripetal force and the change in its angular momentum. Let us now see what happens to roller 3 .

Roller 3: If roller 3 turns left, the centripetal force will be provided correctly if N₁ > N₂ . This however gives a torque about the CM that is going into the page. On the other hand, during left turn the change in the angular momentum comes out of the page. Thus the torque and the change in angular momentum are in opposite directions. Exactly the same situation arises for a right turn. Because of this inconsistency, the roller fails to turn at any of the curves. This example teaches us about the centre of mass motion combined with angular momentum changes about the CM. We now move on to discuss the general form of the equation relating the torque and the angular momentum.