Angular Momentum

To find the direction of the angular momentum we use the right hand rule in the same way we used it for angular velocity. Thus, if we look down on the skater, and curl our fingers around in the counterclockwise direction, our thumb points towards us. Thus the angular momentum of the skater is pointing upwards.

One might think that the maximum angular momentum will be when the object is traveling in the tangential direction with respect to the radius. However, notice that the radius is smallest at the point when the object is traveling in the tangential direction. Since angular momentum varies with radius, it cannot be maximal at this point. We will show that at all points, the angular momentum of the particle is the same. Let's take another look at the figure, and calculate the angular momentum at some arbitrary point, P:

At this point P, the particle is a distance from the origin. In addition, the component of the velocity in the tangential direction at P is given by 3 cosθ . Thus the angular momentum at this point is:

Notice that the thetas cancel, and this answer is valid for P anywhere on the line of travel of the particle. Thus we have shown that the angular momentum of the particle is the same in all places. This agrees with our theorem that a net torque is required to change the angular momentum of a particle.

It can be easily shown, and has been established in other sections, that the moment of inertia of a thin hoop is simply MR ² . Thus the angular momentum is easily calculable:

Quite simply, the total angular momentum is zero. At every point while the two particles are traveling, one particle moves in the clockwise direction with respect to O and one moves in the counterclockwise direction. Also, at every point, both particles have the same distance to the axis, and angle between the radius and the velocity of the particle. Thus the two particles always have equal and opposite angular momenta, and the total momentum of the system is zero.

We begin by drawing a diagram of the spinning top:

If we can find the torque acting on the top we can also find the direction of the change in linear momentum, as τ = . To find the net torque on the top, we look at the forces acting upon the top. Where the top is in contact with the ground, a normal force acts in the vertical direction. Also, a gravitational force acts from the center of mass of the top. Let's take our origin to be the point at which the top is in contact with the ground. The gravitational force, then, exerts a torque of magnitude mg sinθ . Since the normal force acts at our origin, it exerts no torque. Thus the net torque on the top has magnitude mg sinθ , and points horizontally, into the page of our figure (by the right hand rule). Since a net torque changes the angular momentum of an object, our change in momentum is in the same direction, resulting in the precession motion of the top.