Suppose the disk in [link] has a mass of 50.0 g and an initial velocity of 30.0 m/s when it strikes the stick that is 1.20 m long and 2.00 kg.

(a) What is the angular velocity of the two after the collision?

(b) What is the kinetic energy before and after the collision?

(c) What is the total linear momentum before and after the collision?

Strategy for (a)

We can answer the first question using conservation of angular momentum as noted. Because angular momentum is $\mathrm{I\omega }$, we can solve for angular velocity.

Solution for (a)

Conservation of angular momentum states

where primed quantities stand for conditions after the collision and both momenta are calculated relative to the pivot point. The initial angular momentum of the system of stick-disk is that of the disk just before it strikes the stick. That is,

where $I$ is the moment of inertia of the disk and $\omega$ is its angular velocity around the pivot point. Now, $I={\text{mr}}^{\mathrm{\text{2}}}$ (taking the disk to be approximately a point mass) and , so that

After the collision,

It is $\omega \prime$ that we wish to find. Conservation of angular momentum gives

Rearranging the equation yields

where $I\prime$ is the moment of inertia of the stick and disk stuck together, which is the sum of their individual moments of inertia about the nail. [link] gives the formula for a rod rotating around one end to be $I={\text{Mr}}^2/3$. Thus,

Entering known values in this equation yields,

The value of $I\prime$ is now entered into the expression for $\omega \prime$, which yields

Strategy for (b)

The kinetic energy before the collision is the incoming disk’s translational kinetic energy, and after the collision, it is the rotational kinetic energy of the two stuck together.

Solution for (b)

First, we calculate the translational kinetic energy by entering given values for the mass and speed of the incoming disk.

After the collision, the rotational kinetic energy can be found because we now know the final angular velocity and the final moment of inertia. Thus, entering the values into the rotational kinetic energy equation gives

Strategy for (c)

The linear momentum before the collision is that of the disk. After the collision, it is the sum of the disk’s momentum and that of the center of mass of the stick.

Solution of (c)

Before the collision, then, linear momentum is

After the collision, the disk and the stick’s center of mass move in the same direction. The total linear momentum is that of the disk moving at a new velocity $v\prime =\mathrm{r\omega }\prime$ plus that of the stick’s center of mass,

which moves at half this speed because $v_{\text{CM}}=\left(\frac{r}{2}\right)\omega \prime =\frac{v\prime }{2}$. Thus,

Gathering similar terms in the equation yields,

so that

Substituting known values into the equation,

Discussion

First note that the kinetic energy is less after the collision, as predicted, because the collision is inelastic. More surprising is that the momentum after the collision is actually greater than before the collision. This result can be understood if you consider how the nail affects the stick and vice versa. Apparently, the stick pushes backward on the nail when first struck by the disk. The nail’s reaction (consistent with Newton’s third law) is to push forward on the stick, imparting momentum to it in the same direction in which the disk was initially moving, thereby increasing the momentum of the system.

Section Summary

• Angular momentum $L$ is analogous to linear momentum and is given by $L=\mathrm{I\omega }$.
• Angular momentum is changed by torque, following the relationship $\text{net}\tau =\frac{\text{Δ}L}{\text{Δ}t}.$
• Angular momentum is conserved if the net torque is zero $L=\text{constant}\left(\text{net}\tau =\text{0}\right)$ or $L=L\prime \left(\text{net}\tau =0\right)$ . This equation is known as the law of conservation of angular momentum, which may be conserved in collisions.

Conceptual Questions

Problems & Exercises

Twin skaters approach one another as shown in [link] and lock hands. (a) Calculate their final angular velocity, given each had an initial speed of 2.50 m/s relative to the ice. Each has a mass of 70.0 kg, and each has a center of mass located 0.800 m from their locked hands. You may approximate their moments of inertia to be that of point masses at this radius. (b) Compare the initial kinetic energy and final kinetic energy.

(a) 3.13 rad/s

(b) Initial KE = 438 J, final KE = 438 J

Suppose a 0.250-kg ball is thrown at 15.0 m/s to a motionless person standing on ice who catches it with an outstretched arm as shown in [link].

(a) Calculate the final linear velocity of the person, given his mass is 70.0 kg.

(b) What is his angular velocity if each arm is 5.00 kg? You may treat the ball as a point mass and treat the person's arms as uniform rods (each has a length of 0.900 m) and the rest of his body as a uniform cylinder of radius 0.180 m. Neglect the effect of the ball on his center of mass so that his center of mass remains in his geometrical center.

(c) Compare the initial and final total kinetic energies.