Part A

From the conservation of angular momentum,

$\begin{array}{rcl}\mathbf{(}{\mathbf{I}}_{\mathbf{t}}\mathbf{+}{\mathbf{I}}_{\mathbf{r}}\mathbf{)}{\mathbf{\omega}}_{\mathbf{f}}& \mathbf{=}& {\mathbf{I}}_{\mathbf{t}}{\mathbf{\omega}}_{\mathbf{i}}\\ {\mathbf{\omega}}_{\mathbf{f}}& \mathbf{=}& \frac{{\mathbf{I}}_{\mathbf{t}}{\mathbf{\omega}}_{\mathbf{i}}}{\mathbf{(}{\mathbf{I}}_{\mathbf{t}}\mathbf{+}{\mathbf{I}}_{\mathbf{r}}\mathbf{)}}\end{array}$

To understand how to use conservation of angular momentum to solve problems involving collisions of rotating bodies.

Consider a turntable to be a circular disk of moment of inertia *I*_{t} rotating at a constant angular velocity *ω*_{i} (note that angular velocities use the Greek letter *omega* and not *double-u*) around an axis through the center and perpendicular to the plane of the disk (the disk's "primary axis of symmetry") as shown in (Figure 1).

1. The axis of the disk is vertical and the disk is supported by frictionless bearings. The motor of the turntable is off, so there is no external torque being applied to the axis.

Another disk (a record) is dropped onto the first such that it lands coaxially (the axes coincide). The moment of inertia of the record is *I*_{r}. The initial angular velocity of the second disk is zero.

There *is* friction between the two disks.

After this "rotational collision," the disks will eventually rotate with the same angular velocity.

Part A

What is the final angular velocity, *ω*_{f}, of the two disks?

Express *ω*_{f} (omega subscript f) in terms of *I*_{t}, *I*_{r}, and *ω*_{i} (omega subscript i).

Part B

Because of friction, rotational kinetic energy is not conserved while the disks' surfaces slip over each other. What is the final rotational kinetic energy, *K*_{f}, of the two spinning disks?

Express the final kinetic energy in terms of *I*_{t}, *I*_{r}, and the initial kinetic energy *K*_{i} of the two-disk system. No angular velocities should appear in your answer.

Frequently Asked Questions

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