## Chapter 8

## Q. 8.17

## Q. 8.17

**THE SPINNING STOOL**

**GOAL** Apply conservation of angular momentum to a simple system.

**PROBLEM** A student sits on a pivoted stool while holding a pair of weights. (See Fig. 8.36.) The stool is free to rotate about a vertical axis with negligible friction. The moment of inertia of student, weights, and stool is 2.25 \mathrm{~kg} \cdot \mathrm{m}^{2}. The student is set in rotation with arms outstretched, making one complete turn every 1.26 \mathrm{~s}, arms outstretched. (**a**) What is the initial angular speed of the system? (**b**) As he rotates, he pulls the weights inward so that the new moment of inertia of the system (student, objects, and stool) becomes 1.80 \mathrm{~kg} \cdot \mathrm{m}^{2}. What is the new angular speed of the system? (**c**) Find the work done by the student on the system while pulling in the weights. (Ignore energy lost through dissipation in his muscles.)

**STRATEGY** (**a**) The angular speed can be obtained from the frequency, which is the inverse of the period. (**b**) There are no external torques acting on the system, so the new angular speed can be found with the principle of conservation of angular momentum. (**c**) The work done on the system during this process is the same as the system’s change in rotational kinetic energy.

## Step-by-Step

## Verified Solution

(**a**) Find the initial angular speed of the system.

Invert the period to get the frequency, and multiply by 2 \pi :

\omega_{i}=2 \pi f=2 \pi / T=4.99 \mathrm{rad} / \mathrm{s}

(**b**) After he pulls the weights in, what’s the system’s new angular speed?

Equate the initial and final angular momenta of the system:

\mathbf{ (1) } \quad L_{i}=L_{f} \rightarrow I_{i} \omega_{i}=I_{f} \omega_{f}

Substitute and solve for the final angular speed \omega_{f} :

\begin{aligned}\mathbf{ (2) } \quad \left(2.25 \mathrm{~kg} \cdot \mathrm{m}^{2}\right)(4.99 \mathrm{rad} / \mathrm{s}) &=\left(1.80 \mathrm{~kg} \cdot \mathrm{m}^{2}\right) \omega_{f} \\\omega_{f} &=6.24 \mathrm{rad} / \mathrm{s}\end{aligned}

(**c**) Find the work the student does on the system.

Apply the work-energy theorem:

\begin{aligned}W_{\text {student }}=& \Delta K_{r}=\frac{1}{2} I_{f} \omega_{f}^{2}-\frac{1}{2} I_{i} \omega_{i}{ }^{2} \\=& \frac{1}{2}\left(1.80 \mathrm{~kg} \cdot \mathrm{m}^{2}\right)(6.24 \mathrm{rad} / \mathrm{s})^{2} \\& -\frac{1}{2}\left(2.25 \mathrm{~kg} \cdot \mathrm{m}^{2}\right)(4.99 \mathrm{rad} / \mathrm{s})^{2} \\W_{\text {student }}=& 7.03 \mathrm{~J}\end{aligned}

**REMARKS** Although the angular momentum of the system is conserved, mechanical energy is not conserved because the student does work on the system.