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## 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.

## 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.