Question 8.17: THE SPINNING STOOL GOAL Apply conservation of angular moment...

(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.