Do A professor sits on a rotating stool that is spinning at $10.0 \mathrm{rpm}$ while she holds a heavy weight in each of her hands. Her outstretched hands are $0.745 \mathrm{~m}$ from the axis of rotation, which passes through her head into the center of the stool. When she symmetrically pulls the weights in closer to her body, her angular speed increases to $24.5 \mathrm{rpm}$. Neglecting the mass of the professor, how far are the weights from the rotational axis after she pulls her arms in? distance: m

Step 1 ：Given that the initial angular speed is \(10.0 \, \text{rpm}\), the initial distance from the axis of rotation is \(0.745 \, \text{m}\), and the final angular speed is \(24.5 \, \text{rpm}\).

Step 2 ：First, we need to convert the angular speed from rpm to rad/s. We know that \(1 \, \text{rpm} = \frac{2\pi}{60} \, \text{rad/s}\). So, \(\omega_1 = 10.0 \times \frac{2\pi}{60} = 1.0471975511965976 \, \text{rad/s}\) and \(\omega_2 = 24.5 \times \frac{2\pi}{60} = 2.5656340004316642 \, \text{rad/s}\).

Step 3 ：According to the conservation of angular momentum, the initial angular momentum is equal to the final angular momentum. That is, \(m r_1^2 \omega_1 = m r_2^2 \omega_2\).

Step 4 ：We can cancel out the mass \(m\) from both sides and solve for \(r_2\), the final distance from the axis of rotation. This gives us the equation \(r_2 = \sqrt{\frac{r_1^2 \omega_1}{\omega_2}}\).

Step 5 ：Substituting the given values into the equation, we get \(r_2 = \sqrt{\frac{(0.745)^2 \times 1.0471975511965976}{2.5656340004316642}} = 0.4759630409249552 \, \text{m}\).

Step 6 ：So, the weights are approximately \(\boxed{0.476 \, \text{m}}\) from the rotational axis after the professor pulls her arms in.