Question

A professor sits on a rotating stool that is spinning at 10.0 rpm while she holds a heavy weight in each of her hands. Her outstretched hands are 0.705 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 20.5 rpm. Neglecting the mass of the professor, how far are the weights from the rotational axis after she pulls her arms in?'

Answer 1

Answer:

0.5864 m

Explanation:

$r_1$ = 0.705 m

$\omega_1$ = 10 rpm

$\omega_2$ = 20.5 rpm

$\left(m r_{1}^{2}\right) \omega_{1} &=\left(m r_{2}^{2}\right) \omega_{2}$

$r_{2}^{2} &=\frac{\left(m r_{1}^{2}\right) \omega_{1}}{\left(\omega_{2}\right)}$

$=\frac{\left(r_{1}^{2}\right) \omega_{1}}{\left(\omega_{2}\right)}$

$=\frac{\left((0.705 \mathrm{m})^{2}\right)(10 \mathrm{rpm})}{(20.5 \mathrm{rpm})}$

$=\sqrt{\frac{\left((0.705 \mathrm{m})^{2}\right)(10 \mathrm{rpm})}{(20.5 \mathrm{rpm})}}$

$r_{2} =0.5864\ m$

The weights are 0.5864 m apart.