Question

A student sits in a chair that can spin without friction. The student has her hands outstretched and starts rotating at 1.9 rev/s. Her initial rotational inertia about the central axis is 12.00 kg m2. She pulls her hands inward and decreases her rotational inertial to 5.30 kg m2. What is her resulting angular speed after she pulls her hands inward?(rad/s)

Answer 1

Answer:

the resulting angular speed after she pulls her hand inwards in (rad/s)  is  27.02 rad/s

Explanation:

Given that :

the initial angular speed $\omega_1 = 1.9 \ \ rev/s$

Initial rotational inertia $I_1 = 12.00 \ \ kg.m^2$

Final angular speed $\omega_2 = ???$

Final rotational inertia $I_2 = 5.3 \ \ kg.m^2$

According to conservation of momentum :

Initial momentum = final momentum

$I_1 \omega_1 = I_2 \omega_2$

$\omega_2 = \frac{I_1 \omega_1}{I_2}$

$\omega_2 = \frac{12.00*1.9}{5.3}$

$\omega_2 = 4,3$ $rev/s$

To rad/s ; we have:

$1 \ rev/s = 2 \pi \ \ rad/s$

$\omega_2 = 4.3 * 2 \pi \ \ rad/s$

$\omega_2 = 27.02 \ \ rad/s$

Therefore the resulting angular speed after she pulls her hand inwards in (rad/s) = 27.02 rad/s