Question

An ice skater is spinning at 5.2 rev/s and has a moment of inertia of 0.32 kg * m2.

Answer 1

Explanation:

The angular momentum is given by the moment of inertia, multiplied by the angular speed of the rotating body:

$L=I\omega$

The angular speed is given by:

$\omega=2\pi f\\\omega=2\pi 5.2\frac{rev}{s}\\\omega=32.67\frac{rad}{s}$

Now, we calculate the angular momentum:

$L=0.32kg\cdot m^2(32.67\frac{rad}{s})\\L=10.45\frac{kg\cdot m^2}{s}$

The average torque is defined as:

$\tau=I\alpha$

$\alpha$ is the angular acceleration, which is defined as:

$\alpha=\frac{\omega_f-\omega_0}{t}$

We have to calculate $\omega_f$:

$\omega_f=2\pi (2.75\frac{rad}{s})\\\omega_f=17.28\frac{rad}{s}$

Now, we calculate the angular acceleration:

$\alpha=\frac{17.28\frac{rad}{s}-32.67\frac{rad}{s}}{12s}\\\alpha=-1.28\frac{rad}{s^2}$

Finally, we can know the average torque:

$\tau=0.32kg\cdot m^2(-1.28\frac{rad}{s^2})\\\tau=-0.41N\cdot m$