Question

An ice skater is spinning at 5.2 rev/s and has a moment of inertia of 0.32 kg * m2.
Calculate the angular momentum, in kilogram meters squared per second, of the ice skater spinning at 5.2 rev/s.
Suppose instead he keeps his arms in and allows friction of the ice to slow him to 2.75 rev/s. What is the magnitude of the average torque that was exerted, in N * m, if this takes 12 s?

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:

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

Answer 2

(a) The angular momentum of the skater is 10.45 kgm²/s

(b) The magnitude of the average torque that was exerted, is 0.41 Nm.

Angular momentum of the ice skater

The angular momentum of the skater is calculated as follows;

L = Iω

where;

• ω is angular speed (rad/s)

ω = 5.2 rev/s x 2π rad = 32.67 rad/s

L = 0.32 x 32.67

L = 10.45 kgm²/s

Angular acceleration of the skater

The angular acceleration is calculated as follows;

$\alpha= \frac{\omega _f - \omega _i}{t}$

• ωf is the final angular speed = 2.75 rev/s x 2π rad = 17.28 rad/s

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

Average torque

The magnitude of the average torque that was exerted, is calculated as;

τ = Iα

τ = 0.32 x (1.28)

τ = 0.41 Nm.

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