Question

A 60.0-kg skater begins a spin with an angular speed of 6.0 rad/s. By changing the position of her arms, the skater decreases her moment of inertia by two times. What is the skater's final angular speed

Answer 1

Answer:

The final angular speed of the skater is 12 radians per second.

Explanation:

Let consider the skater as a rotating system, given the absence of external forces, the Principle of Angular Momentum Conservation is applied:

$I_{o}\cdot \omega_{o} = I_{f}\cdot \omega_{f}$

Where:

$I_{o}$, $I_{f}$ - Initial and final moment of inertia, measured in $kg \cdot m^{2}$.

$\omega_{o}$, $\omega_{f}$ - Angular speed, measured in radians per second.

The final angular speed is cleared afterwards:

$\omega_{f} = \frac{I_{o}}{I_{f}} \cdot \omega_{o}$

Given that $I_{f} = \frac{1}{2}\cdot I_{o}$ and $\omega_{o} = 6\,\frac{rad}{s}$, the final angular speed is:

$\omega_{f} = \frac{I_{o}}{\frac{1}{2}\cdot I_{o} } \cdot \omega_{o}$

$\omega_{f} = 2 \cdot \omega_{o}$

$\omega_{f} = 2 \cdot \left(6\,\frac{rad}{s} \right)$

$\omega_{f} = 12\,\frac{rad}{s}$

The final angular speed of the skater is 12 radians per second.