Question

An ice skater has a moment of inertia of 5.0 kg.M^2 when her arms are outstretched , and at the time she is spinning at 3.0 rev/s . If she pulls in her arms and decreases her moment of inertia to 2.0 kg.M^2 , how fast will she be spinning

Answer 1

Answer:

The ice skater will spin at an angular speed of 7.5 revolutions per second.

Explanation:

From statement we notice that ice skater can be considered as a system, on which no external forces are exerted, such that Principle of Angular Momentum can be applied:

$I_{1}\cdot \omega_{1} = I_{2}\cdot \omega_{2}$ (Eq. 1)

Where:

$I_{1}$ - Moment of inertia of the ice skater when her arms are outstretched, measured in kilogram-square meters.

$\omega_{1}$ - Angular speed when arms are outstretched, measured in revolutions per second.

$I_{2}$ - Moment of inertial of the ice skater when her arms are pulled, measured in kilogram-square meters.

$\omega_{2}$ - Angular speed when arms are pulled, measured in revolutions per second.

If $I_{1} = 5\,kg\cdot m^{2}$, $\omega_{1} = 3\,\frac{rev}{s}$ and $I_{2} = 2\,kg\cdot m^{2}$, the angular speed when arms are pulled is:

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

$\omega_{2} = \left(\frac{5\,kg\cdot m^{2}}{2\,kg\cdot m^{2}} \right)\cdot \left(3\,\frac{rev}{s} \right)$

$\omega_{2} = 7.5\,\frac{rev}{s}$

The ice skater will spin at an angular speed of 7.5 revolutions per second.