Question

# An ice skater is spinning on frictionless ice with her arms extended outward. She then pulls her arms in toward her body, reducing her moment of inertia. Her angular momentum is conserved, so as she reduces her moment of inertia, her angular velocity increases and she spins faster. Compared to her initial rotational kinetic energy, her final rotational kinetic energy is (a) the same (b) larger, because her angular speed is larger (c) smaller, because her moment of inertia is smaller.

Solution

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The rotational kinetic energy $KE_{rot}$ of a rigid object rotating with an angular speed $\omega$ about a fixed axis and having a moment of inertia $I$ is given as:

$KE_{rot}=\frac{1}{2} I \omega^2,$

The angular momentum $L$ of a rigid body rotating with an angular velocity $\omega$ about a fixed axis and having a moment of inertia $I$ with respect to that axis is given as:

$L=I \omega.$

So, the equation of rotational kinetic energy $KE_{rot}$ of a rigid object rotating becomes:

\begin{align*} KE_{rot} &=\frac{1}{2} \ \dfrac{L}{\omega} \omega^2 \\ KE_{rot} &=L \ \omega \\ \end{align*}

Since the angular momentum L is conserved, and the angular velocity increases, hence, the kinetic energy increases as well, $\omega \uparrow \ \text{and} \ EK_{rot} \uparrow.$ This increase in kinetic energy is given to the skater as a result of the work done by her muscles in pulling the arm.

So, the answer is: b) Larger, because her angular speed is larger.

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