Conservation of Angular Momentum: Spinning Skater

A spinning ice skater pulls her arms in close to her body and dramatically speeds up. The reason is conservation of angular momentum: in the absence of external torques, the product $L = I\omega$ stays constant.

When she pulls her arms in, her moment of inertia $I$ decreases (because mass is now closer to the rotation axis). To keep $L = I\omega$ constant, her angular velocity $\omega$ must increase.

The Conservation Law

$L_{\text{initial}} = L_{\text{final}}$

$I_1\,\omega_1 = I_2\,\omega_2$

Solving for the new angular velocity:

$\omega_2 = \dfrac{I_1}{I_2}\,\omega_1$

If she halves her moment of inertia, her angular velocity doubles.

Step-by-Step Solution

Given: A skater modeled as a central rod (mass $M$, fixed) plus two arms outstretched at radius $R_1$, then pulled in to radius $R_2$. Initial angular velocity $\omega_1 = 2\;\text{rad/s}$. For simplicity, treat the arms as point masses each of mass $m$.

Find: The new angular velocity $\omega_2$ after pulling the arms in, and the change in kinetic energy.

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Step 1 — Compute the initial moment of inertia.

If the central body has moment of inertia $I_{\text{body}}$ and each arm contributes $m R_1^{2}$:

$I_1 = I_{\text{body}} + 2 m R_1^{2}$

Numerically (suppose $I_{\text{body}} = 1\;\text{kg}\cdot\text{m}^{2}$, $m = 5\;\text{kg}$, $R_1 = 0.8\;\text{m}$):

$I_1 = 1 + 2(5)(0.64) = 1 + 6.4 = 7.4\;\text{kg}\cdot\text{m}^{2}$

Step 2 — Compute the initial angular momentum.

$L_1 = I_1\,\omega_1 = (7.4)(2) = 14.8\;\text{kg}\cdot\text{m}^{2}/\text{s}$

Step 3 — Compute the final moment of inertia (arms pulled in to $R_2 = 0.2\;\text{m}$).

$I_2 = I_{\text{body}} + 2 m R_2^{2} = 1 + 2(5)(0.04) = 1 + 0.4 = 1.4\;\text{kg}\cdot\text{m}^{2}$

That's about a 5× decrease in $I$.

Step 4 — Apply conservation of angular momentum.

$L_2 = L_1 = 14.8\;\text{kg}\cdot\text{m}^{2}/\text{s}$

$\omega_2 = \dfrac{L_2}{I_2} = \dfrac{14.8}{1.4} \approx 10.571\;\text{rad/s}$

That's about 5× faster than the initial 2 rad/s — which makes sense given $I$ decreased by 5×.

Step 5 — Compute the kinetic energy before and after.

$\text{KE}_1 = \tfrac{1}{2}I_1\,\omega_1^{2} = \tfrac{1}{2}(7.4)(4) = 14.8\;\text{J}$

$\text{KE}_2 = \tfrac{1}{2}I_2\,\omega_2^{2} = \tfrac{1}{2}(1.4)(111.75) \approx 78.225\;\text{J}$

The kinetic energy increased by about 63 J! Where did this energy come from?

Step 6 — The energy source.

The skater's muscles do work pulling her arms in against the centrifugal effect (the apparent outward force in her rotating frame). That work goes into the rotational kinetic energy. So although angular momentum is conserved, kinetic energy is not — the skater added energy by exerting muscular effort.

The general formula for the work done:

$W = \text{KE}_2 - \text{KE}_1 = \dfrac{L^{2}}{2}\left(\dfrac{1}{I_2} - \dfrac{1}{I_1}\right)$

For our numbers: $W = 78.225 - 14.8 = 63.425\;\text{J}$.

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Answer:

• Initial: $I_1 = 7.4$, $\omega_1 = 2\;\text{rad/s}$, $L = 14.8$, KE = $14.8\;\text{J}$
• Final: $I_2 = 1.4$, $\boxed{\omega_2 \approx 10.57\;\text{rad/s}}$, $L = 14.8$ (conserved), KE ≈ $78.23\;\text{J}$

The angular velocity increases by the same factor that $I$ decreased — about 5×. The kinetic energy also increases by 5× because $\text{KE} = L^{2}/(2I)$, and $I$ shrunk. The skater's muscles supplied the extra energy.

Try It

• Slide the arm radius to pull the arms in or out.
• Watch the angular velocity respond — and the spin rate visibly speed up or slow down.
• The HUD shows that $L = I\omega$ stays constant while $\omega$ and $I$ change in lockstep.
• The energy bar shows kinetic energy growing as the arms come in (from muscular work).