An ice dancer with her arms stretched out starts into a spin with an angular velocity of 1.00 rad/s. Her moment of inertia with her arms stretched out is 2.48 kg m2. What is the increase in her rotational kinetic energy when she pulls in her arms to make her moment of inertia 1.40 kg m2?
A) 0.957 JB) 0.902 JC) 0.870 JD) 0.750 JE) 0.690 J

The increase in her rotational kinetic energy when she pulls in her arms to make her moment of inertia 1.40 kg.m² is; Option A; 0.957 J

We are given;

Angular velocity 1; ω1 = 1 rad/s

Initial moment of inertia; I1 = 2.48 kg.m²

Final moment of inertia; I2 = 1.4 kg.m²

Now,formula for rotational kinetic energy is;

KE = ½Iω²

Thus;

KE_1 = ½ × 2.48 × 1²

KE_1 = 1.24 J

Now, from conservation of angular momentum we know that;

I1•ω_1 = I2•ω_2

Thus,

2.48 × 1 = 1.4 × ω_2

ω_2 = 2.48/1.4

ω_2 = 1.7714 rad/s

Thus;

KE_2 = ½ × 1.4 × 1.7714²

KE_2 = 2.197 J

Thus, increase is;

ΔKE = KE_2 - KE_1

ΔKE = 2.197 - 1.24

ΔKE = 0.957J

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A) 0.957 J

Explanation:detailed explanation and calculation is shown in the image below

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