Conceptual Example 14 provides useful background for this problem. A playground

Question

Conceptual Example 14 provides useful background for this problem. A playground carousel is free to rotate about its center on frictionless bearings, and air resistance is negligible. The carousel itself (without riders) has a moment of inertia of 125 kg·m2. When one person is standing on the carousel at a distance of 1.50 m from the center, the carousel has an angular velocity of 0.600 rad/s. However, as this person moves inward to a point located 0.870 m from the center, the angular velocity increases to 0.800 rad/s. What is the person’s mass?

Explanation / Answer

Carousel's moment of inertia I = 125 kg.m/s

Mass of person = m

Initial radius R1 = 1.5 m

Final radius R2 = 0.87 m

Initial angular velocity w0 = 0.6 rad/s

Final angular velocity w1 = 0.8 rad/s

Initially, the angular momentum is thus

L = (I + m*R1^2)*w0

Finally, it is

L = (I + m*R2^2)*w1

But L must be unchanged, so

(I + m*R1^2)*w0 = (I + m*R2^2)*w1

I*w0 + m*w0*R1^2 = I*w1 + m*w1*R2^2

m*w0*R1^2 - m*w1*R2^2 = I*w1 - I*w0

m*(w0*R1^2 - w1*R2^2) = J*(w1 - w0)

m = I*(w1 - w0)/(w0*R1^2 - w1*R2^2)

m = 125*(0.8 - 0.6)/(0.6*1.5^2 - 0.8*0.87^2)

m = 33.58 kg

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