A horizontal circular platform (m = 108.1 kg, r = 3.25m) rotates about a frictionless vertical axle. A student (m = 85.3kg) walks slowly from the rim of the platform toward the center. The angular velocity, omega, of the system is 2.5 rad/s when the student is at the rim. Find omega when the student is 2.29m from the center.

asked by Jackie

13 years ago

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1 answer

moment of inertia is the sum of the platform and the student (mr^2).

I1*w1=I2*w2

answered by bobpursley

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Question

A horizontal circular platform (m = 108.1 kg, r = 3.25m) rotates about a frictionless vertical axle. A student (m = 85.3kg) walks slowly from the rim of the platform toward the center. The angular velocity, omega, of the system is 2.5 rad/s when the student is at the rim. Find omega when the student is 2.29m from the center.

1 answer

moment of inertia is the sum of the platform and the student (mr^2).

I1*w1=I2*w2

bobpursley · Answer

moment of inertia is the sum of the platform and the student (mr^2). I1*w1=I2*w2