A circular platform of radius Rp = 3.51 m and mass Mp = 353 kg rotates on fricti

Question

A circular platform of radius Rp = 3.51 m and mass Mp = 353 kg rotates on frictionless air bearings about its vertical axis at 4.35 rpm. An 68.5-kg man standing at the very center of the platform starts walking (at t = 0) radially outward at a speed of 0.703 m/s with respect to the platform. Approximating the man by a vertical cylinder of radius Rm = 0.221 m, determine an equation (specific expression) for the angular velocity of the platform as a function of time. What is the angular velocity when the man reaches the edge of the platform?

Explanation / Answer

sol: suppose that w = angular velocity I stands for Inertia I1w1 = I2w2 so Inertia 1 times angular velocity 1 equals Inertia 2 times angular velocity 2 When the person walks to the edge, they have a moment of inertia due to a point mass I = mr2 = (75 kg)(3.1 m)^2 = 720.75 kgm2 the total moment of inertia after they walk to the edge is I2 = 1500 kgm2 + 720.75 kgm2 = 2220.75 kgm2 I1w1 = I2w2 (1500 kgm2)(2.5 rad/s) = (2220.75kgm2)w2 w2 = 1.689 rad/s the angular velocity when the person reaches the edge is 1.689 rad/s answer

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