A circular platform of radius R p = 4.83 m and mass M p = 463 kg rotates on fric

Question

A circular platform of radius Rp = 4.83 m and mass Mp = 463 kg rotates on frictionless air bearings about its vertical axis at 6.07 rpm. An 68.3-kg man standing at the very center of the platform starts walking (at t = 0) radially outward at a speed of 0.495 m/s with respect to the platform. Approximating the man by a vertical cylinder of radius Rm = 0.247 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

moment of inertia of disc with respect to axis passing through centre of disc = 463 x 4.83^2/2

= 5400.64

of the man = mr^2/2 +mx^2(x is the distance from centre)

x = 0.495t

= 68.3(0.247^2/2 + (0.495t)^2)

= 2.08 + 16.74t^2

( 5400.64+2.08+16.74t^2)w= 5400.64 x 2pi x 0.101

( 5400.64+2.08+16.74t^2)w= 3434.3

when the person reaches the edge time = 4.83/0.495 = 9.76

w = 0.49

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