Question

A man of mass 75kg stands at the center of a rotating merry-go-round platform of radius 3.0 m and moment of inertia 920 kgm^2. The platform rotates without friction with angular velocity 2.0 rad/s. The man walks radially to the edge of the platform. Calculate: the angular velocity of the system when the man reaches the edge, and the change in the kinetic energy of the system.

Answer 1

here,

mass , m = 75 kg

radius , r = 3 m

the moment of inertia of ground , I = 920 kg.m^2

initial angular velocity , w0 = 2 rad/s

let the final angular velocity be w

using conservation of angular momentum

(I * w0) = ( I + m * r^2 ) * w

( 920 * 2) = ( 920 + 75 * 3^2) * w

w = 1.15 rad/s

the final angular speed is 1.15 m/s

the change in kinetic energy , KE = KEf - KEi

KE = 0.5 * ( I + m * r^2 ) * w^2 - 0.5 * I * w0^2

KE = 0.5 * ( (920 + 75*9)*1.15^2 - 920 * 2^2)

KE = - 785.3 J

the chnage in kinetic energy is - 785.3 J