A person of mass 80.0kg stands at the center of a rotating merry-go-round platfo

Question

A person of mass 80.0kg stands at the center of a rotating merry-go-round platform of radius 2.50m and moment of inertia 910kg*m^2 . The platform rotates without friction with angular velocity 2.20rad/s . The person walks radially to the edge of the platform. A. Calculate the angular velocity when the person reaches the edge. B. Calculate the rotational kinetic energy of the system of platform plus person before and after the person's walk.

Explanation / Answer

By the law of angular momentum conservation:- =>L(initial) = L(final) =>I1 x ?1 = I2 x ?2 =>980 x 2.1 = [980 + mr^2] x ?2 =>2058 = [980 + 73 x (2.9)^2] x ?2 =>?2 = 1.29 rad b)K(initial) = 1/2I?^2 = 1/2 x 980 x (2.1)^2 = 2160.90 J & K(final) = 1/2I?^2 = 1/2 x [980 + 73 x (2.9)^2] x (1.29)^2 = 1326.23 J In hope that all the units of data you have provided are in MKS system.

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