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A carnival merry-go-round has a large disk-shaped platform of mass 120 kg that can rotate about a center axle. A 60-kg student stands at rest at the edge of the platform 4.0 m from its center. The platform is also at rest. The student starts running clockwise around the edge of the platform and attains a speed of 2.0 m/s relative to the ground.

(a) determine the rotational velocity of the platform.
(b) determine the change of kinetic enrgy of the system consisting of the platform and the student.

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Answer

given,

radius of merry - go - round = 4 m

mass of the disk = 120 kg

speed of the merry- go-round = 0 rad/s

speed = 2 m/s

mass of student = 60 kg


I_(disk) = (1)/(2)MR^2


I_(disk) = (1)/(2)* 120 * 4^2


I_(disk) = 960 kg.m^2

initial angular momentum of the system


L_i = I\omega_i + mvR


L_i =960 * 0 + 60 * 2 * 4


L_i =480\ kg.m^2/s

final angular momentum of the system


L_f = (960 + 60* 4^2)\omega_(f)

from conservation of angular momentum


L_i = L_f


480 = (1920)\omega_(f)

now,

change in kinetic energy of the system

initially the merry-go-round was on rest KE_i = 0


\Delta KE = KE_f- KE_i


\Delta KE = ((1)/(2)mv^2 + (1)/(2)I\omega^2)- 0


\Delta KE = (1)/(2)* 60 * 2^2 + (1)/(2)* 960 *( {(v)/(r))^2


\Delta KE =120 + 480 * (4)/(16)


\Delta KE = 240\ J

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