A person of mass 75kg stands at the center of a rotating merry-go-round platform
ID: 1378418 • Letter: A
Question
A person of mass 75kg stands at the center of a rotating merry-go-round platform of radius 2.6m and moment of inertia 940kg?m2 . The platform rotates without friction with angular velocity 2.5rad/s . The person walks radially to the edge of the platform.
1.)Calculate the angular velocity when the person reaches the edge.
2.)Calculate the rotational kinetic energy of the system of platform plus person before and after the person's walk.
Express your answers using two significant figures. Enter your answers numerically separated by a comma.
Explanation / Answer
Given
Mass of the person, m = 75 kg
Radius of the merry go round , r = 2.6 m
Moment of inertia of the merry go round , I1 = 940 kg m2
The angular velocity of the platform, ?1 = 2.5 rad/s
Moment of inertia of the person is , I2 = I1 + mr2
= 940 kg m2+75kg (2.6m)2
I2 = 1447 kg m2
a) Using conservationof angular momentum
I1?1 = I2?2
(940 kg m2)2.5 rad/s = ( 1447 kg m2) ?2
?2 =1.62 rad/s
b) Initial rotational kinetic energy , K1 = 1/2 I1?12
K1 = 0.5 (940 kg m2)(2.5 rad/s)2
= 2937.5 J
Final rotational kinetic energy , K2 = 1/2 I2?22 K1 = 0.5 (1447 kg m2)(1.62 rad/s)2
= 1898.75 J
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