A 50-kg boy running at 4.4 m/s jumps tangentially onto a small circular merry-go

Question

A 50-kg boy running at 4.4 m/s jumps tangentially onto a small circular merry-go-round of radius 2.0 m and rotational inertia 2.0×102 kgm2 pivoting on a frictionless bearing on its central shaft. Merry-go-round initially rotating at 1.1 rad/s in the same direction that the boy is running.

A) Determine the rotational speed of the merry-go-round after the boy jumps on it.

B) Find the change in kinetic energy of the system consisting of the boy and the merry-go-round.

C) Find the change in the boy's kinetic energy.

D) Find the change in the kinetic energy of the merry-go-round.

Explanation / Answer

from conservation of angular momentum

total intial ang momentum = total final ang momentum

I W1 + mV1 R = I W2 + m r^2 W2

(200 *1.1) + (50 * 4.4 * 2) = (200 * W2) + ( 50 * 2^2 * W2)

W2 = 1.65 rad/s

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part B :

Initial KE = KE rot + KE trans

intial KE = 0.5 I W1^2 + 0.5 m v1^2

intial KE = 0.5 * 200 * 1^2 + (0.5 * 50 * 4.4^2) = 584 J

final KE = 0.5 * 200 * 1.65+ (0.5 * 50 *2^2 * 1.53^2) = 399.09 J

KE change =- 584 - 399.09 = 184.91 Joules

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part C:

change in boys KE = 0.5 m v1^2 - m r^2 W2^2

KE = 0.5 * 50 * 4.4^2 - (50 * 2^2* 1.65^2)

KE = -60.5 Joules
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change in KE of merry go round

dKE = 0.5 * I w2^2 - 0.5 I W1^2

dKE = 0.5 * 200 * (1.65^2 - 1.1^2)

dKE = 151.25 Joules

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