A 40-kg boy running at 4.0 m/s jumps tangentially onto a small stationary circul

Question

A 40-kg boy running at 4.0 m/s jumps tangentially onto a small stationary circular merry-go-round of radius 2.0 m and rotational inertia 20 kg middot m^2 pivoting on a frictionless bearing on its central shaft, Determine the rotational velocity of the merry-go-round after the boy jumps on it. Find the change in kinetic energy of the system consisting of the boy and the merry-go-round, Find the change in the boy's kinetic energy, Find the change in the kinetic energy of the merry-go-round, Compare the kinetic energy changes in parts (b) through (d).

Explanation / Answer

R = radius of merry-go-round = 2 m

m = mass of boy = 40 kg

V = speed of boy = 4 m/s

I = moment of inertia of merry-go-round = 20 kgm2

W = angular speed

using conservation of angular momentum

mVR = (I + mR2) W

40 (4) (2) = (20 + 40 (2)2) W

W = 1.8 rad/s

b)

change in KE = (0.5) (I + mR2) W2 - (0.5) mV2 = (0.5) (20 + 40 (2)2) (1.8)2 - (0.5) (40) (4)2 = - 28.4 J

c)

change in KE = (0.5) (mR2) W2 - (0.5) mV2 = (0.5) (40 (2)2) (1.8)2 - (0.5) (40) (4)2 = - 60.8 J

d)

change in KE = (0.5) (I ) W2 = (0.5) (20) (1.8)2 = 32.4 J

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