A 45-kg boy running at 3.6 m/s jumps tangentially onto a small stationary circul

Question

A 45-kg boy running at 3.6 m/s jumps tangentially onto a small stationary 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.

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

Express your answer to two significant figures and include the appropriate units.

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

3. Find the change in the boy's kinetic energy.

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

Explanation / Answer

1) using law of conservation of angular momentum

(I1*w1)+(m*v*r) = I2*w2

w1 =0 rad/s

I1 = 2*10^2 kg.m^2

m = 45 kg

v = 3.6 m/s

r = 2 m

I2 = (2*10^2)+(45*2^2) = 380 kg-m^2

w2 = (m*v*r)/(I2) = (45*3.6*2)/(380) = 0.852 rad/s

2) initial KE = Ki = 0.5*m*v^2 = 0.5*45*3.6^2 = 291.6 J

Final KE = Kf = 0.5*I2*w2^2 = 0.5*380*0.852^2 = 137.92 J

change in KE = 137.92 - 291.6 = -153.68 J

3) boy's KE = 0.5*m*v^2 - 0.5*m*u^2 = 0.5*45*((2*0.852)^2-3.6^2) = -226.26 J

4) change in KE of the meery go round is 0.5*I1*w2^2 = 0.5*200*0.852^2 = 72.6 J

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