6. A child with mass m = 40 kg runs along a line tangent to the edge of a statio

Question

6. A child with mass m = 40 kg runs along a line tangent to the edge of a stationary merry-go-round (with frictionless bearings) of radius R0 =2 meters, mass 40 kg and moment of inertia 100 kg m^2. The child moving at a speed of 5 m/s tangent to the edge, jumps on the merry-go-round and stays right at the edge. Treating the child as a point mass at 2 meters from the center of rotation, find: a) the initial angular momentum, L0, of the merry-go-round and child system before the child jumps on. b) the angular speed of the child and merry-go-round after the child jumps on. c) the initial kinetic energy of the child. d) the final kinetic energy of the child and merry-go-round.

Explanation / Answer

mass of child = mc = 40 kg

speed of child = Vc = 5 m/s

distance of child from axis of rotation = r = 2 m

moment of inertia of merry-go-round = Im = 100 kg m2

initial angular velocity of merry go round = Wi =0

final angular velocity = Wf

moment of inertia of child = Ic = mc r2 = (40) (2)2 = 160 kg m2

A)

initial angular momentum of child and merry-go-round ::

Li = initial angular momentum of child + initial angular momentum of merry-go-round

Li = mc Vc r + Im Wi

Li = (40) (5) (2) + (100) (0)

Li = 400 kgm2/s

B)

Using conservation of angular momentum ::

final angular momentum = initial angular momentum

Lf = Li

final angular momentum of child + final angular momentum of merry-go-round = initial angular momentum of child + initial angular momentum of merry-go-round

Ic Wf + Im Wf = 400

160 Wf + 100 Wf = 400

Wf =1.54 rad/s

C)

initial kinetic energy = (0.5) mc Vc2 = (0.5) (40) (5)2 = 500 J

D)

Final kinetic energy = kinetic energy of child + kinetic energy of merry-go-round

Final kinetic energy = (0.5) Ic Wf2 + (0.5) Im Wf2

final kinetic energy = (0.5) (Ic + Im ) Wf2 = (0.5) (160 + 100) (1.54)2 = 308.31 J

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