Question

Three children are riding on the edge of a merry-go-round that is 122 kg, has a 1.60 m radius, and is spinning at 15.3 rpm. The children have masses of 19.9, 29.0, and 38.8 kg. If the child who has a mass of 29.0 kg moves to the center of the merry-go-round, what is the new angular velocity in rpm?

Answer 1

Mass of the marry go round M = 122 kg

Radius of the marry go round R = 1.6 m

Angular velocity of marry go round w = 15.3 rpm

w1= 0.255 rev/sec

w1= 0.255*2π rad/sec

w1= 1.6022 rad /sec

Now three children having mass 19.9 , 29 , 38.8 are on the edge of the marry go round

Hence the total mass of the system will be

M' = M + 19.9+29+38.8

M' = 122+19.9+29.38.8

M' = 209.7 kg

Hence the momentum of inertia of the system will be

Angular momentum of the system p1

P1 = I1 w1

P1 = ( M' * R^2 w1)/2

Now in second case the boy having mass 29 kg is move towards the center of the marry go round hence there will be no contribution in moment of interia of system by 29 kg child

Hence the total effective mass of the system will be

M'' = 122 + 19.9 + 38.8

M" = 180.7 kg

Hence the momentum of inertia of system will be I2

Now the angular momentum will be P2

P2 = I2 *w2

P2 = ( M"*R^2*w2)/2

Now by the conservation of momentum

P1 = p2

(M' *R^2 * w1 )/2 = ( M"*R^2*w2)/2

209.7 *w1 = 180.7*w2

w2 =( 209.7*w1)/180.7

w2 = 1.1604 w1

w2 = 1.1604 *15.3

w2= 17.7541 RPM