A merry-go-round is essentially a solid disk with a mass of 120 kg and a radius

Question

A merry-go-round is essentially a solid disk with a mass of 120 kg and a radius of 1.6 m. A 30-kg child sits on the outside edge of the merry-go-round. The merry-go-round is rotating at 4.8 rad/s. A )What is the rotational kinetic energy of the merry-go-round and child together? B) The child wants to stop the merry-go-round, so he drags his foot along the ground. This applies a tangential force of 29 N to the edge of the merry-go-round. How many complete rotations does the merry-go-round make before it comes to rest?

Explanation / Answer

The rotational kinetic energy of the system is

K = [ ( IM + ICh ) 2 ]/2

where
•IM is the moment of inertia of the merry-go-round relative to and axis at its center

IM = ( mM R2 )/2 = ( 120 Kg * ( 1.6 m )2 )/2 = 153.6 Kg.m2
•ICh is the moment of inertia of the child relative to and axis at its center of the merry-go-round

ICh = mCh R2 = 30 Kg * ( 1.6 m )2 = 48 Kg.m2

Then

K = [ ( IM + ICh ) 2 ]/2

K = [ ( 153.6 Kg.m2 + 48 Kg.m2 ) ( 4.8 Rad/s )2 ]/2 = 2322.4 J

from work energy theorem

W = K

F . d = K ---> d = K / F

d = ( 2322.4 J ) / ( 29 N ) = 80 m

if you divide the above result by the length of the circle you findthe number of turn made by the merry-go

n = d / ( 2 R )

n = ( 80 m ) / ( 2 * 1.6 m ) = 8.95

the merry-go-round makes 8 complete rotations

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