A child of mass 46.3 kg sits on the edge of a merry-go-round with radius 2.6 m a

Question

A child of mass 46.3 kg sits on the edge of a merry-go-round with radius 2.6 m and mo- ment of inertia 253.52 kgm2. The merry- go-round rotates with an angular velocity of 1.9 rad/s. The child then walks towards the center of themerry-go-round and stops at a distance 0.936 m from the center. Now what is the angular velocity of the merry-go-round? Answer in units of rad/s. A child of mass 46.3 kg sits on the edge of a merry-go-round with radius 2.6 m and mo- ment of inertia 253.52 kgm2. The merry- go-round rotates with an angular velocity of 1.9 rad/s. The child then walks towards the center of themerry-go-round and stops at a distance 0.936 m from the center. Now what is the angular velocity of the merry-go-round? Answer in units of rad/s.

Explanation / Answer

the mass of merry go round=253.52/2.6*2.6=37.50kg now I1=[37.5+46.3]*2.6*2.6=566.48kgm2 the new moment of inertiaI2=[46.3+37.5]*0.936*0.936=73.41kgm2 we have the principle I11=I22 2=566.48*1.9/73.41=14.66rad/sec

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