A playground merry-go-round of radius R=2 m has a moment of inertia I=250 kgm^2 and is rotating at 10 rev/min about a frictionless, vertical axle. Facing the axle, a 25 kg child hops onto the merry-go-round and manages to sit down on the edge. The new angular speed of the merry-go-round is A.10 rev/min B.7.14 rev/min C.3.14 rev/min D.8.45 rev/min

Question

A playground merry-go-round of radius R=2 m has a moment of inertia I=250 kgm^2  and is rotating at 10 rev/min about a frictionless, vertical axle. Facing the axle, a 25 kg child hops onto the merry-go-round and manages to sit down on the edge. The new angular speed of the merry-go-round is A.10 rev/min B.7.14 rev/min C.3.14 rev/min D.8.45 rev/min

driwant9HB · Accepted Answer

The correct option is  B 7.14 rev/minGiven: R=2 m,   I  =  250  k  g      /        m    2  , ω=10 rev/min, m=25kgWhen the boy jumped onto the merry-go-round, its moment of inertia changed. In order to maintain the same angular momentum, the round then decreased its rotational speed. Moment of inertia of boy about centre will be    =  m      R    2  From conservation of angular momentum,Iiωi=Ifωf250×10=[250+(25)(2)2]ωfωf=7.14 rev/min

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