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Question

In a playground, there is a small merry-go-round of radius $$1.20\ m$$ and mass $$180\ kgs$$. Its radius of gyrations $$91.0\ cm$$. A child of mass $$44.0\ kg$$ runs at the initially stationary merry-go-round and then jumps on. Neglect friction between the bearings and the shaft of the merry-go-round. Calculate the angular speed of the merry-go-round and child after the child has jumped onto the merry-go-round. 


Solution

No external torques act on the system consisting of the child and the merry-go-round, so the total angular momentum of the system is conserved.
An object moving a straight line has angular momentum about any point that is not on the line. The magnitude of the angular momentum of the child the center of the merry-go-round is given by Eq. $$11-21,mvR$$, where $$R$$ is the radius of the merry-go-round.
The initial angular momentum is given by $$ L_f=L_{child} = mvR$$
; the final angular momentum is given by $$L_f=(I+mR^2)w$$, where $$w$$ is the final common angilar velocity of the merry-go-round and child. Thus $$mvR=(I+mR^2)w$$ and
$$w=\dfrac{mvR}{I+mR^2}=\dfrac{158\ kg.m^2/s}{149\ kg.m^2+(44.0\ kg)(1.20\ m)^2}=0.744\ rod/s$$.
Note:The child initially had an angular velocity of 
$$w=\dfrac{v}{R}=\dfrac{3.00\ m/s}{1.20\ m}=2.5\ rod/s$$.
After he jumped onto the merry-go-round, the rotational inertia of the system (merry-go-round+child)increases, so the angular velocity decreases.

Physics

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