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Question

In figure, a 30 kg child stands on the edge of a stationary merry-go-round of. radius 2 m The rotational inertia of the merry-go-round about its rotation axis is 196 kgm2 The child catches a ball of mass 1 kg thrown by a friend. Just before the ball is caught, it has a horizontal velocity v of magnitude 10 m/s, at angle ϕ=37 with a line tangent to the outer edge of the merry-go-round, as shown. What is the angular speed of the merry-go-round just after the ball is caught?

A
0.05 rad/s
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B
0.25 rad/s
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C
0.75 rad/s
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D
0.10 rad/s
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Solution

The correct option is A 0.05 rad/s
Formula used : L=Iω and Lball=mvr

Given:
mman=30kg,R=2m,Iplatform=196kgm2,mball=1kg,vball=10m/sϕ=37

Let mball and v be the mass and initial speed of the ball and R the radius of the merry-go-round. The initial angular momentum is given by

Li=mball(vcosϕ)R=1×10cos37×2=16units

=(196+30×22+1×22)ω

Li=Lf

16=320ω

ω=0.05 rad/s

FINAL ANSWER: (a)

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