HRW7 11-58 A 27.2 kg child stands on the edge of a stationary merry-go-round of

Question

HRW7 11-58 A 27.2 kg child stands on the edge of a stationary merry-go-round of mass 117. kg and radius 1.62 m. The rotational inertia of the merry-go-round about its axis of rotation is 257. kg*m2. The child catches a ball of mass 1.73 kg thrown by a friend. Just before the ball is caught, it has a horizontal velocity of 12.0 m/s that makes an angle of 37.0° with a line tangent to the outer edge of the merry-go-round, as shown in the overhead view of figure 12-44. What is the angular speed of the merry-go-round just after that ball is caught?
rad/s

Explanation / Answer

Here the angular momentum will remain conserved therefore
initial angular momentum = mball*Vball*r = 1.73*(12Cos37)*1.62 = 26.86
Final angular momentum = IW
where I is moment of inertia ans W is angular velocity of the merry go rounnd
I = Imerry + Iboy+ Iball = 257 + (27.2*(1.622)) + (1.73*1.622) = 332.92 kg-m2
Now conservation of angular momentum
W = 26.86/332.92 = 0.0807 rad/s

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