A 81.5kg cadet stands at the edge of a stationary merry-go-round at rest (assume

Question

A 81.5kg cadet stands at the edge of a stationary merry-go-round at rest (assume the disk to be uniform cylindrical disk of radius 2.06m and mass of 145kg, where I= 1/2MR^2). The merry-go-round is mounted on frictionless axle. The cadet throws a shot-put of mass Ms=7.27kg with speed of 18.5m/s relative to the ground and tangent to the edge of the merry-go-round. A). Calculate the moment of inertia of the merr-go-round/cadet system after the shot-put is thrown (without the shot-put) relative to the center of the merry-go-round. treat cadet as point of mass. B). What is the initial angular momentum of the merry-go-round/cadet/shot-put combination before the shot-put was thrown? C). Calculate the angular momentum of the shot-put at the moment it leaves the merry-go-round/cadet system relative to the center of the merry-go-round. D). Calculate the angular momentum of the merry go round/cadet system at the moment that the shot-put leaves the system. E). Determine the velocity of the cadet immediately after the shot-put is thrown. Assume the cadet is on the edge and doesnt slide on the surface.

Explanation / Answer

A)

moment of inertia of system , I = 0.5*145 * 2.06^2 + 7.27 * 2.06^2

I = 338.51 Kg.m^2

B)initial angular momentum = 0

C) angular momentum of shot-put = m*v*r

angular momentum of shot-put = 7.27 * 18.5 * 2.06

angular momentum of shot-put = 277.1 Kg.m/s

D)

after the shot put leaves the system ,

as the total angualar momentum is zero initally ,

angular momentum of the merry go round/cadet system = 277.1 Kg.m/s

E)

Now, If*w = 277.1

338.51 * wf = 277.1

wf = 0.819 rad/s

Now , velocity of cadet = 0.819 * 2.06

velocity of cadet = 1.69 m/s

the velocity of cadet is 1.69 m/s

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