Consider the two people on the merry-go-round to be 85 kg point masses located a

Question

Consider the two people on the merry-go-round to be 85 kg point masses located at the edge of a 200 kg solid disk with a radius of 1.5m. a) Calculate the moment of inertia for the system of two 85 kg point masses at 1.5 m and a 200 kg solid disk of radius 1.5m. b) After 6 seconds of the motorcycle applying a force at the radius of the merry-go-round, the system rotates at 1 revolution per second. Calculate the average force applied by the motorcycle on the system during the s second interval. c) Once the unfortunate person is released (without putting a net torque on the remaining parts of the system), they leave with linear momentum that is tangent to the merry-go-round thus leaving with angular momentum. Does the remaining person/merry-go-round rotational velocity speed up, slow down or remain the same?

Explanation / Answer

Given

merry go round radius r = 1.5 m, maass M = 200 kg

mass of the two point masses each m = 85 kg

the moment of inertia of solid disc is I = MR^2/2

a)
here the moment of inertia of merry go round is I = (0.5*M*R^2+ 2mr^2)

       I = (0.5*200*1.5^2+2*85*1.5^2) kg m2

       I 607.5 kgm2

b)

   the angular speed of the merry go round is 1 revolution per sec = 2pi rad /s

the torque acting on the merry-go-round is T = I*alpha

                   T = 607.5*2pi/6 Nm

                   T = 636.173 Nm

and from the relation torque T = r*F sin theta

           F = T/(r sin theta)

               = 636.173 /(1.5sin90) N

               = 424.12 N

c) when a person released the moment of inertia becomes less when compared with previous case and there by the rotational speed increases

answert is rotational speed up.

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