A large horizontal circular platform (M=91.7 kg, r=3.52 m) rotates about a frict

Question

A large horizontal circular platform (M=91.7 kg, r=3.52 m) rotates about a frictionless vertical axle. A student (m=60.99 kg) walks slowly from the rim of the platform toward the center. The angular velocity of the system is 3.9 rad/s when the student is at the rim. Find the moment of inertia of platform through the center with respect to the z-axis. Tries 0/5 Find the moment of inertia of the student about the center axis (while standing at the rim) of the platform. Tries 0/5 Find the moment of inertia of the student about the center axis while the student is standing 1.82 m from the center of the platform. Tries 0/5 Find the angular speed when the student is 1.82 m from the center of the platform. Tries 0/5

Explanation / Answer

M=91.7kg, r =3.52m, m= 60.99kg, w =3.9rad/s

1) M.I of circular platform(disk) =1/2*M*R^2 ==> I= 1/2*91.7*3.52^2 =568.0998 kg*m^2

2)M.I of student =m*r^2 =60.99*3.52^2 =755.69kg.m^2

3)M.I of student =m*r^2 =60.99*1.82^2 =202.0232kg.m^2

4)M.I of system = M.I of circular platform(disk) + M.I of student = 1/2*M*R^2 + m*r^2

when student at rim: I1= 1/2*91.7*3.52^2 + 60.99*3.52^2 = 1323.79kg.m^2

When student at r= 1.82m

I2= 1/2*91.7*3.52^2 + 60.99*1.82^2 = 770.12kg.m^2

As no external forces involved when student walked, angular momentum is conserved

=> I1*W1 = I2*W2 => 1323.79*3.9 =770.12*W2

=> W2 = 6.703 rad/s

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