A large horizontal circular platform (M=70.1 kg, r=3.31 m) rotates about a frictionless vertical axle. A student (m=72.3 kg) walks slowly from the rim of the platform toward the center. The angular velocity ω of the system is 1.90 rad/s when the student is at the rim. Find ω (in rad/s) when the student is 1.45 m from the center.
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The angular momentum at the start is:
L = (I + m R^2) ω ,
where I is the moment of inertia of the platform (m R^2 is the moment of inertia of the (point mass) student when he is at the rim (r=R) ).
Because angular momentum is conserved we have
(I + m R^2) ω = (I + m r^2) w ,
where r is the final position of the student (r = 1.45 m) and w the final angular velocity.
For a uniform disk, the moment of inertia relative to the central axes is
I = 1/2 M R^2.
So
w = ω (1/2 M R^2 + m R^2) / (1/2 M R^2 + m r^2)
= 1.90 rad/s *( 1/2 * 70.1kg*(3.31m)^2 + 72.3 kg * (3.31 m)^2 ) /
( 1/2 * 70.1kg*(3.31m)^2 + 72.3 kg * (1.45 m)^2
= 4.17 rad/s
L = (I + m R^2) ω ,
where I is the moment of inertia of the platform (m R^2 is the moment of inertia of the (point mass) student when he is at the rim (r=R) ).
Because angular momentum is conserved we have
(I + m R^2) ω = (I + m r^2) w ,
where r is the final position of the student (r = 1.45 m) and w the final angular velocity.
For a uniform disk, the moment of inertia relative to the central axes is
I = 1/2 M R^2.
So
w = ω (1/2 M R^2 + m R^2) / (1/2 M R^2 + m r^2)
= 1.90 rad/s *( 1/2 * 70.1kg*(3.31m)^2 + 72.3 kg * (3.31 m)^2 ) /
( 1/2 * 70.1kg*(3.31m)^2 + 72.3 kg * (1.45 m)^2
= 4.17 rad/s