Angular Momentum for general physics one
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Angular Momentum for general physics one

[From: ] [author: ] [Date: 12-04-05] [Hit: ]
45 m from the center.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) ).(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.......
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
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