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| helokit |
Posted: Mar 3 2006, 04:04 AM
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Member ![]() ![]() Group: Members Posts: 70 Joined: 13-December 05 Positive Feedback: 100% Feedback Score: 1 |
A person stands, hands at the side, on a platform that is rotating at a rate of 1 = 1.00 rev/s. If the person now raises his arms to a horizontal position, Fig. 10-59, the speed of rotation decreases to 2 = 0.90 rev/s.
(a) Why does this occur? As moment of inertia increases, angualr velocity must decrease ( i don't know how to start the problem. What i do with the rev/s. I know the inertia of a point mass is I=mr^2. does that work or no? |
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| UCBearCat14159 |
Posted: Mar 3 2006, 06:02 AM
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this is an application of conservation of angular momentum. It occurs simply because angular momentum is conserved throughout the change in the moment of inertia... this is the same principle that is applied when an ice skaters spin faster and faster as they pull their arms in
L= I * w where L is angular momentum, I is the moment of inertia, and w is the angular rate of rotation. If momentum is conserved, L1 = L2, in other words I1 * w1 = I2 * w2 I2/I1 = w1/ w2 I2/I1 is the factor by which the moment of inertia has changed. |
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