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posted by  chiefdog on 7/22/2008 7:59:09 PM  |  status: Live  

impulse-momentum

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  1. Inside radius is 0.4m and outside radius is 0.75m. Mass = 100kg and radius of gyration for wheel about its center is 0.35m. The system is initially at rest and then rolls without slipping when the force P = (t + 10) N is applied. Find angular velocity at t=5sec. Answer: 1.05 rad/s. Give direction.
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posted by zsm28 on 7/23/2008 2:07:20 AM  |  status: Live
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Response Details:
P = t + 10, r = 0.40 m, R = 0.75 m, m = 100 kg, k = 0.35 m, moment of inertia about the center = mk2, moment of inertia about the point A on the wheel contacting the ground I = mk2 + mR2 = m(k2 + R2) = 100*(0.352 + 0.752) = 68.5 kg-m2
The torque about A is T = P*(r + R) = (t + 10)*(0.40 + 0.75) = 1.15*(t + 10)
 
68.5 ω = 1.15*(52/2 + 10*5)
∴ω = 1.05 rad/s
direction: CW
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