A student holds a bike wheel and starts it spinning with an initial angular spee

Question

A student holds a bike wheel and starts it spinning with an initial angular speed of 9.0 rotations per second. The wheel is subject to some friction, so it gradually slows down. In the 10-s period following the inital spin, the bike wheel undergoes 77.5 complete rotations.

Assuming the frictional torque remains constant, how much more time ?ts will it take the bike wheel to come to a complete stop?

The bike wheel has a mass of 0.825 kg and a radius of 0.385 m. If all the mass of the wheel is assumed to be located on the rim, find the magnitude of the frictional torque ?f that was acting on the spinning wheel.

sapling learning A student holds a bike wheel and starts it spinning with an initial angular speed of 9.0 rotations per second. The wheel is subject to some friction, so it gradually slows down. In the 10-s period following the inital spin, the bike wheel undergoes 77.5 complete rotations. Assuming the frictional torque remains constant, how much more time Ats will it take the bike wheel to come to a complete stop? Number The bike wheel has a mass of 0.825 kg and a radius of 0.385 m. If all the mass of the wheel is assumed to be located on the rim, find the magnitude of the frictional torque that was acting on the spinning wheel. Number N·m Im

Explanation / Answer

Frictional torque = T
w1 = 9 * 2pi = 56.52 rad s-1
I = mr^2 = 0.825*0.385^2 = 0.122 kg m2
After 10 seconds
theta = w1t - 0.5*T*t^2/I = 56.52*10 - 0.5*T*100/0.122 = 565.2 - 409.83*T
but theta = 77.5*2*pi = 486.7 rad = 566.2 - 409.83T
T = 0.1939 Nm
w2 = w1 - at = 40.619 rad/s
angular acc, a = 0.1939/0.122 = 1.59 rad/s/s

Now w3 = 0
2*1.59*theta = 40.619^2
theta = 518.85 rad = n*2*pi
1. n = 82.62 turns more
2. T = 0.1939 Nm

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