A 43 kg flywheel with a radius of gyration of 0.20 m is supported on a single be

Question

A 43 kg flywheel with a radius of gyration of 0.20 m is supported on a single bearing at its center of rotation. The flywheel is spinning at a speed of 16 rads/s when a loss of lubrication causes the coefficient of friction between the 100 mm diameter bearing and flywheel to increase to mu(k)=0.30. Determine the time required for the friction to stop the flywheel from rotating. Also determine the angular displacement in this time period.

t=______________________________

theta=________________________

Explanation / Answer

inertia of flywheel = 43*0.2^2 = 1.72 kg m^2

intial angular speed , u = 16 rad/sec

frictional force = 43*9.8*0.3 = 126.42 N

frictional torque = 126.42*0.1 = 12.642 Nm

so... angular retardation (a) = torque / inertia = - 12.642/1.72 = -7.35 rad/sec2

final angular speed , v = 0

v = u + a*t

0 = 16 - 7.35*t

so... t = 16/7.35 = 2.176871 secs

also

v^2 = u^2 + 2*a*s

0 = 16^2 - (2*7.35*s)

so. angle s = 17.414966 radians = 17.414966 * 180 / ( pi ) = 997.804 degrees

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