A thin, light wire is wrapped around the rim of a wheel. The wheel rotates about

Question

A thin, light wire is wrapped around the rim of a wheel. The wheel rotates about a stationary horizontal axle that passes through the center of the wheel. The wheel has radius 0.190m and moment of inertia for rotation about the axle of 0.555kg?m2 . A small block with mass 0.420kg is suspended from the free end of the wire. When the system is released from rest, the block descends with constant acceleration. The bearings in the wheel at the axle are rusty, so friction there does -5.00J of work as the block descends 3.25m . What is the magnitude of the angular velocity of the wheel after the block has descended 3.25m?

Explanation / Answer

by conservation law,
PEblock + Fbearing = KEwheel + KEblock
mgh + F = mv2/2 + I?2/2
at any moment the the point of ocntact of string anf wheel is stationary i.e v=r?
mgh + F = m(rw)2/2 + I?2/2
mgh + F = [mr2/2 + I/2]?2
?2 = [mgh + F]/ [mr2/2 + I/2]
=> ? =? ([mgh + F]/ [mr2/2 + I/2]) = 5.02

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