Q) A fly wheel rotates about its central axis . Due to friction at the axis, it experiences angular retardation proportional to its angular velocity. If its angular velocity falls to half its initial value while its makes n revolutions , how many more revolutions will it make before coming to rest?

A) - a = k.w .........(k is a constant of proportionality)
(dw/dQ)(dQ/dt) = k.(dw/dt)
dw = - k.dQ
w = -kQ + c
Initially Q = 0, and w = W0
W0 = C

w = -kQ + W0
After n revolutions,
W0 / 2 = -k(2.n.pi) + W0
W0 / 2 = 2.k.n.pi
k = W0 / 4.n.pi

For final stop, w = 0
w = -kQ + W0
0 = -(W0 / 4.n.pi). Q + W0
Q = 4.n.pi
means 2n revolutions.
And given that n revolutions have already occurred. So number of revolutions that will further occur before the flywheel stops = 2n - n = n revolutions.

Thus, the flywheel will complete 'n' more revolutions before it stops.
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