We wrap a light, flexible cable around a solid cylinder with mass 10.0 kg and diameter 30.0 cm....

We wrap a light, flexible cable around a solid cylinder with mass 10.0 kg and diameter 30.0 cm. We tie the free end of the cable to a block of mass 15.0 kg. The 15.0 kg mass is released from rest and falls, causing the cylinder to turn about a frictionless axle through its center. As the block falls, the cable unwinds without stretching or slipping, truing the cylinder.

(a) How far will the mass have to descend to give the cylinder 270J of kinetic energy?

moment of inertia of solid cylinder = MoI
= mr^2/2 = 10 * 0.15^2/ 2 = 0.1125
rotational KE of solid cylinder
= 0.5 * MoI * w^2
= 0.5 * 0.1125 * w^2
= 270
w = 69.28 rad /s
force = 15 * 9.81 = 147.15N
torque = force * radius = 147.15 * 0.15 = 22.07
= MoI * angular acc

angular acc = 22.07 / 0.1125 = 196.18 rad/s2
v^2 =u^2 + 2as
69.28^2 = 0 + 2 * 196.18 *s
s = 12.2329 rad = 1.947 turn
the mass have to descend
= 1.947 * pi 0.3
= 1.835m
answer