A 4.00 m length of light nylon cord is wound around a uniform cylindrical spool

Question

A 4.00 m length of light nylon cord is wound around a uniform cylindrical spool of radius 0.500 m and mass 1.00 kg. The spool is mounted on a frictionless axle and is initially at rest. The cord is pulled from the spool with a constant acceleration of magnitude 3.00 m/s2.

(a) How much work has been done on the spool when it reaches an angular speed of 7.50 rad/s?
J
(b) Assuming that there is enough cord on the spool, how long does it take the spool to reach this angular speed?
s
(c) What length of cord is pulled from the spool when the angular speed is reached?
m

Explanation / Answer

a) work done = 1/2 I w2 = 1/2 * (1/2 m r2) * w2

= 1/4 * 1.00 * 0.5002 * 7.52

work done = 3.51 J

b) w = a t / r

time t = r w / a = 0.500 * 7.5 / 3.00

time t = 1.25 s

c) thetaf = thetai + wi t + 1/2 alpha t2

thetaf = 1/2 alpha t2

= 1/2 * (3.00 / 0.500) * 1.252 = 4.69 rad

s = r theta = 0.5 * 4.69

length of cord = 2.34 m

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