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 8.40 rad/s?
__________________J

(b) How long does it take the spool to reach this angular speed?
__________________s

(c) How much cord is left on the spool when it reaches this angular speed?
_________________m

Explanation / Answer

a) Work = Change in angular momentum = I

I = MR*R/2 = 0.125

Work = 0.125 * 8.4 = 1.05 J

b) = a/R = 3/0.5 = 6 rad/s2

t = / = 1.4 sec

c) = t*t/2 = 6*1.4*1.4/2 = 5.88 rad

l = R = 5.88*0.5 = 2.94 m

Remaining = 4 - 2.94 = 1.06 m

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