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

Question

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 2.94 m/s_2. (a) How much work has been done on the spool when it reaches an angular speed of 6.95 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 done = change in KE

W = KEf - KEi = Iw^2 /2 - 0

W = ( M R^2 /2 ) ( w^2 ) /2 = (1 x 0.5^2 /2 ) (6.95^2 ) /2

W = 3.02 J

b) Using wf = wi + alpha*t

and alpha = a/r = 2.94/0.5 = 5.88 rad/s^2

6.95 = 0 + 5.88t

t = 1.18 sec

c) linear acceleration of cord = 2.94 m/s^2

using d = ut + at^2 /2

d = 0 + (2.94 x 1.18^2 /2 ) = 2.05 m

length of cord on spool = 4 - 2.05 =1.95 m

so no cord is left on spool.

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