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 2.92 m/s2 (a) How much work has been done on the spool when it reaches an angular speed of 6.30 rad/s? (b) How long does it take the spool to reach this angular speed? (c) How much cord is left on the spool when it reaches this angular speed? Need Help? Read N

Explanation / Answer

given

m = 1 kg

r = 0.5 m

a = 2.92 m/s2

w = 6.30 rad/s

(a)

Work done = energy gained = 1/2*I*w2

where I = 1/2*m*r2 = 1/2*1*(0.50)2

Work done = 1/2* 1/2*1*(0.50)2 *(6.30)2

Work done = 2.48 J

(b)

t =v/a = w*r/a = 6.30*0.5/2.92 = 1.07 s

(c)

d = u*t +1/2*a*t2

given that spool is initially at rest . so, u =0

d = 1/2*a*t2 = 1/2*2.92*(1.07)2 = 1.69 m

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