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 2.74 m/s2.

(a) How much work has been done on the spool when it reaches an angular speed of 6.80 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

Here,

let the velocity of the spool is v

a)

work done on the spool = kinetic energy of spool

work done on the spool = 0.5 * m * v^2 + 0.5 * I * w^2

work done on the spool = 0.5 * 1 * (6.8/.5)^2 + 0.5 * 0.5 * 0.50^2 * 1 * 6.8^2

work done on the spool = 95.4 J

the work done on the spool is 95.4 J

b)

let the time taken is t

angular acceleration = at/R

angular acceleration = 2.74/0.5 rad/s

Using first equation of motion

w = angular acceleration* time

6.80 = 2.74/0.5 * t

t = 1.241 s

the time taken is 1.241 s

c)

For the cord left is L

using second equation of motion

4 - L = 0.5 * at * t^2

4 - L = 0.5 * 2.74 * 1.241^2

L = 1.891 m

1.891 m cord is left on the spool

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