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 2.82 m/s2. (a) How much work has been done on the spool when it reaches an angular speed of 7.45 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

Answer 1

a) I = 0.50 M * R^2

I = 0.50 * 1 * 0.50^2

I = 0.125 Kg.m^2

work done on spool = increase in kinetic energy

work done on spool = 0.50 * I * w^2

work done on spool = 0.50 * 0.125 * 7.45^2

work done on spool = 3.47 J

b)

angular acceleration = acceleration/radius

angular acceleration = 2.82/0.50

angular acceleration = 5.64 rad/s^2

let the time taken is t

7.45 = 5.64 * t

t = 1.32 s

the time taken is 1.32 s

c)

the length of cord released, L

L = 0.50 at^2

L = 0.50 * 2.82 * 1.32^2

L = 2.46 m

length of string left = 4 - 2.46 = 1.54 m

the length of string is left is 1.54 m