Question

A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N ? What is the linear acceleration of the rope ? Assume that there is no slipping.

Answer 1

Given, the mass of the hollow cylinder is 3 kg, radius of the cylinder is 40 cm and the applied force on the rope is 30 N.

Write the equation for the centripetal force acting on the hollow cylinder.

F c =mrα

Here, m is the mass of the hollow cylinder, r is the radius of the cylinder and α is the angular acceleration of the cylinder.

Under equilibrium condition, the centripetal force balances the tension force applied by the rope.

The equilibrium equation of forces acting on the cylinder is,

F c =T

Here, T is the tension force applied by the rope.

Substitute the value of F c in the above equation.

mrα=T

Substitute the values in the above equation.

( 3 kg )( 40 cm )α=30 N ( 3 kg )( 40 cm )( 1 m 100 cm )α=30 N α= 30 N ( 3 kg )( 0.40 m ) =25 rad⋅ s −2

Thus, the angular acceleration of the cylinder is 25 rad⋅ s −2 .

The equation for the linear acceleration of the rope is,

a c =rα

Substitute the values in the above equation.

a c =( 40 cm )( 25  rad/ s 2 ) =( 40 cm )( 1 m 100 cm )( 25  rad/ s 2 ) =( 0.40 m )( 25  rad/ s 2 ) =10 m⋅ s −2

Thus, the linear acceleration of the rope is 10 m⋅ s −2 .