Question

A solid cylinder of mass 50 kg and radius $ 0.5{\text{ }}m $ is free to rotate about the horizontal axis. A massless string is wound around the cylinder with one end attached to it and the other hanging freely. Tension in the string required to produce an angular acceleration of 2 revolutions $ {s^{ - 2}} $ is:-
A) 25 N
B) 50 N
C) $ 78.5N $
D) 157 N

Answer 1

Hint: Since the string is wound around the cylinder when it is pulled, it rotates the cylinder about its axis. The tension needed in the string is equal to the force required to act tangentially on the surface of the cylinder to produce enough angular acceleration of 2 revolutions $ {s^{ - 2}} $.

Formula used: In this question, we will use the following formula
Torque acting on a body:
 $\Rightarrow \tau = F.r = I\alpha $ where $ F $ is the force acting on the rod, $ r $ is the distance of the centre of mass of the rod from the hinge, $ I $ is the moment of inertia of the rod about its end, and $ \alpha $ is the angular acceleration of the rod.

Complete step by step answer
We’ve been given that a solid cylinder that rotates about its horizontal axis has a string wound around the cylinder. To find the tension required to rotate the cylinder, we have to equalize the torque generated by the string with the torque required by the cylinder to rotate at an angular acceleration of 2 revolutions $ {s^{ - 2}} $
This can be written as:
 $\Rightarrow \tau = F.r = I\alpha $
Where $ F.r $ is the torque provided by the string where $ F = T $ is the tension provided by the string, and $ I\alpha $ is the torque needed by the cylinder to rotate with an angular acceleration $ \alpha $ such that its moment of inertia is $ I $.
For a solid cylinder, $ I = \dfrac{{M{R^2}}}{2} $. Now since the cylinder has two revolutions $ {s^{ - 2}} $, its angular acceleration will be
 $\Rightarrow \alpha = \dfrac{{2 \times 2\pi }}{1} $
 $ \Rightarrow \alpha = 4\pi \,rad/{s^2} $
Then substituting all these values in the formula $ F.r = I\alpha $, we get
 $\Rightarrow T \times 0.5 = \dfrac{{(50){{(0.5)}^2}}}{2} \times 4\pi $

$ \therefore T = 157\,N $ which corresponds to option (D).

Note
Here we have assumed that the string is being pulled completely tangentially to the string as it wound around the string otherwise, we would have to take into account the direction of the force acting on the cylinder in the direction tangential to the cylinder. We must be careful to find the angular acceleration of the cylinder is calculated in the units of degrees per seconds squared and not revolutions per second.