Question

A one-piece cylinder has a core section protruding from the larger drum and is free to rotate around its central axis. A rope wrapped around the drum of radius 1.33 m exerts a force of 4.35 N to the right on the cylinder. A rope wrapped around the core of radius 0.51 m exerts a force of 6.62 N downward on the cylinder. What is the magnitude of the net torque acting on the cylinder about the rotation axis? Answer in units of N m

Answer 1

Magnitude the net torque about its axis of rotation is 2.41 Nm

Solution:

As per the question:

The radius of the wrapped rope around the drum, r = 1.33 m

Force applied to the right side of the drum, F = 4.35 N

The radius of the rope wrapped around the core, r' = 0.51 m

Force on the cylinder in the downward direction, F' = 6.62 N

Now, the magnitude of the net torque is given by:

`\tau_(net) = \tau + \tau'`

where

`\tau` = Torque due to Force, F

`\tau'` = Torque due to Force, F'

`tau = F* r`

`tau' = F'* r'`

Now,

`\tau_(net) = - F* r + F'* r'`

`\tau_(net) = - 4.35* 1.33 + 6.62* 0.51 = - 2.41\ Nm`

The net torque comes out to be negative, this shows that rotation of cylinder is in the clockwise direction from its stationary position.

Now, the magnitude of the net torque:

`|\tau_(net)| = 2.41\ Nm`