A light rope is wound around a hollow cylinder of mass $$5$$ kg and radius $$70$$ cm. The rope is pulled with a force of $$52.5$$ N. The angular acceleration of the cylinder will be _____ rad s$$^{-2}$$.

A light rope is wound around a hollow cylinder and pulled with a force. We need to find the angular acceleration.

We are given that Mass $$m = 5$$ kg, radius $$R = 70$$ cm $$= 0.7$$ m, force $$F = 52.5$$ N.

To begin,

A hollow cylinder (thin-walled cylindrical shell) has all its mass concentrated at the radius $$R$$. Its moment of inertia about its central axis is:

$$ I = mR^2 $$

This differs from a solid cylinder, which has $$I = \frac{1}{2}mR^2$$. For a hollow cylinder, since all mass is at distance $$R$$ from the axis, we simply have $$I = mR^2$$.

$$ I = 5 \times (0.7)^2 = 5 \times 0.49 = 2.45\;\text{kg m}^2 $$

Next,

The rope is wound around the cylinder at radius $$R$$, so the force $$F$$ acts tangentially at a perpendicular distance $$R$$ from the axis. The torque is:

$$ \tau = F \times R = 52.5 \times 0.7 = 36.75\;\text{N m} $$

From this,

The rotational analogue of Newton's second law states:

$$ \tau = I\alpha $$

where $$\alpha$$ is the angular acceleration. Solving for $$\alpha$$:

$$ \alpha = \frac{\tau}{I} = \frac{36.75}{2.45} = 15\;\text{rad/s}^2 $$

The angular acceleration of the cylinder is 15 rad s$$^{-2}$$.