A light rope is wound around a hollow cylinder of mass 5kg and...

The formula to calculate the moment of inertia $\left(I\right)$ of the cylinder about its radial axis is given by

$I = M{r}^{2} . . . \left(1\right)</p><p>Also, the torque \left(τ\right) on the cylinder about the central axis because of the application of the external force is given by</p><p>$τ = Fr . . . \left(2\right)$</p><p>Also, the torque can be expressed as</p><p>$τ = I\alpha . . . \left(3\right)

Substitute the expressions from equation (1) and (2) into equation (3) and simplify to obtain the angular acceleration.

$\begin{cases}Fr & = & M{r}^{2}\alpha \\ & ⇒ & \alpha = \dfrac{F}{Mr} . . . \left(4\right)\end{cases}$

Substitute the values of the known parameters into equation (4) to calculate the required angular acceleration.

$\begin{cases}\alpha & = & \dfrac{52 . 5N}{5kg \times 0 . 70m} \\ & = & 15rad{s}^{- 2}\end{cases}$