91Ó°ÊÓ

The mass of the cylinder = 5kg
The diameter of the cylinder = 60 cm
So radius = 30 cm = 0.3 m

Lets draw the FBD as below

Angular acceleration is calculated by

$\mathrm{Î\pm }=\frac{\mathrm{Ï„}}{I}.............................\left(1\right)$

Torque can be calculated by

$\mathrm{τ}=F×l................................\left(2\right)$

Where, F is the net force acting on the cylinder and l is the length of arm.

Only force acting is gravitational force

Substitute in equation(2) G = mg

$\mathrm{Ï„}=mgR......................\left(3\right)$

This is in the shape of disc so use the moment of inertia of disc.

$I=I_{C}M+M{R}^2......................\left(4\right)$

I_CM is the moment of inertia about it center of mass $\frac{M{R}^2}{2}$

Substitute in equation(4) we get

$I=\frac{3M{R}^2}{3}$

Substitute values in equation (1) to get angular acceleration

$$\begin{gathered} \mathrm{Î\pm }=\frac{MgR}{\frac{3}{2}M{R}^2} \\ \mathrm{Î\pm }=\frac{2g}{3R}………………………….\left(5\right) \end{gathered}$$

Substitute given values in the equation (5) we get

$$\begin{gathered} \mathrm{Î\pm }=\frac{2×9.8{\mathrm{ms}}^{-2}}{3×0.3\mathrm{m}} \\ =21.8{\mathrm{rads}}^{-2} \end{gathered}$$

The mass of the cylinder = 5kg
The diameter of the cylinder = 60 cm
So radius = 30 cm = 0.3 m

To calculate the angular speed use the formula below

${\mathrm{Ó\neg }}_f^2-{\mathrm{Ó\neg }}_i^2=2\mathrm{Î\pm }\mathrm{θ},…………………………..\left(6\right)$

Where

Ӭ_f= final velocity, Ӭ_i = initial velocity ,θ is angular displacement and α is angular displacement.

Substitute values: s the cylinder is below axle so angular displacement ,θ =π/2

Ó¬_i =0 and α =21.8 rad/s²

$$\begin{gathered} {\mathrm{Ó\neg }}_f^2-(0{)}^2=2×(21.8rad/s^2)×(\mathrm{Ï€}/2) \\ {\mathrm{Ó\neg }}_f=8.25rad/s \end{gathered}$$