A homogeneous solid cylindrical roller of radius $$R$$ and mass $$M$$ is pulled on a cricket pitch by a horizontal force. Assuming rolling without slipping, angular acceleration of the cylinder is:

Let us denote the horizontal pulling force by $$F$$, the mass of the solid cylinder by $$M$$ and its radius by $$R$$. Because the surface offers sufficient grip, the motion is pure rolling without slipping. For pure rolling we always have the kinematic relation

$$a = \alpha R,$$

where $$a$$ is the linear (translational) acceleration of the centre of mass and $$\alpha$$ is the angular acceleration we have to find.

The two forces that produce horizontal translation are the external pulling force $$F$$ (towards the right, say) and the static frictional force $$f$$ exerted by the ground (towards the left, opposite to the motion, because the ground resists the tendency of the point of contact to slide forward). Applying Newton’s second law for translation,

$$F - f = M a.$$

Next we need the rotational equation about the centre of mass. The pulling force $$F$$ acts through the centre, so it produces no torque. The only horizontal torque comes from the frictional force $$f$$. The magnitude of this torque is

$$\tau = f R.$$

For rotation we apply Newton’s second law for rotation, $$\tau = I\alpha$$. For a homogeneous solid cylinder, the moment of inertia about its central axis is

$$I = \dfrac{1}{2} M R^{2}.$$

Therefore,

$$f R = I\alpha = \left(\dfrac{1}{2} M R^{2}\right)\alpha.$$

Simplifying,

$$\alpha = \dfrac{f R}{\dfrac{1}{2} M R^{2}} = \dfrac{2f}{M R}.$$

Now relate $$a$$ and $$\alpha$$ using the rolling condition $$a = \alpha R$$:

$$a = \alpha R = \left(\dfrac{2f}{M R}\right) R = \dfrac{2f}{M}.$$

Substituting this value of $$a$$ back into the translational equation $$F - f = M a$$, we obtain

$$F - f = M\left(\dfrac{2f}{M}\right) = 2f.$$

Hence

$$F = 3f \quad\Longrightarrow\quad f = \dfrac{F}{3}.$$

Finally, substitute this value of $$f$$ into the expression for $$\alpha$$:

$$\alpha = \dfrac{2f}{M R} = \dfrac{2}{M R}\left(\dfrac{F}{3}\right) = \dfrac{2F}{3 M R}.$$

Replacing $$M$$ by the symbol $$m$$ used in the options, we have

$$\alpha = \dfrac{2F}{3 m R}.$$

Hence, the correct answer is Option C.