Question

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:

• A. $\frac{2F}{3mR}$
• B. $\frac{3F}{2mR}$
• C. $\frac{F}{3mR}$
• D. $\frac{F}{2mR}$

Answer 1

The correct option is A

$\frac{2F}{3mR}$

Step 1. Given data

Radius of the cylindrical roller is $R$

Mass of the cylindrical roller is $m$

Step 2. Finding the angular acceleration of the cylinder:

By using the formula of torque,$\tau$

$\tau =I\alpha$ [Where, $I$ is moment of inertia, $\alpha$is angular acceleration]

Also, $\tau =FR$ [Where, $F$ is force]

Moment of inertia about an axis passing through the center of cylinder is

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

Substituting this value,

$\tau =\frac{3}{2}m{R}^2\alpha$

Now, we can equate the two values of the torque

$\frac{3}{2}m{R}^2\alpha =FR$

$\alpha =\frac{2F}{3mR}$

Therefore, the angular acceleration is $\frac{2F}{3mR}$

Hence, option$\left(A\right)$ is correct.