Question

A uniform solid cylinder of mass \( m \) and radius \( r \) rolls along an inclined rough plane of inclination \( 45^\circ \). If it starts to roll from rest from the top of the plane, then the linear acceleration of the cylinder axis will be:

• A. \( \frac{1}{\sqrt{2}} g \)
• B. \( \frac{1}{3\sqrt{2}} g \)
• C. \( \frac{\sqrt{2} g}{3} \)
• D. \( \sqrt{2} g \)

Hint:

For rolling motion without slipping, use both the translational and rotational equations of motion. The frictional force causes the rotation, and the rolling condition \( a = r\alpha \) relates the linear and angular accelerations.

Answer 1

The Correct Option is C

Step 1: Find the acceleration of the cylinder. For a solid cylinder rolling down an incline without slipping, the acceleration $a$ is given by: \[ a = \frac{g \sin \theta}{1 + \frac{I}{mr^2}} \] where $I$ is the moment of inertia of the cylinder about its axis of rotation, $m$ is the mass, $r$ is the radius, and $\theta$ is the angle of inclination. 
For a solid cylinder, $I = \frac{1}{2}mr^2$. So, \[ a = \frac{g \sin \theta}{1 + \frac{\frac{1}{2}mr^2}{mr^2}} = \frac{g \sin \theta}{1 + \frac{1}{2}} = \frac{2}{3}g \sin \theta \] Step 2: Substitute the given values. In this case, $\theta = 45^\circ$, so $\sin \theta = \sin 45^\circ = \frac{1}{\sqrt{2}}$. \[ a = \frac{2}{3}g \left( \frac{1}{\sqrt{2}} \right) = \frac{2}{3\sqrt{2}}g = \frac{2\sqrt{2}}{3 \cdot 2}g = \frac{\sqrt{2}}{3}g \] Therefore, the linear acceleration of the cylinder axis is $\frac{\sqrt{2}}{3}g$. The correct answer is (3).