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

To determine the linear acceleration of the cylinder's axis as it descends the inclined plane, an analysis of the forces acting upon it is required. This analysis will employ Newton's second law of motion and rotational dynamics. It is stipulated that the inclined plane possesses sufficient roughness to permit the cylinder to roll without slippage.

Given:

• Angle of inclination, \(\theta = 45^\circ\)
• Gravitational acceleration, \(g\)

Concept: For a cylinder undergoing rolling without slipping, the relationship between its linear acceleration \(a\) and angular acceleration \(\alpha\) is defined as:

\(a = r \alpha\)

The moment of inertia \(I\) of a solid cylinder about its central axis is given by:

\(I = \frac{1}{2} m r^2\)

Application of Newton's second law along the incline yields:

\(m g \sin\theta - f = m a\)

Herein,  \(f\) represents the frictional force.

For the rotational dynamics:

\(f \cdot r = I \alpha = \frac{1}{2} m r^2 \alpha\)

By substituting \(a = r \alpha\), the frictional force can be expressed as:

\(f = \frac{1}{2} m a\)

Substituting this expression for \(f\) into the linear motion equation results in:

\(m g \sin\theta - \frac{1}{2} m a = m a\)

Simplification of this equation leads to:

\(m g \sin\theta = \frac{3}{2} m a\)

Upon canceling \(m\) and solving for \(a\), we obtain:

\(a = \frac{2}{3} g \sin\theta\)

Substituting the given angle \(\theta = 45^\circ\):

\(a = \frac{2}{3} g \sin 45^\circ = \frac{2}{3} \cdot \frac{g}{\sqrt{2}}\)

The resulting linear acceleration is:

\(a = \frac{\sqrt{2} g}{3}\)

This result is validated.

Conclusion: The linear acceleration of the cylinder's axis as it rolls down the inclined plane is \(\frac{\sqrt{2} g}{3}\).