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 find the linear acceleration of the cylinder axis as it rolls down the inclined plane, we need to analyze the forces acting on the cylinder and apply Newton's second law of motion along with the rotational dynamics. We are given that the inclined plane is rough, allowing the cylinder to roll without slipping. 

Given:

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

Concept: For a cylinder rolling without slipping, the relationship between linear acceleration \(a\) and angular acceleration \(\alpha\) is given by:

\(a = r \alpha\)

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

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

Applying Newton's second law along the incline, we have:

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

where \(f\) is the frictional force.

For rotational motion:

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

Substituting \(a = r \alpha\) gives:

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

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

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

Simplifying this equation, we get:

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

Canceling \(m\) and solving for \(a\):

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

Substituting \(\theta = 45^\circ\):

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

Hence, the linear acceleration is:

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

This is the correct answer.

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

Answer 2

Step 1: Find the acceleration of the cylinder. For a solid cylinder rolling down an incline without slipping, the acceleration \(a\) is given by the formula:

\[ a = \frac{g \sin \theta}{1 + \frac{I}{mr^2}}, \] 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, the moment of inertia \(I\) is given by: \[ I = \frac{1}{2}mr^2. \] Substituting this into the formula for acceleration: \[ a = \frac{g \sin \theta}{1 + \frac{\frac{1}{2}mr^2}{mr^2}} = \frac{g \sin \theta}{1 + \frac{1}{2}} = \frac{2}{3}g \sin \theta. \]

Step 2:

Substitute the given values. In this case, \(\theta = 45^\circ\), so \(\sin \theta = \sin 45^\circ = \frac{1}{\sqrt{2}}\). \[ a = \frac{2}{3}g \left( \frac{1}{\sqrt{2}} \right) = \frac{2}{3\sqrt{2}}g = \frac{2\sqrt{2}}{3 \cdot 2}g = \frac{\sqrt{2}}{3}g. \]

Final Answer:

The linear acceleration of the cylinder's axis is \( \frac{\sqrt{2}}{3}g \). The correct answer is option (3).