Question

A solid cylinder of mass \( M \) and radius \( R \) rolls down a smooth inclined plane about its own axis and reaches the bottom with velocity \( v \). The height of the inclined plane is

• A. \( \frac{3v^2}{4g} \)
• B. \( \frac{4v^2}{5g} \)
• C. \( \frac{7v^2}{9g} \)
• D. \( \frac{2v^2}{3g} \)

Hint:

For rolling motion, the total kinetic energy is the sum of translational and rotational kinetic energies. Use energy conservation to solve problems involving motion down an inclined plane.

Answer 1

The Correct Option is A

Step 1: Energy considerations.
The solid cylinder rolls down the inclined plane, so its total mechanical energy is conserved. Initially, the cylinder has potential energy \( mgh \), and at the bottom, this energy is converted into kinetic energy, which consists of both translational and rotational kinetic energy. The total kinetic energy is given by: \[ K.E. = \frac{1}{2} M v^2 + \frac{1}{2} I \omega^2 \] For a solid cylinder, the moment of inertia \( I \) is \( \frac{1}{2} M R^2 \), and the angular velocity \( \omega \) is related to the linear velocity \( v \) by \( \omega = \frac{v}{R} \).

Step 2: Equating energy.
Equating the potential energy to the total kinetic energy, we have: \[ mgh = \frac{1}{2} M v^2 + \frac{1}{2} \cdot \frac{1}{2} M v^2 = \frac{3}{4} M v^2 \] Thus, \[ mgh = \frac{3}{4} M v^2 \]
Step 3: Solving for height.
Now, using \( mgh = Mgh \) (since the mass of the cylinder is \( M \)), we can solve for the height \( h \) as: \[ h = \frac{3v^2}{4g} \]
Step 4: Conclusion.
The correct answer is (A) \( \frac{3v^2}{4g} \).