Question

A solid cylinder of radius \(R\) is at rest at a height \(h\) on an inclined plane. If it rolls down, then its velocity on reaching the ground is

• A. \(\sqrt{\frac{5gh}{3}}\)
• B. \(\sqrt{\frac{2h}{3g}}\)
• C. \(\sqrt{\frac{2gh}{3}}\)
• D. \(\sqrt{\frac{4gh}{3}}\)

Hint:

For rolling bodies, always include both translational and rotational kinetic energies: \[ K=\frac12 mv^2+\frac12 I\omega^2 \] and use the rolling condition \[ v=R\omega. \]

Answer 1

The Correct Option is D

Step 1: Apply conservation of mechanical energy.
The solid cylinder starts from rest at height $h$. All potential energy converts to kinetic energy (translational + rotational): \[ mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \] where $v$ is the translational speed at the bottom and $\omega$ is the angular velocity.
Step 2: Use the moment of inertia of a solid cylinder.
For a solid cylinder: \[ I = \frac{1}{2}mR^2 \]
Step 3: Apply the rolling without slipping condition.
For rolling without slipping: \[ v = R\omega \implies \omega = \frac{v}{R} \] Substituting: \[ \frac{1}{2}I\omega^2 = \frac{1}{2} \cdot \frac{1}{2}mR^2 \cdot \frac{v^2}{R^2} = \frac{1}{4}mv^2 \]
Step 4: Substitute back into the energy equation.
\[ mgh = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2 \] Cancelling $m$: \[ gh = \frac{3}{4}v^2 \] \[ v^2 = \frac{4gh}{3} \] \[ v = \sqrt{\frac{4gh}{3}} \]
Step 5: Compare with the options.
$v = \sqrt{\frac{4gh}{3}}$ matches option 4. Note how the rotational kinetic energy reduces the final translational speed compared to a sliding object (which would give $v = \sqrt{2gh}$).
Step 6: State the final answer.
The velocity of the solid cylinder on reaching the ground is $\sqrt{\dfrac{4gh}{3}}$. \[ \boxed{\sqrt{\dfrac{4gh}{3}}} \]