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 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
Show 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
\[
v=R\omega.
\]
Updated On: Jun 22, 2026
- \(\sqrt{\frac{5gh}{3}}\)
- \(\sqrt{\frac{2h}{3g}}\)
- \(\sqrt{\frac{2gh}{3}}\)
- \(\sqrt{\frac{4gh}{3}}\)
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The Correct Option is D
Solution and Explanation
Step 1: Apply conservation of mechanical energy.
Initially, the solid cylinder is at height \(h\), so its total energy is purely gravitational potential energy: \[ PE=mgh \] When it reaches the ground, the energy is converted into: (i) Translational kinetic energy and (ii) Rotational kinetic energy Therefore, \[ mgh=\frac12 mv^2+\frac12 I\omega^2 \]
Step 2: Use moment of inertia of a solid cylinder.
For a solid cylinder, \[ I=\frac12 mR^2 \] Also, for rolling without slipping, \[ v=R\omega \] Hence, \[ \omega=\frac{v}{R} \] Substitute these into the energy equation: \[ mgh=\frac12 mv^2+\frac12\left(\frac12 mR^2\right)\left(\frac{v}{R}\right)^2 \]
Step 3: Simplify the equation.
\[ mgh=\frac12 mv^2+\frac14 mv^2 \] \[ mgh=\frac34 mv^2 \] Cancel \(m\): \[ gh=\frac34 v^2 \] Thus, \[ v^2=\frac{4gh}{3} \] Therefore, \[ v=\sqrt{\frac{4gh}{3}} \]
Step 4: Final conclusion.
Hence, the velocity of the cylinder on reaching the ground is \[ \boxed{\sqrt{\frac{4gh}{3}}} \]
Initially, the solid cylinder is at height \(h\), so its total energy is purely gravitational potential energy: \[ PE=mgh \] When it reaches the ground, the energy is converted into: (i) Translational kinetic energy and (ii) Rotational kinetic energy Therefore, \[ mgh=\frac12 mv^2+\frac12 I\omega^2 \]
Step 2: Use moment of inertia of a solid cylinder.
For a solid cylinder, \[ I=\frac12 mR^2 \] Also, for rolling without slipping, \[ v=R\omega \] Hence, \[ \omega=\frac{v}{R} \] Substitute these into the energy equation: \[ mgh=\frac12 mv^2+\frac12\left(\frac12 mR^2\right)\left(\frac{v}{R}\right)^2 \]
Step 3: Simplify the equation.
\[ mgh=\frac12 mv^2+\frac14 mv^2 \] \[ mgh=\frac34 mv^2 \] Cancel \(m\): \[ gh=\frac34 v^2 \] Thus, \[ v^2=\frac{4gh}{3} \] Therefore, \[ v=\sqrt{\frac{4gh}{3}} \]
Step 4: Final conclusion.
Hence, the velocity of the cylinder on reaching the ground is \[ \boxed{\sqrt{\frac{4gh}{3}}} \]
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