Question

A solid cylinder rolls down on an inclined plane of height \( h \) and inclination \( \theta \). The speed of the cylinder at the bottom is:

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

Hint:

When solving for the speed of a rolling object, remember to account for both translational and rotational kinetic energy. Use conservation of energy to relate potential energy and kinetic energy.

Answer 1

The Correct Option is D

Step 1: Use energy conservation principle For a solid cylinder rolling down an inclined plane, the total energy is conserved. The potential energy at the top of the incline is converted into kinetic energy at the bottom. The potential energy at height \( h \) is: \[ PE = mgh \] At the bottom, the kinetic energy of the cylinder is a combination of translational and rotational energy. The total kinetic energy is: \[ KE = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 \] For a solid cylinder, the moment of inertia \( I \) is given by \( I = \frac{1}{2} m r^2 \) and the relation between linear velocity \( v \) and angular velocity \( \omega \) is \( v = r \omega \). Substituting this into the expression for kinetic energy: \[ KE = \frac{1}{2} m v^2 + \frac{1}{2} \times \frac{1}{2} m r^2 \times \left(\frac{v}{r}\right)^2 = \frac{1}{2} m v^2 + \frac{1}{4} m v^2 = \frac{3}{4} m v^2 \] Step 2: Equating potential and kinetic energy From conservation of energy: \[ mgh = \frac{3}{4} m v^2 \] Solving for \( v \): \[ v^2 = \frac{4gh}{3} \] Thus, the speed of the cylinder at the bottom is: \[ v = \sqrt{\frac{4gh}{3}} \]