Question

A solid cylinder rolls down an inclined plane without slipping. If the translational kinetic energy of the cylinder is 140 J, the total kinetic energy of the cylinder is

• A. 105 J
• B. 70 J
• C. 210 J
• D. 280 J

Hint:

For rolling objects, the total kinetic energy is the sum of both translational and rotational kinetic energies. For a solid cylinder, the rotational kinetic energy is half of the translational kinetic energy.

Answer 1

The Correct Option is C

Step 1: When an object rolls without slipping, its total kinetic energy is the sum of its translational kinetic energy and its rotational kinetic energy. The total kinetic energy is: \[ K_{\text{total}} = K_{\text{trans}} + K_{\text{rot}} \] Step 2: For a solid cylinder rolling without slipping, the rotational kinetic energy is related to the translational kinetic energy by: \[ K_{\text{rot}} = \frac{1}{2} I \omega^2 \] where \( I = \frac{1}{2} m r^2 \) is the moment of inertia for a solid cylinder and \( \omega = \frac{v}{r} \) is the angular velocity. Thus, \[ K_{\text{rot}} = \frac{1}{2} m v^2 \] Therefore, the total kinetic energy becomes: \[ K_{\text{total}} = K_{\text{trans}} + \frac{1}{2} K_{\text{trans}} = \frac{3}{2} K_{\text{trans}} \] Step 3: Given that \( K_{\text{trans}} = 140 \, \text{J} \), the total kinetic energy is: \[ K_{\text{total}} = \frac{3}{2} \times 140 = 210 \, \text{J} \] Thus, the total kinetic energy of the cylinder is 210 J.