Question

The translational kinetic energy of a circular disc of mass 1.5 kg rolling on a horizontal surface without slipping is 24 J. The total kinetic energy of the disc is:

• A. 24 J
• B. 12 J
• C. 36 J
• D. 30 J

Hint:

For a rolling object, the total kinetic energy is the sum of translational and rotational kinetic energy. For a disc, the rotational kinetic energy is half of the translational kinetic energy.

Answer 1

The Correct Option is C

For a disc rolling without slipping, the total kinetic energy is the sum of the translational and rotational kinetic energy. The rotational kinetic energy is half of the translational kinetic energy, so: \[ T_{\text{total}} = T_{\text{trans}} + \frac{1}{2} T_{\text{trans}} = \frac{3}{2} T_{\text{trans}} \] Substitute the given value of \( T_{\text{trans}} = 24 \, \text{J} \): \[ T_{\text{total}} = \frac{3}{2} \times 24 = 36 \, \text{J} \]
Hence, the total kinetic energy of the disc is 36 J.