Question

A thin uniform circular disc rolls with a constant velocity without slipping on a horizontal surface. Its total kinetic energy is

• A. three times its rotational kinetic energy
• B. three times its translational kinetic energy
• C. one and half times its rotational kinetic energy
• D. twice its translational kinetic energy

Hint:

When a disc rolls without slipping, total KE = translational KE + rotational KE about the center of mass.

Answer 1

The Correct Option is A

Step 1: Understand the motion
For a disc rolling without slipping, total kinetic energy is the sum of translational and rotational kinetic energies.
Step 2: Use the formulas
For a disc of mass \( m \), radius \( R \), and center of mass velocity \( v \):
\[ \text{Translational KE} = \frac{1}{2}mv^2 \]
\[ \text{Rotational KE about CM} = \frac{1}{2}I\omega^2 = \frac{1}{2} \cdot \frac{1}{2}mR^2 \cdot \left(\frac{v}{R}\right)^2 = \frac{1}{4}mv^2 \]
Step 3: Total kinetic energy
\[ \text{Total KE} = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2 \]
Step 4: Rotational kinetic energy
\[ \text{Rotational KE} = \frac{1}{4}mv^2 \]
Step 5: Ratio
\[ \text{Total KE} = 3 \times (\text{Rotational KE}) \]