A disc is rolling without slipping. The ratio of its rotational kinetic energy and translational kinetic energy would be -

Options

A. `1:2`
B. `2:1`
C. `1:1`
D. `1:4`

Answer

To find the ratio of the rotational kinetic energy to the translational kinetic energy of a disc rolling without slipping, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Kinetic Energies:**
- The **translational kinetic energy (TKE)** of the disc is given by the formula:
\[ \text{TKE} = \frac{1}{2} m v^2 \]
- The **rotational kinetic energy (RKE)** of the disc is given by the formula:
\[ \text{RKE} = \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

2. **Moment of Inertia for a Disc:**
- The moment of inertia \( I \) for a solid disc about its center is:
\[ I = \frac{1}{2} m r^2 \]

3. **Relationship Between Linear and Angular Velocity:**
- For a disc rolling without slipping, the relationship between the linear velocity \( v \) and the angular velocity \( \omega \) is:
\[ v = r \omega \quad \Rightarrow \quad \omega = \frac{v}{r} \]

4. **Substituting \( \omega \) in RKE:**
- Substitute \( \omega \) into the RKE formula:
\[ \text{RKE} = \frac{1}{2} I \omega^2 = \frac{1}{2} \left(\frac{1}{2} m r^2\right) \left(\frac{v}{r}\right)^2 \]
- Simplifying this gives:
\[ \text{RKE} = \frac{1}{2} \cdot \frac{1}{2} m r^2 \cdot \frac{v^2}{r^2} = \frac{1}{4} m v^2 \]

5. **Finding the Ratio:**
- Now, we can find the ratio of RKE to TKE:
\[ \text{Ratio} = \frac{\text{RKE}}{\text{TKE}} = \frac{\frac{1}{4} m v^2}{\frac{1}{2} m v^2} \]
- The \( m v^2 \) terms cancel out:
\[ \text{Ratio} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{4} \cdot \frac{2}{1} = \frac{1}{2} \]

6. **Final Result:**
- Therefore, the ratio of the rotational kinetic energy to the translational kinetic energy is:
\[ \text{Ratio} = 1 : 2 \]

### Conclusion:
The ratio of the rotational kinetic energy to the translational kinetic energy of a disc rolling without slipping is \( 1 : 2 \).

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