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

Options

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

Answer

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

### Step 1: Understand the Kinetic Energies
The total kinetic energy of a rolling disc consists of two parts:
1. **Translational Kinetic Energy (TKE)**: This is given by the formula:
\[ \text{TKE} = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the disc and \( v \) is its linear velocity.

2. **Rotational Kinetic Energy (RKE)**: This 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.

### Step 2: Determine the Moment of Inertia for a Disc
For a solid disc, the moment of inertia \( I \) about its central axis is:
\[ I = \frac{1}{2} mr^2 \]
where \( r \) is the radius of the disc.

### Step 3: Relate Angular Velocity to Linear Velocity
Since the disc rolls without slipping, the relationship between linear velocity \( v \) and angular velocity \( \omega \) is given by:
\[ v = r\omega \quad \Rightarrow \quad \omega = \frac{v}{r} \]

### Step 4: Substitute Angular Velocity into the Rotational Kinetic Energy Formula
Substituting \( \omega \) into the RKE formula gives:
\[ \text{RKE} = \frac{1}{2} I \left(\frac{v}{r}\right)^2 = \frac{1}{2} \left(\frac{1}{2} mr^2\right) \left(\frac{v^2}{r^2}\right) \]
This simplifies to:
\[ \text{RKE} = \frac{1}{4} mv^2 \]

### Step 5: Calculate the Ratio of RKE to TKE
Now, we can find the ratio of the rotational kinetic energy to the translational kinetic energy:
\[ \text{Ratio} = \frac{\text{RKE}}{\text{TKE}} = \frac{\frac{1}{4} mv^2}{\frac{1}{2} mv^2} \]
The \( mv^2 \) terms cancel out:
\[ \text{Ratio} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{4} \times \frac{2}{1} = \frac{1}{2} \]

### Final Answer
The ratio of the rotational kinetic energy to the translational kinetic energy for a disc rolling without slipping is:
\[ \frac{1}{2} \]

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