A disc is rolling on a surface without slipping. What is the ratio of its translational to rotational kinetic energies ?

Options

A. `5 : 2`
B. `1 : 1`
C. `3 : 2`
D. `2 : 3`

Answer

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

### Step 1: Understand the Kinetic Energies
- The translational kinetic energy (TKE) of the disc 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.

### Step 2: Determine the Rotational Kinetic Energy
- The rotational kinetic energy (RKE) 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 3: Find the Moment of Inertia for a Disc
- For a solid disc, the moment of inertia \( I \) about its central axis is:
\[ I = \frac{1}{2} m r^2 \]
where \( r \) is the radius of the disc.

### Step 4: Relate Linear and Angular Velocity
- Since the disc is rolling without slipping, the relationship between linear velocity \( v \) and angular velocity \( \omega \) is:
\[ v = r \omega \quad \Rightarrow \quad \omega = \frac{v}{r} \]

### Step 5: Substitute \( \omega \) in RKE
- Substitute \( \omega \) in 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:
\[ \text{RKE} = \frac{1}{2} \left(\frac{1}{2} m r^2\right) \left(\frac{v^2}{r^2}\right) = \frac{1}{4} mv^2 \]

### Step 6: Calculate the Ratio of TKE to RKE
- Now, we can find the ratio of TKE to RKE:
\[ \text{Ratio} = \frac{\text{TKE}}{\text{RKE}} = \frac{\frac{1}{2} mv^2}{\frac{1}{4} mv^2} \]
The \( mv^2 \) terms cancel out:
\[ \text{Ratio} = \frac{\frac{1}{2}}{\frac{1}{4}} = \frac{1}{2} \times \frac{4}{1} = 2 \]

### Final Answer
Thus, the ratio of translational kinetic energy to rotational kinetic energy is:
\[ \text{Ratio} = 2:1 \]

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