If a sphere is rolling, the ratio of the translation energy to total kinetic energy is given by

Options

A. `7 : 10`
B. `2 : 5`
C. `10 : 7`
D. `5 : 7`

Answer

To solve the problem of finding the ratio of translational kinetic energy to total kinetic energy for a rolling sphere, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Translational Kinetic Energy (TKE)**:
The translational kinetic energy (TKE) of a sphere is given by the formula:
\[ TKE = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the sphere and \( v \) is its linear velocity.

2. **Identify the Rotational Kinetic Energy (RKE)**:
The rotational kinetic energy (RKE) of the sphere is given by the formula:
\[ RKE = \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

3. **Moment of Inertia for a Sphere**:
The moment of inertia \( I \) for a solid sphere is:
\[ I = \frac{2}{5} m r^2 \]
where \( r \) is the radius of the sphere.

4. **Relate Angular Velocity to Linear Velocity**:
The angular velocity \( \omega \) is related to the linear velocity \( v \) by:
\[ \omega = \frac{v}{r} \]
Therefore, \( \omega^2 = \frac{v^2}{r^2} \).

5. **Substituting \( I \) and \( \omega \) into RKE**:
Substitute the expressions for \( I \) and \( \omega^2 \) into the RKE formula:
\[ RKE = \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v^2}{r^2}\right) \]
Simplifying this gives:
\[ RKE = \frac{1}{2} \cdot \frac{2}{5} m v^2 = \frac{1}{5} mv^2 \]

6. **Calculate Total Kinetic Energy (TKE + RKE)**:
The total kinetic energy (TKE + RKE) is:
\[ \text{Total KE} = TKE + RKE = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 \]
To combine these, find a common denominator:
\[ \text{Total KE} = \frac{5}{10} mv^2 + \frac{2}{10} mv^2 = \frac{7}{10} mv^2 \]

7. **Calculate the Ratio of TKE to Total KE**:
Now, we can find the ratio of the translational kinetic energy to the total kinetic energy:
\[ \text{Ratio} = \frac{TKE}{\text{Total KE}} = \frac{\frac{1}{2} mv^2}{\frac{7}{10} mv^2} \]
The \( mv^2 \) terms cancel out:
\[ \text{Ratio} = \frac{\frac{1}{2}}{\frac{7}{10}} = \frac{1}{2} \cdot \frac{10}{7} = \frac{5}{7} \]

### Final Answer:
The ratio of the translational kinetic energy to the total kinetic energy for a rolling sphere is:
\[ \frac{5}{7} \]