When a sphere rolls without slipping the ratio of its kinetic energy of translation to its total kinetic energy is.

Options

A. `1 : 7`
B. `1 : 2`
C. `1 : 1`
D. `5 : 7`

Explanation

To find the ratio of the kinetic energy of translation to the total kinetic energy of a sphere rolling without slipping, we can follow these steps:

### Step 1: Understand the Kinetic Energies Involved
When a sphere rolls without slipping, it has two types of kinetic energy:
1. Kinetic energy of translation (due to its center of mass moving).
2. Kinetic energy of rotation (due to it spinning about its center of mass).

### Step 2: Write the Expression for Kinetic Energy of Translation
The kinetic energy of translation (K.E. translational) of the sphere can be expressed as:
\[ K.E._{\text{translational}} = \frac{1}{2} m v^2 \]
where \( m \) is the mass of the sphere and \( v \) is its linear velocity.

### Step 3: Write the Expression for Kinetic Energy of Rotation
The kinetic energy of rotation (K.E. rotational) can be expressed as:
\[ K.E._{\text{rotational}} = \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

### Step 4: Determine the Moment of Inertia for a Sphere
For a solid sphere, the moment of inertia \( I \) about its center is given by:
\[ I = \frac{2}{5} m r^2 \]
where \( r \) is the radius of the sphere.

### Step 5: Relate Angular Velocity to Linear Velocity
Since the sphere rolls without slipping, we have the relationship:
\[ v = r \omega \implies \omega = \frac{v}{r} \]

### Step 6: Substitute \( \omega \) into the Rotational Kinetic Energy
Substituting \( \omega \) into the rotational kinetic energy expression gives:
\[ K.E._{\text{rotational}} = \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v}{r}\right)^2 \]
This simplifies to:
\[ K.E._{\text{rotational}} = \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v^2}{r^2}\right) = \frac{1}{5} m v^2 \]

### Step 7: Calculate Total Kinetic Energy
The total kinetic energy \( K.E._{\text{total}} \) is the sum of the translational and rotational kinetic energies:
\[ K.E._{\text{total}} = K.E._{\text{translational}} + K.E._{\text{rotational}} = \frac{1}{2} m v^2 + \frac{1}{5} m v^2 \]

### Step 8: Combine the Energies
To combine these, we need a common denominator:
\[ K.E._{\text{total}} = \frac{5}{10} m v^2 + \frac{2}{10} m v^2 = \frac{7}{10} m v^2 \]

### Step 9: Find the Ratio of Kinetic Energies
Now, we can find the ratio of the kinetic energy of translation to the total kinetic energy:
\[ \text{Ratio} = \frac{K.E._{\text{translational}}}{K.E._{\text{total}}} = \frac{\frac{1}{2} m v^2}{\frac{7}{10} m v^2} \]
The \( m v^2 \) cancels out:
\[ \text{Ratio} = \frac{\frac{1}{2}}{\frac{7}{10}} = \frac{1}{2} \times \frac{10}{7} = \frac{5}{7} \]

### Conclusion
Thus, the ratio of the kinetic energy of translation to the total kinetic energy of a sphere rolling without slipping is:
\[ \frac{5}{7} \]