The ratio rotational and translational kinetic energies of a sphere is

Options

A. `2//9`
B. `2//7`
C. `2//5`
D. `7//2`

Explanation

To find the ratio of rotational and translational kinetic energies of a sphere, we can follow these steps:

### Step 1: Write the formulas for kinetic energies
- The rotational kinetic energy \( K_1 \) is given by the formula:
\[ K_1 = \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

- The translational kinetic energy \( K_2 \) is given by the formula:
\[ K_2 = \frac{1}{2} m v^2 \]
where \( m \) is the mass and \( v \) is the linear velocity.

### Step 2: Find the 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.

### Step 3: Substitute the moment of inertia into the rotational kinetic energy formula
- By substituting the expression for \( I \) into the equation for \( K_1 \):
\[ K_1 = \frac{1}{2} \left(\frac{2}{5} m r^2\right) \omega^2 = \frac{1}{5} m r^2 \omega^2 \]

### Step 4: Relate angular velocity to linear velocity
- We know that the angular velocity \( \omega \) is related to the linear velocity \( v \) by:
\[ \omega = \frac{v}{r} \]
Therefore, substituting \( \omega \) in terms of \( v \) into the equation for \( K_1 \):
\[ K_1 = \frac{1}{5} m r^2 \left(\frac{v}{r}\right)^2 = \frac{1}{5} m r^2 \frac{v^2}{r^2} = \frac{1}{5} m v^2 \]

### Step 5: Substitute into the translational kinetic energy formula
- Now, we can express \( K_2 \) in terms of \( v \):
\[ K_2 = \frac{1}{2} m v^2 \]

### Step 6: Find the ratio of rotational to translational kinetic energy
- Now we can find the ratio \( \frac{K_1}{K_2} \):
\[ \frac{K_1}{K_2} = \frac{\frac{1}{5} m v^2}{\frac{1}{2} m v^2} \]
- The \( m v^2 \) terms cancel out:
\[ \frac{K_1}{K_2} = \frac{1/5}{1/2} = \frac{2}{5} \]

### Final Result
The ratio of rotational to translational kinetic energy of a sphere is:
\[ \frac{K_1}{K_2} = \frac{2}{5} \]