The ratio of rotational and translatory kinetic energies of a solid sphere is

Options

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

Answer

To find the ratio of rotational kinetic energy to translational kinetic energy of a solid sphere, we will follow these steps:

### Step 1: Write the formula for translational kinetic energy.
The translational kinetic energy (K_trans) of an object is given by the formula:
\[ K_{\text{trans}} = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the sphere and \( v \) is its linear velocity.

### Step 2: Write the formula for rotational kinetic energy.
The rotational kinetic energy (K_rot) of an object is given by the formula:
\[ K_{\text{rot}} = \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

### Step 3: Determine the moment of inertia for a solid sphere.
For a solid sphere rotating about its center of mass, the moment of inertia \( I \) is given by:
\[ I = \frac{2}{5} m r^2 \]
where \( r \) is the radius of the sphere.

### Step 4: Relate angular velocity to linear velocity.
The relationship between angular velocity \( \omega \) and linear velocity \( v \) is given by:
\[ \omega = \frac{v}{r} \]

### Step 5: Substitute the moment of inertia and angular velocity into the rotational kinetic energy formula.
Substituting \( I \) and \( \omega \) into the rotational kinetic energy formula, we get:
\[ K_{\text{rot}} = \frac{1}{2} \left( \frac{2}{5} m r^2 \right) \left( \frac{v}{r} \right)^2 \]
This simplifies to:
\[ K_{\text{rot}} = \frac{1}{2} \left( \frac{2}{5} m r^2 \right) \left( \frac{v^2}{r^2} \right) = \frac{1}{2} \cdot \frac{2}{5} m v^2 = \frac{1}{5} mv^2 \]

### Step 6: Find the ratio of rotational to translational kinetic energy.
Now we can find the ratio of rotational kinetic energy to translational kinetic energy:
\[ \text{Ratio} = \frac{K_{\text{rot}}}{K_{\text{trans}}} = \frac{\frac{1}{5} mv^2}{\frac{1}{2} mv^2} \]
The \( mv^2 \) terms cancel out, leading to:
\[ \text{Ratio} = \frac{\frac{1}{5}}{\frac{1}{2}} = \frac{2}{5} \]

### Final Answer:
The ratio of rotational to translational kinetic energy of a solid sphere is:
\[ \frac{2}{5} \]

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