A solid sphere of radius r and mass m rotates about an axis passing through its centre with angular velocity omega. Its K.E. Is

A. `mr^2ω^2`
B. `2/3` `mr^2ω^2`
C. `1/2` `mr^2ω^2`
D. `1/5` `mr^2ω^2`

Options

A. `mr^2ω^2`
B. `2/3` `mr^2ω^2`
C. `1/2` `mr^2ω^2`
D. `1/5` `mr^2ω^2`

Answer

To find the kinetic energy of a solid sphere rotating about an axis through its center, we can follow these steps:

### Step 1: Identify the Moment of Inertia
The moment of inertia (I) of a solid sphere about an axis through its center is given by the formula:
\[ I = \frac{2}{5} m r^2 \]
where \( m \) is the mass of the sphere and \( r \) is its radius.

### Step 2: Use the Kinetic Energy Formula
The kinetic energy (K.E.) of a rotating object can be calculated using the formula:
\[ K.E. = \frac{1}{2} I \omega^2 \]
where \( \omega \) is the angular velocity.

### Step 3: Substitute the Moment of Inertia into the Kinetic Energy Formula
Now, substitute the expression for the moment of inertia into the kinetic energy formula:
\[ K.E. = \frac{1}{2} \left( \frac{2}{5} m r^2 \right) \omega^2 \]

### Step 4: Simplify the Expression
Now, simplify the expression:
\[ K.E. = \frac{1}{2} \cdot \frac{2}{5} m r^2 \cdot \omega^2 \]
\[ K.E. = \frac{1}{5} m r^2 \omega^2 \]

### Step 5: Conclusion
Thus, the kinetic energy of the solid sphere is:
\[ K.E. = \frac{1}{5} m r^2 \omega^2 \]

From the given options, the correct answer is:
**D. \( \frac{1}{5} m r^2 \omega^2 \)**

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