A sphere of mass m rolls on a plane surface. Find its kinetic energy at an instant when its centre moves with speed v.

Explanation

To find the kinetic energy of a sphere of mass \( m \) that rolls on a plane surface with its center moving at a speed \( v \), we need to consider both its translational and rotational kinetic energy.

### Step-by-Step Solution:

1. **Identify the Types of Kinetic Energy**:
- The total kinetic energy (KE) of the rolling sphere consists of two components:
- Translational kinetic energy (TKE) due to the motion of the center of mass.
- Rotational kinetic energy (RKE) due to the rotation about its center of mass.

2. **Write the Expressions for Kinetic Energy**:
- The translational kinetic energy is given by:
\[ \text{TKE} = \frac{1}{2} m v^2 \]
- The rotational kinetic energy is given by:
\[ \text{RKE} = \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia of the sphere about its center of mass and \( \omega \) is the angular velocity.

3. **Relate Angular Velocity to Linear Velocity**:
- For a sphere rolling without slipping, the relationship between angular velocity \( \omega \) and linear velocity \( v \) is:
\[ \omega = \frac{v}{r} \]
where \( r \) is the radius of the sphere.

4. **Calculate the Moment of Inertia**:
- The moment of inertia \( I \) for a solid sphere about its center of mass is:
\[ I = \frac{2}{5} m r^2 \]

5. **Substitute \( \omega \) in the RKE Expression**:
- Substitute \( \omega \) into the rotational kinetic energy expression:
\[ \text{RKE} = \frac{1}{2} I \left(\frac{v}{r}\right)^2 = \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v^2}{r^2}\right) \]
- Simplifying this gives:
\[ \text{RKE} = \frac{1}{2} \cdot \frac{2}{5} m \cdot \frac{v^2}{r^2} \cdot r^2 = \frac{1}{5} m v^2 \]

6. **Combine the Kinetic Energies**:
- Now, add the translational and rotational kinetic energies:
\[ \text{Total KE} = \text{TKE} + \text{RKE} = \frac{1}{2} m v^2 + \frac{1}{5} m v^2 \]

7. **Find a Common Denominator and Simplify**:
- To combine the two fractions, find a common denominator:
\[ \text{Total KE} = \frac{5}{10} m v^2 + \frac{2}{10} m v^2 = \frac{7}{10} m v^2 \]

8. **Final Result**:
- Therefore, the total kinetic energy of the sphere is:
\[ KE = \frac{7}{10} m v^2 \]