A loop of mass M and radius R is rolling on a smooth horizontal surface with speed 'v'. It's total kinetic energy is

Options

A. `1//2Mv^(2)`
B. `3//2 Mv^(2)`
C. `Mv^(2)`
D. `1//2MR^(2)omega^(2)`

Answer

To find the total kinetic energy of a loop of mass \( M \) and radius \( R \) rolling on a smooth horizontal surface with speed \( v \), we need to consider both the translational and rotational kinetic energies.

### Step-by-Step Solution:

1. **Identify the Components of Kinetic Energy**:
- The total kinetic energy (\( KE_{total} \)) of the loop consists of two parts: translational kinetic energy (\( KE_{trans} \)) and rotational kinetic energy (\( KE_{rot} \)).
\[ KE_{total} = KE_{trans} + KE_{rot} \]

2. **Calculate Translational Kinetic Energy**:
- The translational kinetic energy of the loop is given by the formula:
\[ KE_{trans} = \frac{1}{2} M v^2 \]

3. **Calculate Rotational Kinetic Energy**:
- The rotational kinetic energy is given by the formula:
\[ KE_{rot} = \frac{1}{2} I \omega^2 \]
- For a loop, the moment of inertia \( I \) about its center of mass is:
\[ I = M R^2 \]
- The angular velocity \( \omega \) can be related to the linear speed \( v \) by the equation:
\[ \omega = \frac{v}{R} \]
- Substituting \( \omega \) into the rotational kinetic energy formula:
\[ KE_{rot} = \frac{1}{2} (M R^2) \left(\frac{v}{R}\right)^2 = \frac{1}{2} M R^2 \cdot \frac{v^2}{R^2} = \frac{1}{2} M v^2 \]

4. **Combine Both Kinetic Energies**:
- Now, we can combine both kinetic energies to find the total kinetic energy:
\[ KE_{total} = KE_{trans} + KE_{rot} = \frac{1}{2} M v^2 + \frac{1}{2} M v^2 = M v^2 \]

### Final Answer:
The total kinetic energy of the loop is:
\[ KE_{total} = M v^2 \]