A unifrom sphere of mass 500 g rolls without slipping on a plants horizontal surface with its center moving at a speed of `5.00cm//s` .It kinetic energy is :

Options

A. `6.25xx 10 ^(-4)J`
B. `8.75xx 10^(-4)J`
C. `8.75 xx 10^(-3)J`
D. `1.13xx 10^(-3)J`

Answer

To find the total kinetic energy of a uniform sphere rolling without slipping, we need to consider both its translational and rotational kinetic energy.

### Step-by-Step Solution:

1. **Identify the mass and speed**:
- Mass of the sphere, \( m = 500 \, \text{g} = 0.5 \, \text{kg} \) (conversion from grams to kilograms).
- Speed of the center of the sphere, \( v = 5 \, \text{cm/s} = 0.05 \, \text{m/s} \) (conversion from centimeters to meters).

2. **Calculate the translational kinetic energy (TKE)**:
\[ \text{TKE} = \frac{1}{2} mv^2 \]
Substituting the values:
\[ \text{TKE} = \frac{1}{2} \times 0.5 \, \text{kg} \times (0.05 \, \text{m/s})^2 \]
\[ = \frac{1}{2} \times 0.5 \times 0.0025 \, \text{m}^2/\text{s}^2 \]
\[ = \frac{1}{2} \times 0.5 \times 2.5 \times 10^{-3} \]
\[ = 0.000625 \, \text{J} \]

3. **Calculate the moment of inertia (I) of the sphere**:
The moment of inertia for a solid sphere is given by:
\[ I = \frac{2}{5} m r^2 \]
However, we do not need the radius \( r \) directly since it will cancel out later.

4. **Relate angular velocity (ω) to linear velocity (v)**:
For rolling without slipping, we have:
\[ v = \omega r \implies \omega = \frac{v}{r} \]

5. **Calculate the rotational kinetic energy (RKE)**:
\[ \text{RKE} = \frac{1}{2} I \omega^2 \]
Substituting \( I \) and \( \omega \):
\[ \text{RKE} = \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v}{r}\right)^2 \]
\[ = \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v^2}{r^2}\right) \]
\[ = \frac{1}{2} \cdot \frac{2}{5} m v^2 \]
\[ = \frac{1}{5} mv^2 \]

6. **Substituting values into RKE**:
\[ \text{RKE} = \frac{1}{5} \times 0.5 \, \text{kg} \times (0.05 \, \text{m/s})^2 \]
\[ = \frac{1}{5} \times 0.5 \times 0.0025 \]
\[ = 0.00025 \, \text{J} \]

7. **Calculate the total kinetic energy (TKE + RKE)**:
\[ \text{Total KE} = \text{TKE} + \text{RKE} \]
\[ = 0.000625 \, \text{J} + 0.00025 \, \text{J} \]
\[ = 0.000875 \, \text{J} \]

### Final Answer:
The total kinetic energy of the sphere is \( 0.000875 \, \text{J} \) or \( 8.75 \times 10^{-4} \, \text{J} \).