A uniform sphere of mass `500 g` rolls without slipping on a plane surface so that its centre moves at a speed of `0.02 m//s`.
The total kinetic energy of rolling sphere would be (in `J`)

Options

A. `1.4 xx 10^(-4)J`
B. `0.75 xx 10^(-3)J`
C. `5.75 xx 10^(-3) J`
D. `4.9 xx 10^(-5)J`

Explanation

To find the total kinetic energy of a rolling sphere, we need to consider both its translational and rotational kinetic energy. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the mass and speed of the sphere
- Given mass \( m = 500 \, \text{g} = 0.5 \, \text{kg} \)
- Given speed of the center of mass \( v_{cm} = 0.02 \, \text{m/s} \)

### Step 2: Write the formula for total kinetic energy
The total kinetic energy \( KE \) of a rolling sphere is the sum of its translational kinetic energy and rotational kinetic energy:
\[ KE = KE_{translational} + KE_{rotational} \]
Where:
- \( KE_{translational} = \frac{1}{2} m v_{cm}^2 \)
- \( KE_{rotational} = \frac{1}{2} I \omega^2 \)

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

### Step 4: Relate linear speed to angular speed
For rolling without slipping, the relationship between linear speed \( v_{cm} \) and angular speed \( \omega \) is:
\[ v_{cm} = r \omega \implies \omega = \frac{v_{cm}}{r} \]

### Step 5: Substitute \( \omega \) into the rotational kinetic energy formula
Substituting \( \omega \) into the rotational kinetic energy:
\[ KE_{rotational} = \frac{1}{2} I \left(\frac{v_{cm}}{r}\right)^2 \]
Substituting \( I \):
\[ KE_{rotational} = \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v_{cm}}{r}\right)^2 = \frac{1}{5} m v_{cm}^2 \]

### Step 6: Combine translational and rotational kinetic energy
Now, substituting both kinetic energy components into the total kinetic energy equation:
\[ KE = \frac{1}{2} m v_{cm}^2 + \frac{1}{5} m v_{cm}^2 \]
Factoring out \( m v_{cm}^2 \):
\[ KE = m v_{cm}^2 \left(\frac{1}{2} + \frac{1}{5}\right) \]
Finding a common denominator (10):
\[ KE = m v_{cm}^2 \left(\frac{5}{10} + \frac{2}{10}\right) = m v_{cm}^2 \left(\frac{7}{10}\right) \]

### Step 7: Substitute the values into the equation
Now substituting the values of \( m \) and \( v_{cm} \):
\[ KE = 0.5 \times (0.02)^2 \times \frac{7}{10} \]
Calculating \( (0.02)^2 = 0.0004 \):
\[ KE = 0.5 \times 0.0004 \times \frac{7}{10} = 0.5 \times 0.0004 \times 0.7 = 0.00014 \, \text{J} \]
Converting to scientific notation:
\[ KE = 1.4 \times 10^{-4} \, \text{J} \]

### Final Answer
The total kinetic energy of the rolling sphere is:
\[ \boxed{1.4 \times 10^{-4} \, \text{J}} \]