A solid sphere of mass 500 gm and radius 10 cm rolls without slipping with the velocity 20cm/s. The total kinetic energy of the sphere will be

Options

A. 0.028 J
B. 280 J
C. 140 J
D. 0.014 J

Answer

To find the total kinetic energy of a solid sphere rolling without slipping, we need to consider both its translational and rotational kinetic energies. Let's break down the solution step by step.

### Step 1: Identify the given values
- Mass of the sphere, \( m = 500 \, \text{g} = 0.5 \, \text{kg} \) (conversion from grams to kilograms)
- Radius of the sphere, \( r = 10 \, \text{cm} = 0.1 \, \text{m} \) (conversion from centimeters to meters)
- Velocity of the sphere, \( v = 20 \, \text{cm/s} = 0.2 \, \text{m/s} \) (conversion from centimeters per second to meters per second)

### Step 2: Calculate the translational kinetic energy
The formula for translational kinetic energy (\( KE_{\text{trans}} \)) is given by:
\[ KE_{\text{trans}} = \frac{1}{2} mv^2 \]
Substituting the values:
\[ KE_{\text{trans}} = \frac{1}{2} \times 0.5 \, \text{kg} \times (0.2 \, \text{m/s})^2 \]
\[ KE_{\text{trans}} = \frac{1}{2} \times 0.5 \times 0.04 = 0.01 \, \text{J} \]

### Step 3: Calculate the rotational kinetic energy
For a solid sphere, the moment of inertia (\( I \)) about its center of mass is given by:
\[ I = \frac{2}{5} m r^2 \]
Substituting the mass and radius:
\[ I = \frac{2}{5} \times 0.5 \, \text{kg} \times (0.1 \, \text{m})^2 \]
\[ I = \frac{2}{5} \times 0.5 \times 0.01 = 0.0001 \, \text{kg m}^2 \]

The angular velocity (\( \omega \)) can be calculated using the relationship:
\[ \omega = \frac{v}{r} = \frac{0.2 \, \text{m/s}}{0.1 \, \text{m}} = 2 \, \text{rad/s} \]

Now, we can calculate the rotational kinetic energy (\( KE_{\text{rot}} \)):
\[ KE_{\text{rot}} = \frac{1}{2} I \omega^2 \]
Substituting the values:
\[ KE_{\text{rot}} = \frac{1}{2} \times 0.0001 \, \text{kg m}^2 \times (2 \, \text{rad/s})^2 \]
\[ KE_{\text{rot}} = \frac{1}{2} \times 0.0001 \times 4 = 0.0002 \, \text{J} \]

### Step 4: Calculate the total kinetic energy
The total kinetic energy (\( KE_{\text{total}} \)) is the sum of translational and rotational kinetic energies:
\[ KE_{\text{total}} = KE_{\text{trans}} + KE_{\text{rot}} \]
Substituting the values:
\[ KE_{\text{total}} = 0.01 \, \text{J} + 0.0002 \, \text{J} = 0.0102 \, \text{J} \]

### Final Answer
The total kinetic energy of the sphere is:
\[ \boxed{0.0102 \, \text{J}} \]