A hollow sphere of diameter `0.2` m and mass 2 kg is rolling on an inclined plane with velocity `v = 0.5` m / s. The kinetic energy of the sphere is

Options

A. `0.1` J
B. `0.3` J
C. `0.5` J
D. `0.42` J

Explanation

To calculate the kinetic energy of a hollow sphere rolling down an inclined plane, we need to consider both its translational and rotational kinetic energy. Here’s the step-by-step solution:

### Step 1: Identify the given values
- Diameter of the sphere, \( d = 0.2 \, \text{m} \)
- Radius of the sphere, \( r = \frac{d}{2} = \frac{0.2}{2} = 0.1 \, \text{m} \)
- Mass of the sphere, \( m = 2 \, \text{kg} \)
- Velocity of the sphere, \( v = 0.5 \, \text{m/s} \)

### Step 2: Calculate the translational kinetic energy
The translational kinetic energy (\( KE_t \)) is given by the formula:
\[ KE_t = \frac{1}{2} m v^2 \]
Substituting the known values:
\[ KE_t = \frac{1}{2} \times 2 \, \text{kg} \times (0.5 \, \text{m/s})^2 \]
\[ KE_t = \frac{1}{2} \times 2 \times 0.25 = \frac{1}{2} \times 0.5 = 0.25 \, \text{J} \]

### Step 3: Calculate the rotational kinetic energy
The moment of inertia (\( I \)) of a hollow sphere about its center is given by:
\[ I = \frac{2}{3} m r^2 \]
Substituting the known values:
\[ I = \frac{2}{3} \times 2 \, \text{kg} \times (0.1 \, \text{m})^2 = \frac{2}{3} \times 2 \times 0.01 = \frac{2}{3} \times 0.02 = \frac{0.04}{3} \, \text{kg m}^2 \]

The rotational kinetic energy (\( KE_r \)) is given by:
\[ KE_r = \frac{1}{2} I \omega^2 \]
Where \( \omega \) (angular velocity) can be calculated using the relationship \( v = r \omega \):
\[ \omega = \frac{v}{r} = \frac{0.5 \, \text{m/s}}{0.1 \, \text{m}} = 5 \, \text{rad/s} \]
Now substituting \( I \) and \( \omega \) into the rotational kinetic energy formula:
\[ KE_r = \frac{1}{2} \left(\frac{0.04}{3}\right) (5)^2 = \frac{1}{2} \left(\frac{0.04}{3}\right) \times 25 = \frac{0.04 \times 25}{6} = \frac{1}{6} \, \text{J} \]

### Step 4: Calculate the total kinetic energy
The total kinetic energy (\( KE \)) is the sum of translational and rotational kinetic energy:
\[ KE = KE_t + KE_r = 0.25 \, \text{J} + \frac{1}{6} \, \text{J} \]
To add these, convert \( 0.25 \) to a fraction:
\[ 0.25 = \frac{1}{4} \]
Finding a common denominator (12):
\[ KE = \frac{3}{12} + \frac{2}{12} = \frac{5}{12} \, \text{J} \]

### Step 5: Convert to decimal form
\[ KE = \frac{5}{12} \approx 0.4167 \, \text{J} \]

### Final Answer
The total kinetic energy of the hollow sphere is approximately \( 0.42 \, \text{J} \).

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