A solid sphere of mass 1 kg and radius 10 cm rolls without slipping on a horizontal surface, with a velocity of 20 cm/s. The total kinetic energy of the sphere is

Options

A. 0.014 J
B. 0.028 J
C. 14 J
D. 28 J

Explanation

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

### Step-by-Step Solution:

1. **Identify the given values:**
- Mass of the sphere (m) = 1 kg
- Radius of the sphere (r) = 10 cm = 0.1 m (conversion from cm to m)
- Velocity of the sphere (v) = 20 cm/s = 0.2 m/s (conversion from cm/s to m/s)

2. **Formulas for kinetic energy:**
- The total kinetic energy (KE_total) of the sphere is the sum of translational kinetic energy (KE_trans) and rotational kinetic energy (KE_rot).
- Translational kinetic energy:
\[ KE_{\text{trans}} = \frac{1}{2} m v^2 \]
- Rotational kinetic energy:
\[ KE_{\text{rot}} = \frac{1}{2} I \omega^2 \]
- For a solid sphere, the moment of inertia (I) is given by:
\[ I = \frac{2}{5} m r^2 \]
- The angular velocity (\(\omega\)) can be related to linear velocity (v) using the equation:
\[ v = r \omega \quad \Rightarrow \quad \omega = \frac{v}{r} \]

3. **Calculate \(\omega\):**
\[ \omega = \frac{v}{r} = \frac{0.2 \, \text{m/s}}{0.1 \, \text{m}} = 2 \, \text{rad/s} \]

4. **Calculate the rotational kinetic energy:**
- Substitute \(I\) and \(\omega\) into the rotational kinetic energy formula:
\[ KE_{\text{rot}} = \frac{1}{2} \left(\frac{2}{5} m r^2\right) \omega^2 = \frac{1}{2} \left(\frac{2}{5} \times 1 \, \text{kg} \times (0.1 \, \text{m})^2\right) (2 \, \text{rad/s})^2 \]
- Calculate:
\[ KE_{\text{rot}} = \frac{1}{2} \left(\frac{2}{5} \times 1 \times 0.01\right) \times 4 = \frac{1}{2} \left(\frac{2}{5} \times 0.04\right) = \frac{1}{2} \times \frac{0.08}{5} = \frac{0.08}{10} = 0.008 \, \text{J} \]

5. **Calculate the translational kinetic energy:**
\[ KE_{\text{trans}} = \frac{1}{2} m v^2 = \frac{1}{2} \times 1 \, \text{kg} \times (0.2 \, \text{m/s})^2 = \frac{1}{2} \times 1 \times 0.04 = 0.02 \, \text{J} \]

6. **Calculate the total kinetic energy:**
\[ KE_{\text{total}} = KE_{\text{trans}} + KE_{\text{rot}} = 0.02 \, \text{J} + 0.008 \, \text{J} = 0.028 \, \text{J} \]

### Final Answer:
The total kinetic energy of the sphere is \(0.028 \, \text{J}\).

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