A solid sphere of mass 1 kg rolls on a table with linear speed 1 m/s. Its total kinetic energy is

Options

A. 1 J
B. 0.5 J
C. 0.7 J
D. 1.4 J

Answer

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

### Step 1: Identify the mass and linear speed of the sphere
Given:
- Mass (m) = 1 kg
- Linear speed (v) = 1 m/s

### Step 2: Calculate the translational kinetic energy
The formula for translational kinetic energy (KE_trans) is given by:
\[ KE_{\text{trans}} = \frac{1}{2} m v^2 \]

Substituting the values:
\[ KE_{\text{trans}} = \frac{1}{2} \times 1 \, \text{kg} \times (1 \, \text{m/s})^2 \]
\[ KE_{\text{trans}} = \frac{1}{2} \times 1 \times 1 = 0.5 \, \text{J} \]

### Step 3: Calculate the rotational kinetic energy
To find the rotational kinetic energy (KE_rot), we need the moment of inertia (I) and angular velocity (ω).

1. **Find ω**:
The relationship between linear speed (v) and angular velocity (ω) is:
\[ v = \omega r \]
Therefore,
\[ \omega = \frac{v}{r} \]

2. **Find the moment of inertia (I)**:
For a solid sphere, the moment of inertia about its center of mass is:
\[ I = \frac{2}{5} m r^2 \]

3. **Substituting ω into the rotational kinetic energy formula**:
The formula for rotational kinetic energy is:
\[ KE_{\text{rot}} = \frac{1}{2} I \omega^2 \]
Substituting I and ω:
\[ KE_{\text{rot}} = \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v}{r}\right)^2 \]
Simplifying this:
\[ KE_{\text{rot}} = \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v^2}{r^2}\right) \]
\[ KE_{\text{rot}} = \frac{1}{2} \cdot \frac{2}{5} m v^2 \]
\[ KE_{\text{rot}} = \frac{1}{5} m v^2 \]

Substituting the values:
\[ KE_{\text{rot}} = \frac{1}{5} \times 1 \, \text{kg} \times (1 \, \text{m/s})^2 \]
\[ KE_{\text{rot}} = \frac{1}{5} \times 1 \times 1 = 0.2 \, \text{J} \]

### Step 4: Calculate the total kinetic energy
Now, we can find the total kinetic energy (KE_total) by adding the translational and rotational kinetic energies:
\[ KE_{\text{total}} = KE_{\text{trans}} + KE_{\text{rot}} \]
\[ KE_{\text{total}} = 0.5 \, \text{J} + 0.2 \, \text{J} \]
\[ KE_{\text{total}} = 0.7 \, \text{J} \]

### Final Answer
The total kinetic energy of the solid sphere is **0.7 Joules**.

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