A circular disc of mass 2 kg and radius 10 cm rolls without slipping with a speed 2 m/s. The total kinetic energy of disc is

Options

A. 10 J
B. 6 J
C. 2 J
D. 4 J

Answer

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

### Step 1: Identify the given values
- Mass of the disc (m) = 2 kg
- Radius of the disc (R) = 10 cm = 0.1 m (convert cm to m)
- Speed of the disc (V) = 2 m/s

### Step 2: Calculate the translational kinetic energy (TKE)
The formula for translational kinetic energy is:
\[ \text{TKE} = \frac{1}{2} m V^2 \]
Substituting the values:
\[ \text{TKE} = \frac{1}{2} \times 2 \, \text{kg} \times (2 \, \text{m/s})^2 \]
\[ \text{TKE} = \frac{1}{2} \times 2 \times 4 = 4 \, \text{J} \]

### Step 3: Calculate the rotational kinetic energy (RKE)
For a disc rolling without slipping, the relationship between linear speed (V) and angular speed (ω) is given by:
\[ V = R \omega \quad \Rightarrow \quad \omega = \frac{V}{R} \]
Now, the moment of inertia (I) of a disc about its center is:
\[ I = \frac{1}{2} m R^2 \]
Substituting the values:
\[ I = \frac{1}{2} \times 2 \, \text{kg} \times (0.1 \, \text{m})^2 = \frac{1}{2} \times 2 \times 0.01 = 0.01 \, \text{kg m}^2 \]
Now, substituting ω in the formula for rotational kinetic energy:
\[ \text{RKE} = \frac{1}{2} I \omega^2 \]
Substituting ω:
\[ \text{RKE} = \frac{1}{2} \times 0.01 \, \text{kg m}^2 \times \left(\frac{2 \, \text{m/s}}{0.1 \, \text{m}}\right)^2 \]
Calculating ω:
\[ \omega = \frac{2}{0.1} = 20 \, \text{rad/s} \]
Now substituting ω back:
\[ \text{RKE} = \frac{1}{2} \times 0.01 \times (20)^2 = \frac{1}{2} \times 0.01 \times 400 = 2 \, \text{J} \]

### Step 4: Calculate the total kinetic energy (TKE + RKE)
\[ \text{Total Kinetic Energy} = \text{TKE} + \text{RKE} = 4 \, \text{J} + 2 \, \text{J} = 6 \, \text{J} \]

### Final Answer
The total kinetic energy of the disc is **6 Joules**.

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