A disc of radius 2 m and mass 100kg rolls on a horizontal floor, its centre of mass has speed of `20 cm//s`. How much work is needed to stop it ?

Options

A. 1J
B. 3J
C. 30 KJ
D. 2J

Explanation

To find the work needed to stop a rolling disc, we can use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy.

### Step-by-Step Solution:

1. **Identify the given values:**
- Radius of the disc (r) = 2 m
- Mass of the disc (m) = 100 kg
- Speed of the center of mass (v) = 20 cm/s = 20 × 10^(-2) m/s = 0.2 m/s

2. **Calculate the moment of inertia (I) of the disc:**
The moment of inertia for a solid disc is given by:
\[ I = \frac{1}{2} m r^2 \]
Substituting the values:
\[ I = \frac{1}{2} \times 100 \, \text{kg} \times (2 \, \text{m})^2 = \frac{1}{2} \times 100 \times 4 = 200 \, \text{kg m}^2 \]

3. **Calculate the angular velocity (ω):**
The angular velocity can be calculated using the relation:
\[ \omega = \frac{v}{r} \]
Substituting the values:
\[ \omega = \frac{0.2 \, \text{m/s}}{2 \, \text{m}} = 0.1 \, \text{rad/s} \]

4. **Calculate the initial kinetic energy (KE_initial):**
The total kinetic energy of the rolling disc is the sum of translational and rotational kinetic energy:
\[ KE_{\text{initial}} = \frac{1}{2} I \omega^2 + \frac{1}{2} mv^2 \]
Substituting the values:
\[ KE_{\text{initial}} = \frac{1}{2} \times 200 \, \text{kg m}^2 \times (0.1 \, \text{rad/s})^2 + \frac{1}{2} \times 100 \, \text{kg} \times (0.2 \, \text{m/s})^2 \]
\[ KE_{\text{initial}} = \frac{1}{2} \times 200 \times 0.01 + \frac{1}{2} \times 100 \times 0.04 \]
\[ KE_{\text{initial}} = 1 + 2 = 3 \, \text{J} \]

5. **Calculate the work done (W):**
Since the final kinetic energy (KE_final) when the disc is stopped is 0, the work done to stop the disc is equal to the initial kinetic energy:
\[ W = KE_{\text{final}} - KE_{\text{initial}} = 0 - 3 \, \text{J} = -3 \, \text{J} \]
The work done to stop the disc is 3 J (the negative sign indicates that work is done against the motion).

### Final Answer:
The work needed to stop the disc is **3 joules**.