Two discs are rotating about their axes, normal to the discs and passing through the centres of the discs. Disc `D_(1)` has 2 kg mass and `0.2` m radius and initial angular velocity of `50rads^(-1)` . Disc `D_(2)` has 4kg mass `0.1m` radius and initial angular velocity of `200rads^(-1)`. The two discs are brought in contact face to face, with their axes of rotation coincident. The final angular velocity (in `rad.s^(-1)`) of the system is

Options

A. 60
B. 100
C. 120
D. 40

Explanation

To solve the problem of finding the final angular velocity of the two discs when they come into contact, we will use the principle of conservation of angular momentum. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the given data
- For Disc \(D_1\):
- Mass \(m_1 = 2 \, \text{kg}\)
- Radius \(r_1 = 0.2 \, \text{m}\)
- Initial angular velocity \(\omega_1 = 50 \, \text{rad/s}\)

- For Disc \(D_2\):
- Mass \(m_2 = 4 \, \text{kg}\)
- Radius \(r_2 = 0.1 \, \text{m}\)
- Initial angular velocity \(\omega_2 = 200 \, \text{rad/s}\)

### Step 2: Calculate the moments of inertia of the discs
The moment of inertia \(I\) for a disc about its central axis is given by the formula:
\[ I = \frac{1}{2} m r^2 \]

- For Disc \(D_1\):
\[ I_1 = \frac{1}{2} m_1 r_1^2 = \frac{1}{2} \times 2 \times (0.2)^2 = \frac{1}{2} \times 2 \times 0.04 = 0.04 \, \text{kg m}^2 \]

- For Disc \(D_2\):
\[ I_2 = \frac{1}{2} m_2 r_2^2 = \frac{1}{2} \times 4 \times (0.1)^2 = \frac{1}{2} \times 4 \times 0.01 = 0.02 \, \text{kg m}^2 \]

### Step 3: Apply the conservation of angular momentum
According to the conservation of angular momentum, the total initial angular momentum of the system must equal the total final angular momentum:
\[ I_1 \omega_1 + I_2 \omega_2 = (I_1 + I_2) \omega_f \]

Substituting the values we calculated:
\[ 0.04 \times 50 + 0.02 \times 200 = (0.04 + 0.02) \omega_f \]

### Step 4: Calculate the left side of the equation
Calculating the left side:
\[ 0.04 \times 50 = 2 \, \text{kg m}^2/s \]
\[ 0.02 \times 200 = 4 \, \text{kg m}^2/s \]
So,
\[ 2 + 4 = 6 \, \text{kg m}^2/s \]

### Step 5: Substitute and solve for \(\omega_f\)
Now we can substitute back into the equation:
\[ 6 = (0.04 + 0.02) \omega_f \]
\[ 6 = 0.06 \omega_f \]
Now, solve for \(\omega_f\):
\[ \omega_f = \frac{6}{0.06} = 100 \, \text{rad/s} \]

### Final Answer
The final angular velocity of the system is:
\[ \omega_f = 100 \, \text{rad/s} \]

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