A thin uniform circular disc of mass M and radius R is rotating in a horizontal plane about an axis passing through its centre and perpendicular to its plane with an angular velocity `omega` . Another disc of same dimensions but of mass `(1)/(4)` M is placed gently on the first disc co-axially. The angular velocity of the system is

Options

A. `(2)/(3) omega`
B. `(4)/(5) omega`
C. `(3)/(4) omega`
D. `(1)/(3) omega`

Explanation

To solve the problem, we need to find the new angular velocity of the system after placing the second disc on the first one. We will use the principle of conservation of angular momentum.

### Step-by-Step Solution:

1. **Identify the Moment of Inertia of the First Disc:**
The moment of inertia \( I_1 \) of a thin uniform circular disc about an axis through its center and perpendicular to its plane is given by:
\[ I_1 = \frac{1}{2} M R^2 \]
where \( M \) is the mass of the first disc and \( R \) is its radius.

2. **Identify the Moment of Inertia of the Second Disc:**
The second disc has a mass of \( \frac{1}{4} M \). Its moment of inertia \( I_2 \) is:
\[ I_2 = \frac{1}{2} \left(\frac{1}{4} M\right) R^2 = \frac{1}{8} M R^2 \]

3. **Calculate the Total Moment of Inertia of the System:**
The total moment of inertia \( I' \) of the system when both discs are present is:
\[ I' = I_1 + I_2 = \frac{1}{2} M R^2 + \frac{1}{8} M R^2 \]
To add these, we need a common denominator:
\[ I' = \frac{4}{8} M R^2 + \frac{1}{8} M R^2 = \frac{5}{8} M R^2 \]

4. **Use Conservation of Angular Momentum:**
The initial angular momentum \( L_i \) of the system is:
\[ L_i = I_1 \omega = \left(\frac{1}{2} M R^2\right) \omega \]
The final angular momentum \( L_f \) after the second disc is placed on the first is:
\[ L_f = I' \omega' = \left(\frac{5}{8} M R^2\right) \omega' \]
According to the conservation of angular momentum:
\[ L_i = L_f \implies \left(\frac{1}{2} M R^2\right) \omega = \left(\frac{5}{8} M R^2\right) \omega' \]

5. **Solve for the New Angular Velocity \( \omega' \):**
Cancel \( M R^2 \) from both sides:
\[ \frac{1}{2} \omega = \frac{5}{8} \omega' \]
Rearranging gives:
\[ \omega' = \frac{1/2}{5/8} \omega = \frac{1}{2} \cdot \frac{8}{5} \omega = \frac{4}{5} \omega \]

### Final Answer:
The angular velocity of the system after placing the second disc is:
\[ \omega' = \frac{4}{5} \omega \]