A thin uniform circular disc of mass M and radius R is rotating in a horizontal plane about an axis perpendicular to the plane at an angular velocity `omega ` . Another disc of mass M / 3 but same radius is placed gently on the first disc coaxially . the angular velocity of the system now is

Options

A. `(4omega)/2`
B. `omega`
C. `(3omega)/4`
D. `(3omega)/8`

Explanation

To solve the problem step by step, we will use the principle of conservation of angular momentum.

### Step 1: Understand the Initial Conditions
We have two discs:
- Disc 1 (mass = M, radius = R) is rotating with an angular velocity \( \omega \).
- Disc 2 (mass = \( \frac{M}{3} \), radius = R) is placed gently on top of Disc 1.

### Step 2: Calculate the Moment of Inertia of Each Disc
The moment of inertia \( I \) for a disc rotating about an axis perpendicular to its plane is given by the formula:
\[ I = \frac{1}{2} m r^2 \]
For Disc 1:
\[ I_1 = \frac{1}{2} M R^2 \]
For Disc 2:
\[ I_2 = \frac{1}{2} \left(\frac{M}{3}\right) R^2 = \frac{1}{6} M R^2 \]

### Step 3: Calculate the Total Moment of Inertia After Placing Disc 2
When Disc 2 is placed on Disc 1, the total moment of inertia \( I_{total} \) of the system becomes:
\[ I_{total} = I_1 + I_2 = \frac{1}{2} M R^2 + \frac{1}{6} M R^2 \]
To combine these, we need a common denominator:
\[ I_{total} = \frac{3}{6} M R^2 + \frac{1}{6} M R^2 = \frac{4}{6} M R^2 = \frac{2}{3} M R^2 \]

### Step 4: Apply Conservation of Angular Momentum
Since no external torque acts on the system, the angular momentum before placing Disc 2 must equal the angular momentum after:
\[ L_{initial} = L_{final} \]
The initial angular momentum \( L_{initial} \) is:
\[ L_{initial} = I_1 \omega = \left(\frac{1}{2} M R^2\right) \omega \]
The final angular momentum \( L_{final} \) is:
\[ L_{final} = I_{total} \omega' = \left(\frac{2}{3} M R^2\right) \omega' \]

Setting these equal gives:
\[ \left(\frac{1}{2} M R^2\right) \omega = \left(\frac{2}{3} M R^2\right) \omega' \]

### Step 5: Solve for the New Angular Velocity \( \omega' \)
We can cancel \( M R^2 \) from both sides (assuming \( M \neq 0 \) and \( R \neq 0 \)):
\[ \frac{1}{2} \omega = \frac{2}{3} \omega' \]
Now, solving for \( \omega' \):
\[ \omega' = \frac{3}{4} \omega \]

### Final Answer
The angular velocity of the system after placing the second disc is:
\[ \omega' = \frac{3}{4} \omega \]