A uniform disc of radius a and mass m, is rotating freely with angular speed `omega` in a horizontal plane about a smooth fixed vertical axis through its centre. A particle, also of mass m, is suddenly attached to the rim of the disc of the disc and rotates with it. The new angular speed is

Options

A. `(omega)/(6)`
B. `(omega)/(3)`
C. `(omega)/(2)`
D. `(omega)/(5)`

Explanation

To solve the problem of finding the new angular speed of a uniform disc after a particle is attached to it, we will follow these steps:

### Step 1: Determine the initial moment of inertia of the disc
The moment of inertia \( I \) of a uniform disc about an axis through its center is given by the formula:
\[ I = \frac{1}{2} m a^2 \]
where \( m \) is the mass of the disc and \( a \) is its radius.

### Step 2: Calculate the initial angular momentum
The initial angular momentum \( L_i \) of the disc can be calculated using the formula:
\[ L_i = I \omega \]
Substituting the expression for \( I \):
\[ L_i = \left(\frac{1}{2} m a^2\right) \omega \]

### Step 3: Determine the moment of inertia after the particle is attached
When a particle of mass \( m \) is attached to the rim of the disc, the new moment of inertia \( I' \) of the system (disc + particle) becomes:
\[ I' = I + I_{\text{particle}} = \frac{1}{2} m a^2 + m a^2 \]
The moment of inertia of the particle about the same axis is \( m a^2 \). Therefore:
\[ I' = \frac{1}{2} m a^2 + m a^2 = \frac{3}{2} m a^2 \]

### Step 4: Apply the conservation of angular momentum
Since there is no external torque acting on the system, the angular momentum is conserved:
\[ L_i = L_f \]
where \( L_f \) is the final angular momentum after the particle is attached, given by:
\[ L_f = I' \omega' \]
Setting the initial and final angular momentum equal gives:
\[ \frac{1}{2} m a^2 \omega = \frac{3}{2} m a^2 \omega' \]

### Step 5: Solve for the new angular speed \( \omega' \)
We can cancel \( m a^2 \) from both sides of the equation:
\[ \frac{1}{2} \omega = \frac{3}{2} \omega' \]
Rearranging gives:
\[ \omega' = \frac{1}{3} \omega \]

### Conclusion
The new angular speed after the particle is attached to the rim of the disc is:
\[ \omega' = \frac{\omega}{3} \]