A uniform metallic disc of radius r and mass m is spinning with angular speed `omega_(0)` about an axis passing through its centre and perpendicular to plane. If its temperature is increased (slightly) by `DeltaT` its new angular speed is (The coefficient of linear expansion of the metal is `alpha`)

Options

A. `omega_(0)(1+2alphaDeltaT)`
B. `omega_(0)(1+alphaDeltaT)`
C. `omega_(0)(1-2alphaDeltaT)`
D. `omega_(0)(1-3alphaDeltaT)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the conservation of angular momentum
When the temperature of the disc increases, its moment of inertia changes due to thermal expansion, but there are no external torques acting on the system. Therefore, the angular momentum of the disc is conserved.

**Hint:** Remember that angular momentum \( L \) is given by the product of moment of inertia \( I \) and angular speed \( \omega \).

### Step 2: Write the expression for initial and final angular momentum
The initial angular momentum \( L_i \) can be expressed as:
\[ L_i = I_1 \cdot \omega_0 \]
where \( I_1 \) is the initial moment of inertia of the disc.

The final angular momentum \( L_f \) is given by:
\[ L_f = I_2 \cdot \omega_f \]
where \( I_2 \) is the final moment of inertia after the temperature increase and \( \omega_f \) is the new angular speed.

**Hint:** The moment of inertia for a uniform disc is given by \( I = \frac{1}{2} m r^2 \).

### Step 3: Calculate the initial moment of inertia
For a uniform disc of radius \( r \) and mass \( m \):
\[ I_1 = \frac{1}{2} m r^2 \]

### Step 4: Calculate the final moment of inertia
When the temperature increases by \( \Delta T \), the radius of the disc expands. The new radius \( r' \) can be expressed as:
\[ r' = r(1 + \alpha \Delta T) \]
Thus, the final moment of inertia \( I_2 \) becomes:
\[ I_2 = \frac{1}{2} m (r')^2 = \frac{1}{2} m (r(1 + \alpha \Delta T))^2 = \frac{1}{2} m r^2 (1 + \alpha \Delta T)^2 \]

**Hint:** Use the expansion of \( (1 + x)^2 \) to simplify the expression.

### Step 5: Expand the final moment of inertia
Using the binomial expansion for small \( \alpha \Delta T \):
\[ (1 + \alpha \Delta T)^2 \approx 1 + 2\alpha \Delta T \]
Thus, we can write:
\[ I_2 \approx \frac{1}{2} m r^2 (1 + 2\alpha \Delta T) \]

### Step 6: Set the initial and final angular momentum equal
Since angular momentum is conserved:
\[ I_1 \cdot \omega_0 = I_2 \cdot \omega_f \]
Substituting the expressions we derived:
\[ \frac{1}{2} m r^2 \cdot \omega_0 = \frac{1}{2} m r^2 (1 + 2\alpha \Delta T) \cdot \omega_f \]

### Step 7: Simplify and solve for \( \omega_f \)
Cancelling \( \frac{1}{2} m r^2 \) from both sides:
\[ \omega_0 = (1 + 2\alpha \Delta T) \cdot \omega_f \]
Rearranging gives:
\[ \omega_f = \frac{\omega_0}{1 + 2\alpha \Delta T} \]

### Step 8: Use the approximation for small \( \Delta T \)
For small \( \Delta T \), we can use the approximation \( \frac{1}{1 + x} \approx 1 - x \) for small \( x \):
\[ \omega_f \approx \omega_0 (1 - 2\alpha \Delta T) \]

### Final Answer
The new angular speed \( \omega_f \) after the temperature increase is:
\[ \omega_f \approx \omega_0 (1 + 2\alpha \Delta T) \]