A uniform brass disc of radius a and mass m is set into spinning with angular speed `omega_0` about an axis passing through centre of disc and perpendicular to the palne of disc. If its temperature increases from `theta_1^@C` to `theta_2^@C` with out disturbing the disc, what will be its new angular speed ?

Options

A. `omega_(0)[ 1+2a(theta_(2)-theta_(1))]`
B. `omega_(0)[1+a(theta_(2)-theta_(1))]`
C. `(omega_(0))/([1+2a(theta_(2)-theta_(1))])`
D. None of these

Explanation

To solve the problem of finding the new angular speed of a uniform brass disc after its temperature increases, we can follow these steps:

### Step 1: Understand the Conservation of Angular Momentum
Since no external torque is applied to the system, we can use the principle of conservation of angular momentum. This means that the initial angular momentum of the disc will be equal to its final angular momentum.

### Step 2: Write the Expression for Angular Momentum
The angular momentum \( L \) of a disc is given by the formula:
\[ L = I \omega \]
where \( I \) is the moment of inertia and \( \omega \) is the angular speed.

### Step 3: Calculate Initial Angular Momentum
For a uniform disc, the moment of inertia \( I_0 \) is given by:
\[ I_0 = \frac{1}{2} m a^2 \]
Thus, the initial angular momentum \( L_0 \) can be expressed as:
\[ L_0 = I_0 \omega_0 = \frac{1}{2} m a^2 \omega_0 \]

### Step 4: Determine the Final Moment of Inertia
When the temperature of the disc increases, it will undergo linear expansion. The new radius \( a' \) of the disc can be expressed as:
\[ a' = a(1 + \alpha (\theta_2 - \theta_1)) \]
where \( \alpha \) is the coefficient of linear expansion.

The new moment of inertia \( I' \) will then be:
\[ I' = \frac{1}{2} m (a')^2 = \frac{1}{2} m (a(1 + \alpha (\theta_2 - \theta_1)))^2 \]
Expanding this gives:
\[ I' = \frac{1}{2} m a^2 (1 + \alpha (\theta_2 - \theta_1))^2 \]

### Step 5: Calculate Final Angular Momentum
The final angular momentum \( L' \) can be expressed as:
\[ L' = I' \omega' = \frac{1}{2} m a^2 (1 + \alpha (\theta_2 - \theta_1))^2 \omega' \]

### Step 6: Set Initial and Final Angular Momentum Equal
Using the conservation of angular momentum:
\[ L_0 = L' \]
This gives us:
\[ \frac{1}{2} m a^2 \omega_0 = \frac{1}{2} m a^2 (1 + \alpha (\theta_2 - \theta_1))^2 \omega' \]

### Step 7: Simplify the Equation
Cancelling \( \frac{1}{2} m a^2 \) from both sides:
\[ \omega_0 = (1 + \alpha (\theta_2 - \theta_1))^2 \omega' \]

### Step 8: Solve for the New Angular Speed
Rearranging the equation to solve for \( \omega' \):
\[ \omega' = \frac{\omega_0}{(1 + \alpha (\theta_2 - \theta_1))^2} \]

### Final Result
Thus, the new angular speed \( \omega' \) of the disc after the temperature increase is:
\[ \omega' = \frac{\omega_0}{(1 + 2\alpha (\theta_2 - \theta_1))} \]