A disc of mement of inertia `I` is rotating due to external torque. Its kinetic energy is equal to `Ktheta^(2)`. Where K is the positive constant. Its angular acceleration at an angle `theta` will be:

Options

A. `(7Ktheta)/(I)`
B. `(6Ktheta)/(I)`
C. `(2Ktheta)/(I)`
D. `(4Ktheta)/(I)`

Answer

To find the angular acceleration \( \alpha \) of a disc with moment of inertia \( I \) and kinetic energy given by \( K \theta^2 \), we can follow these steps:

### Step 1: Write the expression for kinetic energy
The kinetic energy \( K \) of a rotating disc is given by:
\[ K = \frac{1}{2} I \omega^2 \]
According to the problem statement, the kinetic energy is also given as:
\[ K = K \theta^2 \]
where \( K \) is a positive constant.

### Step 2: Set the two expressions for kinetic energy equal
Equating the two expressions for kinetic energy, we have:
\[ \frac{1}{2} I \omega^2 = K \theta^2 \]

### Step 3: Rearrange the equation to solve for \( \omega^2 \)
Multiplying both sides by 2 gives:
\[ I \omega^2 = 2K \theta^2 \]
Now, dividing both sides by \( I \):
\[ \omega^2 = \frac{2K \theta^2}{I} \]

### Step 4: Differentiate both sides with respect to \( \theta \)
To find the angular acceleration \( \alpha \), we need to differentiate \( \omega^2 \) with respect to \( \theta \):
\[ \frac{d}{d\theta}(\omega^2) = \frac{d}{d\theta}\left(\frac{2K \theta^2}{I}\right) \]
Using the chain rule on the left side:
\[ 2 \omega \frac{d\omega}{d\theta} = \frac{2K \cdot 2\theta}{I} \]

### Step 5: Simplify the equation
This simplifies to:
\[ 2 \omega \frac{d\omega}{d\theta} = \frac{4K \theta}{I} \]
Now, we can express \( \frac{d\omega}{d\theta} \) as:
\[ \frac{d\omega}{d\theta} = \frac{4K \theta}{2I \omega} = \frac{2K \theta}{I \omega} \]

### Step 6: Relate angular acceleration to the derivative
We know that angular acceleration \( \alpha \) is defined as:
\[ \alpha = \omega \frac{d\omega}{d\theta} \]
Substituting our expression for \( \frac{d\omega}{d\theta} \):
\[ \alpha = \omega \cdot \frac{2K \theta}{I \omega} = \frac{2K \theta}{I} \]

### Final Answer
Thus, the angular acceleration \( \alpha \) at an angle \( \theta \) is:
\[ \alpha = \frac{2K \theta}{I} \]