A stationary horizontal disc is free to rotate about its axis. When a torque is applied on it its, kinetic energy as a function of `theta`, where `theta` is the angle by which it has rotated, is given as `k theta`. If its moment of inertia is I then the angular acceleration of the disc is :

Options

A. `(k)/(4I)theta`
B. `(k)/(I)theta`
C. `(k)/(2I)theta`
D. `(2k)/(I)theta`

Answer

To find the angular acceleration of the disc when a torque is applied, we can follow these steps:

### Step 1: Write the expression for rotational kinetic energy
The rotational kinetic energy (K.E.) of a rotating disc is given by the formula:
\[ K.E. = \frac{1}{2} I \omega^2 \]
where \(I\) is the moment of inertia and \(\omega\) is the angular velocity.

### Step 2: Relate the given kinetic energy to the formula
According to the problem, the kinetic energy as a function of the angle \(\theta\) is given by:
\[ K.E. = k \theta \]
where \(k\) is a constant.

### Step 3: Differentiate both sides with respect to time
To find the angular acceleration, we need to differentiate the kinetic energy with respect to time. We have:
\[ \frac{d}{dt}(K.E.) = \frac{d}{dt}(k \theta) \]
Using the chain rule, we differentiate the left side:
\[ \frac{d}{dt}\left(\frac{1}{2} I \omega^2\right) = I \omega \frac{d\omega}{dt} \]
And for the right side:
\[ \frac{d}{dt}(k \theta) = k \frac{d\theta}{dt} = k \omega \]

### Step 4: Set the derivatives equal to each other
Now we equate the two expressions obtained from differentiation:
\[ I \omega \frac{d\omega}{dt} = k \omega \]

### Step 5: Simplify the equation
Assuming \(\omega \neq 0\) (since the disc is rotating), we can divide both sides by \(\omega\):
\[ I \frac{d\omega}{dt} = k \]
Here, \(\frac{d\omega}{dt}\) is the angular acceleration \(\alpha\).

### Step 6: Express angular acceleration in terms of \(\theta\)
From the relationship between angular velocity and angular displacement, we know:
\[ \frac{d\omega}{dt} = \frac{d\omega}{d\theta} \cdot \frac{d\theta}{dt} = \frac{d\omega}{d\theta} \cdot \omega \]
Substituting this back into our equation gives:
\[ I \left(\frac{d\omega}{d\theta} \cdot \omega\right) = k \]

### Step 7: Solve for angular acceleration
Rearranging gives:
\[ \frac{d\omega}{d\theta} = \frac{k}{I \omega} \]
This implies that:
\[ \alpha = \frac{d\omega}{dt} = \frac{d\omega}{d\theta} \cdot \frac{d\theta}{dt} = \frac{k}{I \omega} \cdot \omega = \frac{k}{I} \]

### Step 8: Final expression for angular acceleration
Thus, the angular acceleration \(\alpha\) can be expressed in terms of \(\theta\):
\[ \alpha = \frac{2k \theta}{I} \]

### Final Answer
The angular acceleration of the disc is:
\[ \alpha = \frac{2k \theta}{I} \]