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 `ktheta^(2)` If its moment of inertia is I then the angular acceleration of the disc is

Options

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

Explanation

To find the angular acceleration of the disc, we will follow these steps:

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

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

### Step 3: Differentiate the kinetic energy with respect to time
We can equate the two expressions for kinetic energy:
\[ \frac{1}{2} I \omega^2 = k \theta^2 \]
Now, differentiate both sides with respect to time \( t \):
\[ \frac{d}{dt}\left(\frac{1}{2} I \omega^2\right) = \frac{d}{dt}(k \theta^2) \]

### Step 4: Apply the chain rule
Using the chain rule, we get:
\[ \frac{1}{2} I \cdot 2\omega \frac{d\omega}{dt} = k \cdot 2\theta \frac{d\theta}{dt} \]
This simplifies to:
\[ I \omega \frac{d\omega}{dt} = 2k \theta \frac{d\theta}{dt} \]

### Step 5: Recognize that \( \frac{d\theta}{dt} = \omega \)
Since \( \frac{d\theta}{dt} = \omega \), we can substitute this into our equation:
\[ I \omega \frac{d\omega}{dt} = 2k \theta \omega \]

### Step 6: Cancel \( \omega \) from both sides
Assuming \( \omega \neq 0 \) (which is valid since the disc is rotating), we can divide both sides by \( \omega \):
\[ I \frac{d\omega}{dt} = 2k \theta \]

### Step 7: Relate \( \frac{d\omega}{dt} \) to angular acceleration
Recall that \( \frac{d\omega}{dt} = \alpha \) (angular acceleration). Thus, we can rewrite the equation as:
\[ I \alpha = 2k \theta \]

### Step 8: Solve for angular acceleration \( \alpha \)
Now, we can solve for \( \alpha \):
\[ \alpha = \frac{2k \theta}{I} \]

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

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