A thin uniform rod of length `2m`,cross sectional area `A` and density `d ` is rotated about an axis passing through the centre and perpendicular to its length with angular velocity `omega`.If value of `omega` in terms of its rotational kinetic energy `E` is `sqrt((`alphaE`)/(Ad))` then value of `a` is

Answer

To solve the problem, we need to find the value of \( \alpha \) in the expression for angular velocity \( \omega \) in terms of the rotational kinetic energy \( E \). The expression given is:

\[ \omega = \sqrt{\frac{\alpha E}{Ad}} \]

### Step 1: Calculate the Moment of Inertia (I) of the Rod
For a thin uniform rod of length \( L \) rotating about an axis through its center and perpendicular to its length, the moment of inertia \( I \) is given by:

\[ I = \frac{1}{12} m L^2 \]

Given that the length \( L = 2 \, \text{m} \), we can substitute this value in:

\[ I = \frac{1}{12} m (2)^2 = \frac{1}{12} m \cdot 4 = \frac{1}{3} m \]

### Step 2: Express Mass (m) in Terms of Density (d) and Cross-Sectional Area (A)
The mass \( m \) of the rod can be expressed in terms of its density \( d \), cross-sectional area \( A \), and volume \( V \). The volume \( V \) of the rod is given by:

\[ V = A \cdot L = A \cdot 2 \]

Thus, the mass \( m \) is:

\[ m = d \cdot V = d \cdot (A \cdot 2) = 2dA \]

### Step 3: Substitute Mass into the Moment of Inertia
Now, substituting \( m \) back into the moment of inertia equation:

\[ I = \frac{1}{3} m = \frac{1}{3} (2dA) = \frac{2dA}{3} \]

### Step 4: Write the Expression for Rotational Kinetic Energy (E)
The rotational kinetic energy \( E \) is given by:

\[ E = \frac{1}{2} I \omega^2 \]

Substituting the expression for \( I \):

\[ E = \frac{1}{2} \left(\frac{2dA}{3}\right) \omega^2 = \frac{dA}{3} \omega^2 \]

### Step 5: Solve for \( \omega^2 \)
Rearranging the expression for \( E \):

\[ \omega^2 = \frac{3E}{dA} \]

### Step 6: Compare with Given Expression
The problem states that:

\[ \omega = \sqrt{\frac{\alpha E}{Ad}} \]

Squaring both sides gives:

\[ \omega^2 = \frac{\alpha E}{Ad} \]

### Step 7: Equate the Two Expressions for \( \omega^2 \)
Now we equate the two expressions for \( \omega^2 \):

\[ \frac{3E}{dA} = \frac{\alpha E}{Ad} \]

### Step 8: Simplify and Solve for \( \alpha \)
Cancelling \( E \) from both sides (assuming \( E \neq 0 \)):

\[ \frac{3}{dA} = \frac{\alpha}{Ad} \]

Cross-multiplying gives:

\[ 3Ad = \alpha dA \]

Dividing both sides by \( dA \) (assuming \( A \neq 0 \) and \( d \neq 0 \)):

\[ \alpha = 3 \]

### Final Answer
Thus, the value of \( \alpha \) is:

\[ \alpha = 3 \]