A uniform thin rod of length l is suspended from one of its ends and is rotated at f rotations per second. The rotational kinetic energy of the rod will be

Options

A. `2mL^(2)pi^(2)n^(2)`
B. `(1)/(2)mL^(2)pi^(2)n^(2)`
C. `(2)/(3)mL^(2)pi^(2)n^(2)`
D. `(1)/(6)mL^(2)pi^(2)n^(2)`

Answer

To find the rotational kinetic energy of a uniform thin rod of length \( l \) suspended from one of its ends and rotated at \( f \) rotations per second, we can follow these steps:

### Step 1: Understand the given parameters
- Length of the rod = \( l \)
- Frequency of rotation = \( f \) rotations per second

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

### Step 3: Calculate the moment of inertia of the rod
For a uniform thin rod of length \( l \) suspended from one end, the moment of inertia \( I \) is given by:
\[ I = \frac{1}{3} m l^2 \]
where \( m \) is the mass of the rod.

### Step 4: Convert frequency to angular velocity
The angular velocity \( \omega \) can be calculated from the frequency \( f \) using the relation:
\[ \omega = 2 \pi f \]

### Step 5: Substitute \( I \) and \( \omega \) into the kinetic energy formula
Substituting the values of \( I \) and \( \omega \) into the kinetic energy formula:
\[ K = \frac{1}{2} \left( \frac{1}{3} m l^2 \right) (2 \pi f)^2 \]

### Step 6: Simplify the expression
Now, simplify the expression:
\[ K = \frac{1}{2} \cdot \frac{1}{3} m l^2 \cdot 4 \pi^2 f^2 \]
\[ K = \frac{2}{3} m l^2 \pi^2 f^2 \]

### Final Result
Thus, the rotational kinetic energy of the rod is:
\[ K = \frac{2}{3} m l^2 \pi^2 f^2 \]