A rod PQ of length L revolves in a horizontal plane about the axis YY´. The angular velocity of the rod is w. If A is the area of cross-section of the rod and `rho` be its density, its rotational kinetic energy is

Options

A. `(1)/(34)AL^(3)rho omega^(2)`
B. `(1)/(2)AL^(3)rho omega^(2)`
C. `(1)/(24)AL^(3)rho omega^(2)`
D. `(1)/(8) AL^(3)rho omega^(2)`

Answer

To find the rotational kinetic energy of a rod PQ of length L revolving about an axis YY' with an angular velocity ω, we can follow these steps:

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

### Step 2: Determine the moment of inertia of the rod
For a uniform rod of length \(L\) rotating about an axis through its center, the moment of inertia \(I\) is given by:
\[ I = \frac{1}{12} m L^2 \]
where \(m\) is the mass of the rod.

### Step 3: Calculate the mass of the rod
The mass \(m\) of the rod can be expressed in terms of its density \(\rho\) and cross-sectional area \(A\):
\[ m = \text{Volume} \times \text{Density} = A \cdot L \cdot \rho \]
Thus, substituting this into the moment of inertia equation gives:
\[ I = \frac{1}{12} (A \cdot L \cdot \rho) L^2 = \frac{1}{12} A \rho L^3 \]

### Step 4: Substitute the moment of inertia into the kinetic energy formula
Now we can substitute the expression for \(I\) back into the kinetic energy formula:
\[ K.E. = \frac{1}{2} \left(\frac{1}{12} A \rho L^3\right) \omega^2 \]

### Step 5: Simplify the expression
Simplifying the expression gives:
\[ K.E. = \frac{1}{24} A \rho L^3 \omega^2 \]

### Final Result
Thus, the rotational kinetic energy of the rod is:
\[ K.E. = \frac{1}{24} A \rho L^3 \omega^2 \]