Two point masses of 0.3 kg and 0.7kg are fixed at the ends of a rod of length 1.4 m and of negligible mass. The rod is set rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum, is located at a distance of

Options

A. 0,42 m from mass of 0.3 kg
B. 0.70 m form mass of 0.7 kg
C. 0.98 m form mass of 0.3 kg
D. 0.98 m from mass of 0.7 kg

Explanation

To solve the problem, we need to find the position along the rod where the axis of rotation should be placed to minimize the work required for rotation. We have two point masses, \( m_1 = 0.3 \, \text{kg} \) and \( m_2 = 0.7 \, \text{kg} \), located at the ends of a rod of length \( L = 1.4 \, \text{m} \).

### Step-by-Step Solution

1. **Define the Position of the Axis**:
Let the distance from the \( 0.3 \, \text{kg} \) mass to the axis of rotation be \( x \). Consequently, the distance from the \( 0.7 \, \text{kg} \) mass to the axis will be \( L - x = 1.4 - x \).

2. **Kinetic Energy Calculation**:
The kinetic energy \( KE \) of the system when rotating about the axis can be expressed as:
\[ KE = \frac{1}{2} m_1 \omega^2 r_1^2 + \frac{1}{2} m_2 \omega^2 r_2^2 \]
where \( r_1 = x \) and \( r_2 = 1.4 - x \). Thus,
\[ KE = \frac{1}{2} (0.3) \omega^2 x^2 + \frac{1}{2} (0.7) \omega^2 (1.4 - x)^2 \]

3. **Simplifying the Kinetic Energy Expression**:
Factor out \( \frac{\omega^2}{2} \):
\[ KE = \frac{\omega^2}{2} \left( 0.3 x^2 + 0.7 (1.4 - x)^2 \right) \]
Expanding the second term:
\[ KE = \frac{\omega^2}{2} \left( 0.3 x^2 + 0.7 (1.96 - 2.8x + x^2) \right) \]
\[ = \frac{\omega^2}{2} \left( 0.3 x^2 + 1.372 - 1.96x + 0.7x^2 \right) \]
\[ = \frac{\omega^2}{2} \left( (0.3 + 0.7)x^2 - 1.96x + 1.372 \right) \]
\[ = \frac{\omega^2}{2} \left( x^2 - 1.96x + 1.372 \right) \]

4. **Finding the Minimum Work**:
To minimize the kinetic energy, we take the derivative of \( KE \) with respect to \( x \) and set it to zero:
\[ \frac{d(KE)}{dx} = \frac{\omega^2}{2} \left( 2x - 1.96 \right) = 0 \]
Solving for \( x \):
\[ 2x - 1.96 = 0 \implies 2x = 1.96 \implies x = 0.98 \, \text{m} \]

5. **Conclusion**:
The axis of rotation should be placed at a distance of \( 0.98 \, \text{m} \) from the \( 0.3 \, \text{kg} \) mass.

### Final Answer:
The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum, is located at a distance of **0.98 m** from the \( 0.3 \, \text{kg} \) mass.