The Question is from neet-physics rotational-motion , and it was created by the author NAGA SAI

Question

Two particles of masses $m_1$ and $m_2$ are connected by a massless rigid rod of length $L$. The system rotates with a constant angular velocity $\omega$ about an axis perpendicular to the rod and passing through its center of mass. The kinetic energy of the system is:

NAGA SAI · Accepted Answer

## CORE INFORMATION
✓ **Answer**: [Option B]
## CONCEPT FRAMEWORK
	**Primary Concept**: Rotational Kinetic Energy
	**Related Topics**: Center of Mass, Moment of Inertia
	**Prerequisites**: Understanding of rotational motion, kinetic energy, center of mass calculation.
## SOLUTION PATHWAY
**Given Information**:
	* Two masses: $m_1$ and $m_2$
	* Length of the rod: $L$
	* Angular velocity: $\omega$
	* Axis of rotation: Perpendicular to the rod and through the center of mass.
**Method & Steps**:
	1. Find the position of the center of mass.
		Reasoning: Let $x_1$ and $x_2$ be the distances of $m_1$ and $m_2$ from the center of mass, respectively. The center of mass is defined by $m_1x_1 = m_2x_2$. Also, $x_1 + x_2 = L$. Solving these gives $x_1 = \frac{m_2 L}{m_1 + m_2}$ and $x_2 = \frac{m_1 L}{m_1 + m_2}$.
		Formula/Rule: Center of mass: $m_1x_1 = m_2x_2$ and $x_1 + x_2 = L$
	2. Calculate the moment of inertia of the system.
		Reasoning: The moment of inertia is $I = m_1 x_1^2 + m_2 x_2^2$.
		Working: Substituting the values of $x_1$ and $x_2$: 
		$I = m_1 \left( \frac{m_2 L}{m_1 + m_2} ight)^2 + m_2 \left( \frac{m_1 L}{m_1 + m_2} ight)^2 = \frac{m_1 m_2^2 L^2 + m_2 m_1^2 L^2}{(m_1 + m_2)^2} = \frac{m_1 m_2 (m_1 + m_2) L^2}{(m_1 + m_2)^2} = \frac{m_1 m_2}{m_1 + m_2} L^2$
		Key Point: Moment of inertia about the center of mass is required.
	3. Calculate the kinetic energy of the system.
		Reasoning: The kinetic energy is given by $K = \frac{1}{2} I \omega^2$.
		Working: Substituting the value of $I$: $K = \frac{1}{2} \left( \frac{m_1 m_2}{m_1 + m_2} L^2 ight) \omega^2 = \frac{1}{2} \frac{m_1 m_2}{m_1 + m_2} L^2 \omega^2$
		Result: $K = \frac{1}{2} \frac{m_1 m_2}{m_1 + m_2} L^2 \omega^2$
		Verification: The result has correct dimensions of energy.
 ## OPTION ANALYSIS
[A] :x: This option is incorrect as it does not consider the reduced mass.
[B] ✓: This option is correct as it matches the calculated kinetic energy.
[C] :x: This option is incorrect as it does not consider the sum of masses in the denominator.
[D] :x: This option is incorrect as it misses the length term and the reduced mass term.
 ## LEARNING AIDS
	**Common Mistakes**:
		1.[Error 1]: Using total mass instead of reduced mass in the formula for KE. [How to avoid]: Always use the reduced mass when dealing with two masses rotating about their center of mass.
		2.[Error 2]: Incorrectly calculating the moment of inertia about an axis that is not through the center of mass. [How to avoid]: Always calculate the moment of inertia about the given axis of rotation.
	**Quick Tips**:
		1.[Helpful shortcut]: Remember the reduced mass formula: $\mu = \frac{m_1 m_2}{m_1 + m_2}$.
		2.[Memory aid]: The kinetic energy formula for rotation is similar to linear kinetic energy, $K = \frac{1}{2}mv^2$, with $I$ replacing $m$ and $\omega$ replacing $v$.

NAGA SAI · Answer

$\frac{1}{2} (m_1 + m_2) L^2 \omega^2$

NAGA SAI · Answer

The correct Option is "$\frac{1}{2} \frac{m_1 m_2}{m_1 + m_2} L^2 \omega^2$"

NAGA SAI · Answer

$\frac{1}{2} m_1 m_2 L^2 \omega^2$

NAGA SAI · Answer

$\frac{1}{2} (m_1 + m_2) \omega^2$

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