The Question is from neet-physics rotational-motion angular-momentum-and-conversation-of-angular-momen, and it was created by the author P.NARASIMHA

Question

A thin uniform rod of mass $M$ and length $L$ is rotating freely with angular speed $\omega_0$ about an axis perpendicular to the rod and passing through one end. A particle of mass $m$ traveling with velocity $v_0$ hits the rod at a distance $L/2$ from the axis and sticks to it. The new angular speed of the system is:

P.NARASIMHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
	**Primary Concept**: Conservation of Angular Momentum
	**Related Topics**: Rotational Motion, Moment of Inertia
	**Prerequisites**: Understanding of angular momentum, moment of inertia of a rod, and concept of inelastic collision.
## SOLUTION PATHWAY
**Given Information**:
	* Mass of rod: $M$
	* Length of rod: $L$
	* Initial angular speed of rod: $\omega_0$
	* Mass of particle: $m$
	* Velocity of particle: $v_0$
	* Distance of impact from axis: $L/2$
	* Particle sticks to the rod (inelastic collision)
**Method & Steps**:
	1. Calculate the initial angular momentum of the rod.
		Reasoning: The angular momentum of the rod is given by the product of its moment of inertia and angular velocity.
		Formula/Rule: $L_{rod} = I_{rod} \omega_0$, where $I_{rod} = \frac{1}{3}ML^2$ for a rod rotating about one end.
	2. Calculate the initial angular momentum of the particle.
		Reasoning: The angular momentum of the particle is given by $r 	imes p$, where $r$ is the position vector from the axis of rotation to the particle, and $p$ is the linear momentum.
		Formula/Rule: $L_{particle} = r m v_0 = (L/2) m v_0$ (since the velocity is perpendicular to the radius).
	3. Calculate the final moment of inertia of the system.
		Reasoning:  The final moment of inertia is the sum of the rod's moment of inertia and the particle's moment of inertia after sticking to the rod.
		Formula/Rule:  $I_{final} = I_{rod} + I_{particle} = \frac{1}{3}ML^2 + m(L/2)^2 = \frac{1}{3}ML^2 + \frac{1}{4}mL^2$
	4. Apply the conservation of angular momentum.
		Reasoning: The total angular momentum before the collision equals the total angular momentum after the collision.
		Formula/Rule: $L_{initial} = L_{final} \implies I_{rod}\omega_0 +  (L/2)mv_0 = I_{final} \omega_f$
	5. Solve for the final angular speed $\omega_f$.
		Working: $(\frac{1}{3}ML^2)\omega_0 + \frac{1}{2}mLv_0 = (\frac{1}{3}ML^2 + \frac{1}{4}mL^2)\omega_f \implies \omega_f = \frac{\frac{1}{3}ML^2\omega_0 + \frac{1}{2}mLv_0}{\frac{1}{3}ML^2 + \frac{1}{4}mL^2} = \frac{\frac{1}{3}ML\omega_0 + \frac{1}{2}mv_0}{\frac{1}{3}ML + \frac{1}{4}mL}$. Multiply by $\frac{12}{L}$ to simplify $\omega_f = \frac{4ML\omega_0 + 6mLv_0/L}{4ML+3mL} = \frac{4M\omega_0 + 6m\frac{v_0}{L}}{4M+3m}$
		Result: $\omega_f = \frac{4M\omega_0 + 6m\frac{v_0}{L}}{4M+3m}$
		Verification: The units of the final answer match with angular velocity.
 ## OPTION ANALYSIS
[A] :x: This option incorrectly combines the angular speed of the rod with the linear velocity of the particle without considering the moment of inertia or the distance of the particle from the axis. It does not account for the rotational inertia of the rod or the particle.
[B] ✓: This option correctly uses the conservation of angular momentum and calculates the new angular speed by considering the initial angular momentum of the rod and the particle, and the combined moment of inertia of the rod and particle after sticking together. The final answer matches the calculated result.
[C] :x: This option makes an error in relating the linear velocity to angular velocity. It does not properly account for the moment of inertia of the rod and the particle.
[D] :x: This option makes an error in the calculation and does not correctly apply conservation of angular momentum. It incorrectly mixes linear and angular variables in the denominator.
 ## LEARNING AIDS
	**Common Mistakes**:
		1. **Incorrect moment of inertia**: Forgetting the correct moment of inertia for a rod about one end. Always use the correct $I = \frac{1}{3}ML^2$.
		2. **Incorrect angular momentum of particle**: Forgetting that the angular momentum of the particle is $rmv$ and that $r$ is the distance from the axis of rotation.
	**Quick Tips**:
		1. **Conservation Principle**: Remember that angular momentum is conserved in the absence of external torques.
		2. **Units Check**: Always check the units of your final answer to make sure it matches the expected quantity.

P.NARASIMHA · Answer

\frac{M\omega_0 + mv_0}{M+m}

P.NARASIMHA · Answer

The correct Option is "\frac{4M\omega_0 + 6m\frac{v_0}{L}}{4M+3m}"

P.NARASIMHA · Answer

\frac{2M\omega_0 + 3mv_0}{2M+3m}

P.NARASIMHA · Answer

\frac{M\omega_0L + 2mv_0}{ML+2m}

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