The Question is from neet-physics rotational-motion centre-of-mass, and it was created by the author P.NARASIMHA

Question

A thin uniform rod of mass $M$ and length $L$ is bent into a semicircle. A particle of mass $m$ is placed at the center of the semicircle. The gravitational force exerted by the rod on the particle is:

P.NARASIMHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: Option D
## CONCEPT FRAMEWORK
	**Primary Concept**: Gravitational Force due to a continuous mass distribution
	**Related Topics**: Integration, Linear Mass Density, Newton's Law of Gravitation
	**Prerequisites**: Newton's Law of Gravitation, Integration techniques
## SOLUTION PATHWAY
**Given Information**:
	* Mass of the rod: $M$
	* Length of the rod: $L$
	* Radius of the semicircle: $R$
	* Mass of the particle: $m$
	*  Rod is bent into a semicircle.
	* Particle is at the center of the semicircle
**Method & Steps**:
	1.  Determine the radius of the semicircle:
		Reasoning: The length of the rod becomes the arc length of the semicircle. The arc length of a semicircle is $\pi R$, where R is the radius.
		Formula/Rule: $L = \pi R$
	2. Calculate the radius:
		Working: $R = \frac{L}{\pi}$
		Key Point: Radius is important for calculating gravitational force. 
	3.  Calculate the linear mass density:
		Reasoning: The linear mass density is the mass per unit length of the rod.
		Formula/Rule: $\lambda = \frac{M}{L}$
	4.  Set up the integral for gravitational force:
		Reasoning: Consider a small element of the semicircle with length $dl = R d	heta$ and mass $dm = \lambda dl$. The gravitational force due to this small element at the center is $dF = \frac{Gmdm}{R^2}$. Since the rod is a semicircle, we need to integrate over the semicircle from $-\pi/2$ to $\pi/2$. Due to symmetry, only the y-component of the force will be non-zero. The y-component of the force is $dF_y = dF cos	heta$.
		Formula/Rule: $dF = \frac{Gmdm}{R^2}$, $dF_y = dF cos	heta$, $dm = \lambda dl = \lambda R d	heta$
	5. Calculate $dF_y$:
		Working: $dF_y = \frac{Gm(\lambda R d	heta)}{R^2}cos	heta = \frac{Gm\lambda}{R} cos	heta d	heta$
	6. Integrate to find the total force:
		Working: $F_y = \int_{-\pi/2}^{\pi/2} \frac{Gm\lambda}{R} cos	heta d	heta = \frac{Gm\lambda}{R} \int_{-\pi/2}^{\pi/2} cos	heta d	heta = \frac{Gm\lambda}{R} [sin	heta]_{-\pi/2}^{\pi/2} = \frac{Gm\lambda}{R} (1 - (-1)) = \frac{2Gm\lambda}{R}$
	7. Substitute values for $\lambda$ and $R$:
		Working: $F_y = \frac{2Gm(M/L)}{L/\pi} = \frac{2\pi GMm}{L^2}$
		Result: $F = \frac{2\pi GMm}{L^2}$
		Verification: The units are correct (force).
 ## OPTION ANALYSIS
[A] :x: Incorrect. The formula does not account for the geometry of the semicircle.
[B] :x: Incorrect. The formula does not account for the geometry of the semicircle.
[C] :x: Incorrect. The formula does not account for the geometry of the semicircle.
[D] ✓: Correct. The calculation matches the result obtained by integration.
 ## LEARNING AIDS
	**Common Mistakes**:
		1. Incorrectly assuming the force is simply $GMm/R^2$: This is incorrect as it does not account for the distribution of mass.
		2. Forgetting to integrate: The gravitational force due to a distributed mass must be found by integrating over the mass distribution.
	**Quick Tips**:
		1. Symmetry can simplify the problem: Use symmetry to determine the direction of the net force.
		2. Always consider linear mass density: Linear mass density helps convert length elements to mass elements for integration.

P.NARASIMHA · Answer

$\frac{GMm}{L^2}$

P.NARASIMHA · Answer

$\frac{2GMm}{L^2}$

P.NARASIMHA · Answer

$\frac{\pi GMm}{L^2}$

P.NARASIMHA · Answer

The correct Option is "$\frac{2\pi GMm}{L^2}$"

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