Question

A metal wire of uniform mass density having length L and mass M is bent to form a semicircular arc and a particle of mass m is placed at the centre of the arc. The gravitational force on the particle by the wire is:

• A. \( \frac{GmM\pi}{2L^2} \)
• B. 0
• C. \( \frac{GmM\pi^2}{L^2} \)
• D. \( \frac{2GmM\pi}{L^2} \)

Answer 1

The Correct Option is D

Analyze the Semi-circular Wire Arc:
The length of the semi-circular arc is given by \( L = \pi R \), where \( R \) represents the radius of the semicircle.
The mass per unit length of the wire is defined as \( \lambda = \frac{M}{L} = \frac{M}{\pi R} \).

Define Gravitational Force Element \( dF \):
Consider an infinitesimal arc segment \( dl \) at an angle \( \theta \) from the center, possessing a mass \( dm \):
\[ dm = \lambda dl = \frac{M}{\pi R} \times R d\theta = \frac{M}{\pi} d\theta \]

The gravitational force \( dF \) exerted by this mass element on a particle of mass \( m \) located at the center is calculated as:
\[ dF = \frac{G \times m \times dm}{R^2} = \frac{Gm \times \frac{M}{\pi} d\theta}{R^2} = \frac{GmM}{\pi R^2} d\theta \] 

Decompose \( dF \) into Components:
Due to symmetry, the horizontal components (along the x-axis) of the gravitational forces exerted by each arc element cancel out. Only the vertical components (along the y-axis) contribute to the net force.
The vertical component of \( dF \) is:
\[ dF_y = dF \cos \theta = \frac{GmM}{\pi R^2} \cos \theta \, d\theta \] 

Integrate \( dF_y \) Over the Semicircle:
The total gravitational force \( F_y \) acting on the particle is obtained by integrating \( dF_y \) over the range of the semicircle, from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \):
\[ F_y = \int_{-\pi/2}^{\pi/2} \frac{GmM}{\pi R^2} \cos \theta \, d\theta \]

\[ F_y = \frac{GmM}{\pi R^2} \int_{-\pi/2}^{\pi/2} \cos \theta \, d\theta \]

\[ F_y = \frac{GmM}{\pi R^2} \left[ \sin \theta \right]_{-\pi/2}^{\pi/2} \]

\[ F_y = \frac{GmM}{\pi R^2} \left( \sin \frac{\pi}{2} - \sin \left( -\frac{\pi}{2} \right) \right) \]

\[ F_y = \frac{GmM}{\pi R^2} (1 + 1) = \frac{2GmM}{\pi R^2} \] 

Substitute \( R \) in terms of \( L \):
Given \( L = \pi R \), the radius can be expressed as \( R = \frac{L}{\pi} \). 

Substitute this expression for \( R \) into the equation for \( F_y \):
\[ F_y = \frac{2GmM}{\pi \left( \frac{L}{\pi} \right)^2} = \frac{2GmM\pi}{L^2} \] 

Final Result:
The net gravitational force exerted by the wire on the particle is:
\[ F = \frac{2GmM\pi}{L^2} \]