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

Consider the Semi-circular Arc of the Wire:
Length of the semi-circular arc: \( L = \pi R \), where \( R \) is the radius of the semicircle.
Mass per unit length of the wire, \( \lambda = \frac{M}{L} = \frac{M}{\pi R} \).

Gravitational Force Element \( dF \):
Consider a small element \( dl \) of the arc at an angle \( \theta \) from the center, with 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 element on the particle of mass \( m \) at the center is:
\[ dF = \frac{G \times m \times dm}{R^2} = \frac{Gm \times \frac{M}{\pi} d\theta}{R^2} = \frac{GmM}{\pi R^2} d\theta \] 

Resolve \( dF \) into Components:
Each element of the arc exerts a gravitational force toward itself. The horizontal components (along the x-axis) will cancel due to symmetry, and only the vertical components (along the y-axis) will add up.
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:
To find the total gravitational force \( F_y \) on the particle, integrate \( dF_y \) 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 \( R = \frac{L}{\pi} \):
Since \( L = \pi R \), we have \( R = \frac{L}{\pi} \). 

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

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