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:
\[ F = \frac{2GmM\pi}{L^2} \]

Answer 2

Step 1: Given data and setup.
A metal wire of length \( L \) and total mass \( M \) is bent into a semicircular arc.
A particle of mass \( m \) is placed at the centre of the semicircle.
We are required to find the net gravitational force on the particle due to the entire wire.

Step 2: Linear mass density.
Let the radius of the semicircle be \( R \).
The length of the semicircular wire is \( L = \pi R \).
Hence, the linear mass density is: \[ \lambda = \frac{M}{L} = \frac{M}{\pi R}. \]

Step 3: Expression for gravitational force due to a small element.
Consider a small element of the wire subtending an angle \( d\theta \) at the centre.
The mass of this small element is: \[ dm = \lambda R \, d\theta = \frac{M}{\pi R} R \, d\theta = \frac{M}{\pi} \, d\theta. \] The gravitational force due to this small element on the particle is: \[ dF = \frac{Gm \, dm}{R^2} = \frac{GmM}{\pi R^2} \, d\theta. \]

Step 4: Resolve components of force.
Each force element \( dF \) makes an angle \( \theta \) with the symmetry axis (the perpendicular bisector of the diameter).
The horizontal components (along the diameter) cancel due to symmetry.
Only the vertical components (along the central perpendicular) add up.

Vertical component of \( dF \): \[ dF_y = dF \cos \theta = \frac{GmM}{\pi R^2} \cos \theta \, d\theta. \]

Step 5: Integrate to find total force.
Integrate from \( \theta = -\pi/2 \) to \( \pi/2 \): \[ F = \int_{-\pi/2}^{\pi/2} \frac{GmM}{\pi R^2} \cos \theta \, d\theta = \frac{GmM}{\pi R^2} [\sin \theta]_{-\pi/2}^{\pi/2}. \] \[ F = \frac{GmM}{\pi R^2} (1 - (-1)) = \frac{2GmM}{\pi R^2}. \]

Step 6: Substitute \( R = \frac{L}{\pi} \).
\[ F = \frac{2GmM}{\pi \left( \frac{L^2}{\pi^2} \right)} = \frac{2GmM\pi}{L^2}. \]

Final Answer:
\[ \boxed{F = \frac{2GmM\pi}{L^2}} \]