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

Answer: (D)

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

Answer 2

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