First slide

Gravitational field intensity

Question

A uniform thin rod of mass m and length l is bent into a semicircle. Gravitational field intensity at the centre O is :

Moderate

A

$\frac{2 πGm}{l^{2}}$ along AOB
B

$\frac{πGm}{l^{2}}$ perpendicular to AOB
C

$\frac{2 πGm}{l^{2}}$ perpendicular to AOB
D

$\frac{πGm}{l^{2}}$ along AOB

Solution

Let the mass per unit length of the wire be $λ = \frac{m}{l} = \frac{m}{π R}$ , where R is the radius of the circle.
Thus horizontal components of the gravitational field intensity at the center cancel each other.
The vertical component of the same = $\int_{- π / 2}^{π / 2} \frac{G (λ Rd θ)}{R^{2}} \cos θ = \frac{2 Gm}{π R^{2}} = \frac{2 π Gm}{l^{2}}$

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