Question

Need answers to this question so that i can compare your solutions to mine. This is a question on Gravitational force.

A uniform wire with a mass of M and a length L is bent into a semicircle. Find the magnitude and direction of the gravitational force this wire exerts on a point with mass m placed at the center of curvature of the semicircle = (the center of the respective full circle).

Please use calculus.
@Physics

Answer 1

\[\int\limits_{0}^{\pi}GMm \times \sin \theta \times \pi \div L ^{2}\]
you get 2piGMm/L^2

Answer 2

First if the radius of the semicircle is r, then \( \pi r = L \) and hence \( r = L/\pi \). Consider the semi-circle to be in the plane from \( \theta = 0 \ to \ \pi. \)

Now consider the force on the mass m from an element of mass dM on the wire. Let the element dM sweep out an angle with respect to the centre of \(d\theta\) and a length dL.
Then
\[ dM/M = dL/L = r d\theta / L = d\theta / \pi \]
and hence \( dM = M/\pi \ d\theta \) and the direction of the force from that mass element is in the negative radial direction: \( \hat{r} = (-\cos \theta, -\sin \theta) \). Therefore the total force on the mass m is
\[ F = \int \frac{Gm}{r^2} \hat{r} dM \ = \ \int_0^{\pi} \frac{GMm\pi}{L^2} (-\cos \theta, -\sin \theta) d\theta = \frac{GMm\pi}{L^2} (0, -2) \]

Hence the force is in the negative y-direction as as all the x-direction forces cancel, as we would expect intuitively. The magnitude of the force is
\[ \frac{2GMm \pi}{L^2} \]

Answer 3

Correction: in the integral, the \( \pi \) should be in the denominator and in the final answer as well:
\[ \frac{2GMm}{\pi L^2} \]