A Semicircular wire has a mass M and length L . A particle of mass m is placed at the centre of the circle.Find the gravitational attraction on the particle due to wire.Ans=2(pie)GMm/L²

Given:

Length of the wire = L = π r  -------------(1)

r be the radius of the semicircle.

Now the gravitational force will act along the radius of the semicircle so the distance between the particle on the wire and the particle on centre will be r only!

Consider an element of rod of length dl and treating it as a point mass then the gravitational force due to this element on the particle will be,

dF = G m (M/L) dl / r2 along radius itself now splitting the components of force we see horizontal components will not contribute only vertical components contribute.

(dF)v = dF sin θ again we can substitute dl in the form of dθ    [ dl = r dθ ] 

Now integrating within the limits 0 to π

We can get the desired answer!

F = (GMm/Lr)∫₀^π sin θ dθ = 2πGMm/L²