Question

A semicircular wire has a length L and mass M. A particle of mass m is placed at the centre of the circle. Find the gravitational attraction on the particle due to the wire.

Answer 1

F = (GMm/Lr)∫0π sin θ dθ = 2πGMm/L2  

Explanation:

Lets say Length of the wire = L = π r  ......(1)

Radius of the semicircle = r

Now the gravitational force will act along the radius therefore the distance between the particle on the wire and the particle on centre is r.

Now find the gravitational force on the particle:

dF = G m (M/L) dl / r2 along radius itself

By splitting the components of force we see that 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)∫0π sin θ dθ = 2πGMm/L2  

Learn more about

What is Gravitational Force?

https://brainly.in/question/6257716

Answer 2

Answer:

Explanation:

In the semicircle, we can consider a small element dthη

then mass of Rdthη=(ML)Rdthη

dF=(Gmrdthetam)/(LR^2)dF1=2dF

since=2GMmLRsinθdθ:. F=int_0^(pi/2) (-2GMm)/(LR) sintheta d theta=−2GMmLR(−1)

2GMmLR=2GMmLLπ

=(2piGMm)/L^2`