A thin rod of mass `M` and length `L` is bent into a semicircle as shown in diagram. What is a gravitational force on a particle with mass `m` at the centre of curvature?

Explanation

consider an elements of rod of length `dl` as shown in diagram and treat it as a small particle of mass `(M//L) dl` situated at a distance `R` from `P`. Then gravitational force due to the element on the particle will be
`dF=(Gm(M//L)(R d theta))/(R^(2))` along `OP`
`[as dl=R d theta]`

so the components of this force along `X` and `Y` axes will be
`dF_(x)=dF cos theta=(GmM cos theta d theta)/(LR)`
`dF_(y)=dF sin theta=(Gm M cos thetad theta)/(LR)`
So that
`F_(x)=(GmM)/(LR)int_(0)^(pi)cos theta d theta=(GmM)/(LR)[sin theta]_(0)^(pi)=0`
`F_(y)=(GmM)/(LR)int_(0)^(pi)sin theta d theta=(GmM)/(LR)[-cos theta]_(0)^(pi)`
`=(2piGmM)/(L^(2))" "[as R=L/pi]`
`F=sqrt(F_(x)^(2)+F_(y)^(2))=F_(y)=(2piGmM)/(L^(2))" "(as F_(x)=0)`