A thin of length `L` is bent to form a semicircle. The mass of rod is M. What will be the gravitational potential at the centre of the circle ?

Explanation

(a). Considering an element of rod of length dl as shown in figure and treating it s a point of mass `(M//L)` dl situated at a distance R from P, the gravitational force due to this element o the particle will be
`dF=(Gm(M//L)(Rdtheta))/(R^(2))` along OP [as `dl=Rdtheta]`
So the component of this force along x and y-axis will be ltbr `dF_(x)=dFcostheta=(GmMcosthetad theta)/(LR),dF_(y)=dFsintheta=(GmMsinthetad theta)/(LR)`
So that `F_(x)=(GmM)/(LR)int_(0)^(pi)costhetad theta=(GmM)/(LR)[sintheta]_(0)^(pi)=0`
and `F_(y)=(GmM)/(LR)int_(0)^(pi)sinthetad theta=(GmM)/(LR)[-cos]_(0)^(pi)=(2piGmM)/(L^(2))[as R=(L)/(pi)]`
So `F=sqrt(F_(x)^(2)+F_(y)^(2))=F_(y)=(2piGmM)/(L^(2))` [as `F_(x) `is zero]
i.e., the resultant force is along the y-axis and has magnitude `[2piGmM//L^(2)]`
(b). If the rod was bent into a complete circle,
`F_(x)=(GmM)/(LR)int_(0)^(2pi)costhetad theta=0` and also, `F_(y)=(GmM)/(LR)int_(0)^(2pi)sintheta d theta=0`
I.e., the resultant force on m at O due to the ring is zero.