(a) A uniform rod of mass `M` and length `L` is placed at distance `L` from a point mass `m` as shown. Find force on `m`

(b) A semiconductor 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.

Explanation

(a)
Mass per unit length of rod `=M/L`
Take an element of length `dx` at distance `x` form `O`
Mass of element `dm=M/L dx`

`dF=(Gmdm)/((2L-x)^(2))=(GMm)/L (dx)/((2L-x)^(2))`
`F=(GMm)/L=int_(0)^(L)(dx)/((2L-x)^(2))`
`=(GMm)/L|((2L-x)^(-1))/((-1)(-1))|_(0)^(L)=(GMm)/L|1/((2L-x))|_(0)^(L)`
`=(GMm)/L|1/((2L-L))-1/(2L)|=(GMm)/(2L^(2))`

`L=piR,R=L//pi`
Mass of element `dn=M/L R d theta=M/(pi) d theta`

`dF=(Gmdm)/(R^(2))=(GmMd theta)/(piR^(2))`

`F_(x)=intdF_(x)=intdF cos theta=(GMm)/(piR^(2))int_(-pi//2)^(pi//2) cos theta d theta`
`=(GMm)/(piR^(2)).2 int_(0)^(pi//2)cos theta d theta`
`=(2GMm)/(piR^(2))|sin theta|_(0)^(pi//2)=(2GMm)/(piR^(2))(sin(pi/2)-sin0)`
`(2GMm)/(piR^(2))=(2GMm)/(piL^(2)//pi^(2))=(2piGMm)/(L^(2))`
`F_(x)=0`
Force on `m,F=F_(x)=(2piGMm)/(L^(2))`