Question

How to solve this ??

A straight rod of length `L` extends from `x = 0` to `x = L`. The linear mass density of the rod varies with x co-ordinate is `lambda = a_"o" + b_"o"x^2`. The gravitational force experienced by a point mass `m` at `x = -a`, is ??

Answer 1

F = `int_0^a Gm(a_o + b_o x^2) /(x+a)^2 dx`

Explanation:

Newton's gravitational force between the point mass m at x=a and the mass of any given point (dm') on the the linear rod is

`G(mdm^')/r^2` , where

dm' = `lamda dx` is a point mass on the rod at point x and

r = x-(-a)= x+a is the distance separating m and dm'

Hence,
`dF = G m(a_o + b_o x^2) /(x+a)^2 dx`
since there are infinite many points between x=0 and x=a,
you will have to integrate it. Ask help from the Calculus section if you need help integrating it.