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 ??

1 Answer
Dec 22, 2017

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.