Question

A straight rod of length L extends form x=a to x=L+a. The gravitational force on a point mass m at x=0 if the mass per unit length of the rod is (A+B$x^2$), is

• A. $GmA(\frac{1}{a+L}-\frac{1}{a}+BL]$
• B. $GmA[\frac{1}{a}-\frac{1}{a+L}+BL]$
• C. $GmA\left[\right(\frac{1}{a+L}-\frac{1}{a})-BL]$
• D. $GmA\left[\right(\frac{1}{a}-\frac{1}{a+L})-BL]$

Answer 1

The correct option is B

$GmA[\frac{1}{a}-\frac{1}{a+L}+BL]$

Mass per unit lenght = A + B $x^2$

So the mass of length dx is

d M = dx(A + b $x^2$)

F = $\sum _a^{a+L}$ Gm ($\frac{1}{x^2}$) dx(A + B $x^2$))

= $\underset{a}{\overset{a+L}{\int }}Gm(\frac{a}{x^2}+B)dx$

= $Gm[\frac{A}{a}-\frac{A}{a+L}+BL]$