Question

A straight rod of length L extends from x=a to x=L +a. The gravitational force is exerted on a point mass 'm' at x=0, if the mass per unit length of the rod is $A+B{x}^2$ is given by.

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

Answer 1

The correct option is B $Gm\left[A\right(\frac{1}{a}-\frac{1}{a+L})+BL$]

$dm=(A+B{x}^2)dx$
$df=\frac{GMdm}{x^2}$
$=F={\int }_a^{a+L}\frac{GM}{x^2}(A+B{x}^2)dx$
$=GM[-\frac{A}{x}+Bx{]}_a^{a+L}$
$Gm\left[A\right(\frac{1}{a}-\frac{1}{a+L})+BL$