Question

A straight rod of length $L$ extends frfom $x=a$ to $x=L+a$. Find the gravitational force it exerts on a point mass $m$ at $x=0$ is(if the linear density of rod $\mu =A+B{x}^2$)

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

Answer 1

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

Let us consider the gravitational force between the point mass and a small element of rod located at a distance x from the mass

$dF=\frac{Gm\left(\mu dx\right)}{x^2}$

Total force between them will be the integral of above equation

$\int dF=Gm{\int }_a^{L+a}(A+B{x}^2)\frac{dx}{x^2}$

$F=Gm{[-A\frac{1}{x}+Bx]}_a^{L+A}$

$F=Gm[-A(\frac{1}{L+a}-\frac{1}{a})+B(L+a-a\left)\right]$

$F=Gm[A(\frac{1}{a}-\frac{1}{L+a})+BL]$