Question

A mass $m$ is at a distance a from one end of a uniform rod of length $l$ and mass $M$. The gravitational force on the mass due to the rod is

• A. $\frac{GMm}{(a+l{)}^2}$
• B. $\frac{GmM}{a(l+a)}$
• C. $\frac{GMm}{a^2}$
• D. $\frac{GmM}{2(l+a{)}^2}$

Answer 1

The correct option is B $\frac{GmM}{a(l+a)}$

Assume a element of mass $dm$ at a distance $X$ from mass $m$

Force due to small element of rod $dF=\frac{Gdm\times m}{X^2}$

$F={\int }_a^{a+l}\frac{Gdm\times m}{X^2}$

$dm=\frac{M}{l}dx$

$F={\int }_a^{a+l}\frac{GM\times m}{l{X}^2}dx$

$F=\frac{GM\times m}{l}[\frac{1}{X}{]}_a^{a+l}$

$F=\frac{GM\times m}{l}[\frac{1}{a+l}-\frac{1}{a}{]}_a^{a+l}$

$F=\frac{GM\times m}{a(l+a)}$