Question

A Straight Rod Of Length L Extends From $\mathrm{X}=\mathrm{a}$ To $\mathrm{X}=\mathrm{L}+\mathrm{a}$. The Gravitational Force Is Exerted On A Point Mass 'm' At $\mathrm{X}=0$, If The Mass Per Unit Length Of The Rod Is $\mathrm{A}+{\mathrm{Bx}}^2$ is Given By.

Answer 1

Step 1: Given

$\text{Lower Limit}=\mathrm{a}$

$\text{Upper limit}=\mathrm{L}+\mathrm{a}$

$\mathrm{dm}=\left(\mathrm{A}+{\mathrm{Bx}}^2\right)\mathrm{dx}$

Step 2: To Find

We have to determine the gravitational force.

Step 3: Calculate gravitational force:

$$\begin{gathered} F={\int }_a^{L+a}\frac{GMdm}{x^2} \\ ={\int }_a^{L+a}\frac{GM\left(A+B{x}^2\right)dx}{x^2} \\ =GM{\int }_a^{L+a}\left(\frac{A}{x^2}+B\right)dx \\ =GM{\left(-\frac{A}{x}+Bx\right)}_a^{L+a} \\ =GM\left(\frac{1}{a}-\frac{1}{a+L}+BL\right) \end{gathered}$$