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.

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Solution

Step 1: Given$\text{LowerLimit}=\mathrm{a}$$\text{Upperlimit}=\mathrm{L}+\mathrm{a}$$\mathrm{dm}=\left(\mathrm{A}+{\mathrm{Bx}}^{2}\right)\mathrm{dx}$Step 2: To FindWe have to determine the gravitational force.Step 3: Calculate gravitational force:$F={\int }_{a}^{L+a}\frac{GMdm}{{x}^{2}}\phantom{\rule{0ex}{0ex}}={\int }_{a}^{L+a}\frac{GM\left(A+B{x}^{2}\right)dx}{{x}^{2}}\phantom{\rule{0ex}{0ex}}=GM{\int }_{a}^{L+a}\left(\frac{A}{{x}^{2}}+B\right)dx\phantom{\rule{0ex}{0ex}}=GM{\left(-\frac{A}{x}+Bx\right)}_{a}^{L+a}\phantom{\rule{0ex}{0ex}}=GM\left(\frac{1}{a}-\frac{1}{a+L}+BL\right)$

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