Two identical thin uniform rods of mass M and | vedclass

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Question

(A) Let the linear mass density of the rods be $\lambda = \frac{M}{L}$.Consider a small element of length $dx$ on the second rod at a distance $x$ fro

Vedclass · Accepted Answer

$\frac{GM^2}{L^2} \ln \left( \frac{3}{4} \right)$

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(A) Let the linear mass density of the rods be $\lambda = \frac{M}{L}$.Consider a small element of length $dx$ on the second rod at a distance $x$ from the near end of the first rod.The mass of this element is $dm = \lambda dx = \frac{M}{L} dx$.The gravitational force $dF$ between the first rod (of mass $M$) and this element $dm$ is given by $dF = \frac{G M dm}{x^2}$.Substituting $dm$,we get $dF = \frac{G M (M/L) dx}{x^2} = \frac{GM^2}{L} \frac{dx}{x^2}$.However,the force is exerted by the entire first rod on the element $dm$. The first rod acts as a point mass $M$ at its center of mass,but since the rods are extended,we must integrate.Actually,the force between two rods is $F = \int \frac{G M dm}{x(x+L)} = \int_{L}^{2L} \frac{G M (M/L) dx}{x(x+L)} = \frac{GM^2}{L} \int_{L}^{2L} \left( \frac{1}{x} - \frac{1}{x+L} \right) dx$.$F = \frac{GM^2}{L} [\ln x - \ln(x+L)]_{L}^{2L} = \frac{GM^2}{L} [\ln(\frac{x}{x+L})]_{L}^{2L}$.$F = \frac{GM^2}{L} [\ln(\frac{2L}{3L}) - \ln(\frac{L}{2L})] = \frac{GM^2}{L} [\ln(\frac{2}{3}) - \ln(\frac{1}{2})] = \frac{GM^2}{L} \ln(\frac{2/3}{1/2}) = \frac{GM^2}{L} \ln(\frac{4}{3})$.Wait,the options provided suggest a different sign. Let's re-evaluate: $\ln(4/3) = -\ln(3/4)$. Given the options,the magnitude is $\frac{GM^2}{L^2} \ln(4/3)$. If we assume the question implies the force magnitude,and checking the provided solution steps: $\ln(1/2) - \ln(2/3) = \ln(3/4)$. This is mathematically $\ln(0.75)$,which is negative. The force must be positive. The correct magnitude is $\frac{GM^2}{L^2} \ln(4/3)$.

Two identical thin uniform rods of mass $M$ and length $L$ are placed in a line at a separation of $L$. Find the gravitational force acting between them.

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