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Question: A small mass and a thin uniform rod each of mass \[m\] are positioned along the same straight line a...

A small mass and a thin uniform rod each of mass mm are positioned along the same straight line as shown. Find the force of gravitational attraction exerted by the rod on the small mass.

Explanation

Solution

First of all, we will find the expression for the force as given by Newton. The gravitational force is directly proportional to the product of the masses and inversely proportional to the square of the distance between them. We will integrate to get the total force taking limits of length.

Complete step by step answer:
In the given question, we are supplied the following information:

There is a small mass and a thin uniform rod each of mass mm are positioned along the same straight line. The length of the rod is 2L2L and the separation distance between the small mass and the rod is LL .We are asked to find the force of gravitational attraction exerted by the rod on the small mass.
To begin with, we will apply the concept here that the mass per unit length of the rod always remains constant.Let us assume that the mass of the segment of the rod is dmdm whose length is dxdx .Now, we will apply the expression for the force to calculate the gravitational force.But we do need to remember that we have to apply integration while calculating the force.The upper limit will be 3L3L while the lower limit is LL .

Now, the force of gravitational attraction on the small mass exerted by the rod is given below:
F=−Gm∫L3Ldmx2 ⇒F=−Gm∫L3Lm2L×dxx2 ⇒F=−Gm×m2L∫L3Ldxx2 ⇒F=−Gm×m2L×[−1x]L3L ⇒F=Gm×m2L×[13L−1L] ∴F=Gm23L2F = - Gm\int_{\text{L}}^{{\text{3L}}} {\dfrac{{dm}}{{{x^2}}}} \\\ \Rightarrow F = - Gm\int_{\text{L}}^{{\text{3L}}} {\dfrac{m}{{2L}} \times \dfrac{{dx}}{{{x^2}}}} \\\ \Rightarrow F = - Gm \times \dfrac{m}{{2L}}\int_{\text{L}}^{{\text{3L}}} {\dfrac{{dx}}{{{x^2}}}} \\\ \Rightarrow F = - Gm \times \dfrac{m}{{2L}} \times \left[ { - \dfrac{1}{x}} \right]_{\text{L}}^{{\text{3L}}} \\\ \Rightarrow F = Gm \times \dfrac{m}{{2L}} \times \left[ {\dfrac{1}{{3L}} - \dfrac{1}{L}} \right] \\\ \therefore F = \dfrac{{G{m^2}}}{{3{L^2}}}

Hence, the gravitational force of attraction is Gm23L2\dfrac{{G{m^2}}}{{3{L^2}}} .

Note: While solving the problem, most of the students seem to have confusion regarding the limits of the integral. The mass responsible for providing the attraction force, on the small mass, is distributed throughout the length of the rod.