© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the Physical Situation
A rod of length $L$ is placed along the x-axis from $x=a$ to $x=a+L$. A point mass $m$ is located at $x=0$. We want to find the gravitational force exerted by this rod on the point mass.
Step 2: Express the Linear Mass Density
We are given that the mass per unit length (linear mass density) of the rod is
$ \lambda(x) = A + B x^2 $.
Each infinitesimal segment $dx$ of the rod has a mass
$ dm = \lambda(x)\,dx = (A + Bx^2)\,dx.
$
Step 3: Write the Expression for the Differential Gravitational Force
The gravitational force $dF$ by a small element $dm$ of the rod on the mass $m$ (both located on the x-axis) is given by Newton’s law of gravitation:
$ dF = \dfrac{G\,m\,dm}{r^2}, $
where $r$ is the distance between the mass element at position $x$ and the point mass at $x=0$. Here, $r = |x - 0| = x$, assuming $x>0$ throughout $[a, a+L]$.
Step 4: Substitute $dm$ and $r$ into the Differential Force
Substituting $dm = (A + Bx^2)\,dx$ and $r = x$, we get:
$ dF = \dfrac{G\,m\,(A + Bx^2)\,dx}{x^2}
= G\,m \left(\dfrac{A}{x^2} + \dfrac{Bx^2}{x^2}\right) dx
= G\,m \left(\dfrac{A}{x^2} + B\right) dx.
$
Step 5: Integrate over the Length of the Rod
To find the total force $F$, integrate $dF$ as $x$ goes from $a$ to $a+L$:
$ F = \int_{x=a}^{x=a+L} G\,m \left(\dfrac{A}{x^2} + B\right) dx
= Gm \int_{a}^{a+L} \left(\dfrac{A}{x^2} + B\right)\,dx.
$
Separate the integral:
$ F = Gm \left[\int_{a}^{a+L} \dfrac{A}{x^2}\,dx + \int_{a}^{a+L} B\,dx\right].
$
1. Integrate $ \dfrac{A}{x^2}$:
$ \int \dfrac{A}{x^2} dx = -\dfrac{A}{x}. $
Evaluating from $a$ to $a+L$ gives
$ \left[-\dfrac{A}{x}\right]_{a}^{a+L} = -\dfrac{A}{a+L} + \dfrac{A}{a}.
$
2. Integrate $ B $:
$ \int B\,dx = Bx. $
Evaluating from $a$ to $a+L$ gives
$ [Bx]_{a}^{a+L} = B(a+L) - B(a) = BL.
$
Step 6: Combine the Terms
Adding these results together:
$ F = Gm \left[\left(-\dfrac{A}{a+L} + \dfrac{A}{a}\right) + BL\right]
= Gm \left[ A\left(\dfrac{1}{a} - \dfrac{1}{a+L}\right) + BL\right].
$
This matches the given correct answer:
$ Gm\Bigl[\,A\Bigl(\dfrac{1}{a} - \dfrac{1}{a+L}\Bigr) + BL\Bigr].
$
Diagram Reference
The diagram below gives a visual representation of the rod along the x-axis and the point mass at the origin:
