Question

A straight rod of length $\ell$ extends from x = $\alpha$ to x = $\ell$ + $\alpha$ as shown in the figure. If the mass per unit length is (a + bx²). The gravitational force it exerts on a point mass m placed at x = 0 is given by

• A. $Gm\left(a\left(\frac{1}{\alpha}-\frac{1}{\alpha+\ell}\right)+b\ell\right)$
• B. $\frac{Gm(a+bx^2)}{\ell^2}$
• C. $Gm\left(\alpha\left(\frac{1}{a}-\frac{1}{a+\ell}\right)+b\ell\right)$
• D. $Gm\left(a\left(\frac{1}{\alpha+\ell}-\frac{1}{\alpha}\right)+b\ell\right)$

Answer 1

Answer: (A)

A

Answer 2

The gravitational force exerted by a continuous mass distribution is found by integrating the differential force exerted by a small mass element.

1. Define a differential mass element:
   Consider a small element of the rod of length $dx$ at a distance $x$ from the origin (where the point mass $m$ is placed).
   The mass per unit length is given as $\lambda = (a + bx^2)$.
   The mass of this differential element $dm$ is $\lambda dx = (a + bx^2) dx$.

2. Calculate the differential gravitational force:
   The gravitational force $dF$ exerted by this element $dm$ on the point mass $m$ at $x=0$ is given by Newton's law of gravitation: $dF = \frac{G m \cdot dm}{x^2}$ Substitute the expression for $dm$: $dF = \frac{G m (a + bx^2) dx}{x^2}$

3. Integrate to find the total force:
   The rod extends from $x = \alpha$ to $x = \ell + \alpha$. To find the total gravitational force $F$, integrate $dF$ over these limits: $F = \int_{\alpha}^{\ell + \alpha} \frac{G m (a + bx^2)}{x^2} dx$ Since $G$ and $m$ are constants, they can be taken out of the integral: $F = Gm \int_{\alpha}^{\ell + \alpha} \left( \frac{a}{x^2} + \frac{bx^2}{x^2} \right) dx$ $F = Gm \int_{\alpha}^{\ell + \alpha} \left( a x^{-2} + b \right) dx$

4. Perform the integration:
   The integral of $a x^{-2}$ is $-a x^{-1} = -\frac{a}{x}$.
   The integral of $b$ is $bx$. $F = Gm \left[ -\frac{a}{x} + bx \right]_{\alpha}^{\ell + \alpha}$

5. Apply the limits of integration: $F = Gm \left[ \left( -\frac{a}{\ell + \alpha} + b(\ell + \alpha) \right) - \left( -\frac{a}{\alpha} + b\alpha \right) \right]$ $F = Gm \left[ -\frac{a}{\ell + \alpha} + b\ell + b\alpha + \frac{a}{\alpha} - b\alpha \right]$ $F = Gm \left[ \frac{a}{\alpha} - \frac{a}{\ell + \alpha} + b\ell \right]$ Factor out $a$ from the first two terms: $F = Gm \left[ a\left(\frac{1}{\alpha} - \frac{1}{\ell + \alpha}\right) + b\ell \right]$

Comparing this result with the given options, it matches option (A).