Question

A straight rod of length \( L \) extends from \( x = a \) to \( x = L + a \). Find the gravitational force it exerts on a point mass \( m \) at \( x = 0 \) if the mass per unit length of the rod is \( \mu = A + Bx^2 \):

• A. \( F = GmA \left[ \frac{1}{a} - \frac{1}{a+L} \right] + BL \)
• B. \( F = Gm \left[ A \left( \frac{1}{a} - \frac{1}{a+L} \right) + B L \right] \)
• C. \( F = Gm \left[ A \left( \frac{1}{a} - \frac{1}{a+L} \right) + B L \right] \)
• D. None

Hint:

To calculate the gravitational force, integrate the force contribution of each small mass element over the length of the rod.

Answer 1

The Correct Option is B

Step 1: An element of the rod at position \(x\) of width \(dx\) has mass \(dm = \mu(x)\,dx = (A+Bx^2)\,dx\), so it attracts the point mass \(m\) at the origin with force \(dF = \dfrac{Gm\,dm}{x^2} = \dfrac{Gm(A+Bx^2)}{x^2}dx\).

Step 2: Split the integrand: \( dF = Gm\left(\dfrac{A}{x^2}+B\right)dx \), then integrate each piece separately from \(x=a\) to \(x=a+L\).

Step 3: The first piece gives \( GmA\displaystyle\int_a^{a+L}\dfrac{dx}{x^2} = GmA\left(\dfrac{1}{a}-\dfrac{1}{a+L}\right)\); the second gives \( GmB\displaystyle\int_a^{a+L}dx = GmBL \).

Step 4: Add the two pieces together, keeping the common factor \(Gm\) outside:\[ \boxed{F = Gm\left[A\left(\frac{1}{a}-\frac{1}{a+L}\right)+BL\right]} \]

Answer 2

Instead of integrating the combined density in one go, split the rod into two overlapping fictitious rods occupying the same region \(x=a\) to \(x=a+L\): one with uniform density \(A\), and one with density \(Bx^2\). By superposition, the total force is just the sum of the forces each would exert on its own.

The uniform-density rod gives the standard result \( F_1 = GmA\left(\dfrac{1}{a}-\dfrac{1}{a+L}\right) \), the textbook result for a point mass gravitating toward a uniform rod.

The \(Bx^2\)-density rod contributes \( F_2 = Gm\displaystyle\int_a^{a+L}\dfrac{Bx^2}{x^2}dx = GmB\displaystyle\int_a^{a+L}dx = GmBL \), since the \(x^2\) in the density exactly cancels the \(x^2\) in the denominator of Newton's law, leaving a flat integrand.

Adding both contributions:\[ \boxed{F = GmA\left(\frac{1}{a}-\frac{1}{a+L}\right) + GmBL = Gm\left[A\left(\frac{1}{a}-\frac{1}{a+L}\right)+BL\right]} \]