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

The force due to a small element of mass \( dm = \mu dx \) at a distance \( x \) from the origin is: \[ dF = \frac{G m \, dm}{x^2} = \frac{G m \, \mu dx}{x^2} \] Integrating this expression for the entire length of the rod gives the total gravitational force: \[ F = Gm \int_a^{a+L} \frac{\mu(x)}{x^2} dx \] Substitute \( \mu(x) = A + Bx^2 \) and solve the integral to get the final result.