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

To find the gravitational force exerted by the rod on the point mass \( m \) at \( x = 0 \), we consider an infinitesimal element of the rod of length \( dx \) at position \( x \) from \( x = a \) to \( x = L + a \). The mass of this element is \( dm = \mu \, dx = (A + Bx^2) \, dx \). The gravitational force \( dF \) between this element and the mass \( m \) is given by \( dF = \frac{Gm \, dm}{x^2} = \frac{G m (A + Bx^2) \, dx}{x^2} \). Integrating \( dF \) from \( x = a \) to \( x = L + a \) gives the total force \( F \):

\( F = \int_{a}^{L+a} \frac{G m (A + Bx^2)}{x^2} \, dx \) 

This integral can be split into two parts:

\( F = Gm \int_{a}^{L+a} \frac{A}{x^2} \, dx + Gm \int_{a}^{L+a} B \, dx \)

The first integral evaluates to:

\( GmA \left[ -\frac{1}{x} \right]_a^{L+a} = GmA \left( -\frac{1}{L+a} + \frac{1}{a} \right) = GmA \left( \frac{1}{a} - \frac{1}{L+a} \right) \)

The second integral evaluates to:

\( GmB \left[ x \right]_a^{L+a} = GmB (L+a-a) = GmBL \)

Combining both results:

\( F = Gm \left[ A \left( \frac{1}{a} - \frac{1}{L+a} \right) + B L \right] \)

Thus, the gravitational force exerted by the rod on the mass \( m \) is:
\( F = Gm \left[ A \left( \frac{1}{a} - \frac{1}{a+L} \right) + B L \right] \)

Answer 2

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.