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.

Answer 3

The rod stretches from \(x=a\) to \(x=a+L\) and has a mass distribution \(\mu(x) = A + Bx^2\) per unit length. Rather than integrating from scratch, a good way to check the options is to test them against the simple limiting case \(B=0\) (a uniform rod of linear density \(A\)), where the force on the point mass is well known:

\[ F_{\text{uniform}} = GmA\int_a^{a+L}\frac{dx}{x^2} = GmA\left(\frac{1}{a}-\frac{1}{a+L}\right) \]

Any candidate formula must collapse to this when \(B=0\).

1. Option 1 \( \left(F = GmA\left[\frac{1}{a}-\frac{1}{a+L}\right]+BL\right)\): setting \(B=0\) leaves \(GmA\left(\frac{1}{a}-\frac{1}{a+L}\right)\), which looks fine at first glance, but the \(BL\) term here carries no factor of \(Gm\) at all, so it can't be added to a gravitational force term that does. This is inconsistent.
2. Option 2 \(\left(F = Gm\left[A\left(\frac{1}{a}-\frac{1}{a+L}\right)+BL\right]\right)\): setting \(B=0\) correctly reduces to \(GmA\left(\frac{1}{a}-\frac{1}{a+L}\right)\), matching the known uniform-rod result, and both terms inside the bracket are properly scaled by \(Gm\). This is consistent.
3. Option 3: written identically to Option 2, so it checks out the same way, but only one listed option can be the intended selection.
4. Option 4 (None): ruled out since Option 2 (and its duplicate) checks out completely.

To confirm fully, carry out the actual integration: the force from an element \(dx\) at position \(x\) is \(dF = \dfrac{Gm(A+Bx^2)}{x^2}dx\). Splitting this into two integrals,

\[ F = Gm\int_a^{a+L}\frac{A}{x^2}dx + Gm\int_a^{a+L}B\,dx = Gm\left[A\left(\frac{1}{a}-\frac{1}{a+L}\right)+BL\right] \]

which is exactly Option 2.

So the correct answer is \(F = Gm\left[A\left(\dfrac{1}{a}-\dfrac{1}{a+L}\right)+BL\right]\).