Question

shows a particle of mass m at distance $x$from the center of a very thin cylinder of mass $M$and length $L$. The particle is outside the cylinder, so $x>L/2$.

a. Calculate the gravitational potential energy of these two masses.

b. Use what you know about the relationship between force and potential energy to find the magnitude of the gravitational force on $m$when it is at position $x.$

Short answer

a. The gravitational potential energy of these two masses $-\frac{GMm}{L}\ln \left(\frac{x+\frac{L}{2}}{x-\frac{L}{2}}\right)$

b. The gravitational force on $m$is$-GMm\frac{4}{4x^2-L^2}$

Step-by-step solution

Given Expression

$U=-\frac{GMm}{r}$

The expression for gravitational P.E. is,

Here, $G$ is universal universal gravitational constant, role="math" localid="1651421308024" $Mm$are masses, and $r$ is distance of separation between the masses.

potential energy (a)

(a) Assume that the rod is split into thin sections each of width $dr$with mass of $dm$ as shown within the figure. because the width of the rod is tiny, all of the mass $dm$is assumed to be at a distance $r$ faraway from the mass $m$.

Gravitation (a)

The fractional of massof every section and total mass is.
$\begin{array}{r}\frac{dm}{M}=\frac{dr}{L}\\ \end{array}$

$dm=\frac{M}{L}dr$

The expression for gravitational P.E. of the masses $dm$and$m$ is

$dU=-\frac{Gmdm}{r}$
Substitute $dm=\frac{M}{L}dr$ within the above equation.
$\begin{array}{r}dU=-\frac{Gm\left(\frac{M}{L}dr\right)}{r}\\ \end{array}$

$\begin{array}{r}\\ =-\frac{GMm}{L}\frac{dr}{r}\end{array}$

Calculate the overall P.E. by integrating the above equation.
$\begin{array}{r}U=\int dU\end{array}$

$\begin{array}{r}=-\frac{GMm}{L}{\int }_{x-\frac{L}{2}}^{x+\frac{L}{2}} \frac{dr}{r}\\ \end{array}$

$=-\frac{GMm}{L}\ln \left(\frac{x+\frac{L}{2}}{x-\frac{L}{2}}\right)$

Therefore, the gravitational mechanical energy of the masses is$-\frac{GMm}{L}\ln \left(\frac{x+\frac{L}{2}}{x-\frac{L}{2}}\right)$

Gravitational Force (b)

(b) Calculate the force on the mass $m$at distance $x$.

$\begin{array}{r}F=-\frac{dU}{dx}\end{array}$

$\begin{array}{r}=\frac{GMm}{L}\frac{d}{dx}\left[\ln \left(x+\frac{L}{2}\right)-\ln \left(x-\frac{L}{2}\right)\right]\end{array}$

$\begin{array}{r}=\frac{GMm}{L}\left(\frac{1}{x+\frac{L}{2}}-\frac{1}{x-\frac{L}{2}}\right)\\ \end{array}$

$=-GMm\frac{4}{4x^2-L^2}\left(x\ge \frac{L}{2}\right)$

Therefore, the force on the mass $m$is $-GMm\frac{4}{4x^2-L^2}$