Rod and Disk Gravity

Classical Mechanics Level pending

A disk has $1$ unit of mass per unit area, and is parametrized as follows:

$x_1 = r \, \cos \theta \\ y_1 = r \, \sin \theta \\ z_1 = 0 \\ 0 \leq r \leq 1 \\ 0 \leq \theta \leq 2 \pi$

A rod has $1$ unit of mass per unit length, and is parametrized as follows:

$-1 \leq x_2 \leq +1 \\ y_2 = 0 \\ z_2 = 1$

What is the magnitude of the gravitational force between them?

Note: Universal gravitational constant $G = 1$ , for simplicity

1 solution

Karan Chatrath
Jun 21, 2019

Consider two points on each body. The position vector of a point on the disk is defined as: $\vec{r}_1 = (r\cos(\theta),r\sin(\theta),0)$

A point on the rod is: $\vec{r}_2 = (x,0,1)$

A vector connecting these two points can be defined as: $\vec{r}_d =\vec{r}_1-\vec{r}_2$

Now, the mass element of the rod is $dM_1=dx$ while that on the disk is defined as $dM_2=rdrd\theta$ .

Using this information, we define the force acting between these two mass elements as:

$d\vec{F} = \frac{G(dM_1)(dM_2)}{\mid\vec{r_d}\mid^3}\vec{r}_d$

By Symmetry, the X and Y components of the resultant force will be zero. The Z Component of the force is computed by solving:

$F_z =\int_{-1}^{1}\int_{0}^{2\pi}\int_{0}^{1}-\frac{r}{{\left(r^2-2rx\cos\left(\mathrm{\theta}\right)+x^2+1\right)}^{3/2}}dr d\theta dx$

The required answer is: $\mid F_z \mid = 3.1669$