- #1
kaze
- 6
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A infinitely long rod, [tex]A[/tex], with linear mass density, say [tex]p[/tex], is placed along the z-axis. Another thin rod, [tex]B[/tex], of length [tex]L[/tex] and mass density, say [tex]p_{2}[/tex], is placed orthogonally to [tex]A[/tex] on the x-axis from [tex]x=a[/tex] to [tex]x=a+L[/tex]. What is the gravitational force experienced by rod [tex]B[/tex] from rod [tex]A[/tex]?
I have solved a similar problem which asks for the gravitational potential given a rod of finite length and a point a perpendicular distance away from the rod's midpoint (thornton and marion problem 5-7). I am trying to extend this but not having much luck. So if we were trying to find the gravitational potential of the end point on rod [tex]B[/tex] closest to [tex]A[/tex] we would get:
[tex] \phi =-G \int_{A} \frac{p}{r}ds =- p G \int_{A} \frac{1}{\sqrt{x^2+a^2}} dx = - p G \lim_{z\rightarrow \infty} \int_{-z}^{z} \frac{1}{\sqrt{x^2+a^2}} dx = - p G \lim_{z \rightarrow \infty} \ln(x+\sqrt{x^2+a^2})|_{-z}^{z}[/tex]
This already goes to infinity. I was thinking that I could make [tex]\phi[/tex] a function of [tex]a[/tex] and then once I find a function for the gravitational potential, I'll integrate over rod [tex]B[/tex] to get the entire potential and hence the force by [tex]F=-\nabla U = -\nabla (m U ) = -m \nabla U[/tex]
However, I guess I am stuck at step one since I cannot get the integral to work out. If you have any suggestions, I'd really appreciate the help. Thanks.
I have solved a similar problem which asks for the gravitational potential given a rod of finite length and a point a perpendicular distance away from the rod's midpoint (thornton and marion problem 5-7). I am trying to extend this but not having much luck. So if we were trying to find the gravitational potential of the end point on rod [tex]B[/tex] closest to [tex]A[/tex] we would get:
[tex] \phi =-G \int_{A} \frac{p}{r}ds =- p G \int_{A} \frac{1}{\sqrt{x^2+a^2}} dx = - p G \lim_{z\rightarrow \infty} \int_{-z}^{z} \frac{1}{\sqrt{x^2+a^2}} dx = - p G \lim_{z \rightarrow \infty} \ln(x+\sqrt{x^2+a^2})|_{-z}^{z}[/tex]
This already goes to infinity. I was thinking that I could make [tex]\phi[/tex] a function of [tex]a[/tex] and then once I find a function for the gravitational potential, I'll integrate over rod [tex]B[/tex] to get the entire potential and hence the force by [tex]F=-\nabla U = -\nabla (m U ) = -m \nabla U[/tex]
However, I guess I am stuck at step one since I cannot get the integral to work out. If you have any suggestions, I'd really appreciate the help. Thanks.