Physics Help Forum Motion of point mass under two central forces -- source of one stationary one moving.

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 Sep 27th 2008, 03:52 AM #1 Junior Member   Join Date: Sep 2008 Location: Bangalore, India Posts: 18 Motion of point mass under two central forces -- source of one stationary one moving. PROBLEM: A point mass ‘A’ is kept at the origin. Another B is kept at the x axis at x = H. Another C is kept at distance H from origin and distance h from B. A, B and C thus form an isosceles triangle with vertex A. Given that A's mass >> B's mass >> C's mass; Newton’s gravitational law governs the bodies. C will have motion because of both A and B while B will have motion because of A only. Find: The equation of path of C in any coordinate system given that B and C have zero initial velocities. Attempted solution: I write, $\displaystyle a_B=\frac{k_a}{r_b^2}$ and solve this by writing, $\displaystyle v\,dv=a\,dr_b$. We get an equation in $\displaystyle v$ and $\displaystyle r_b$. In this equation, we write: $\displaystyle v=dr_b/dt$ and solve for $\displaystyle r_b$ and get it in terms of $\displaystyle t$. We finally arrive at: $\displaystyle r_b=\left(H^{3/2}-k_at\right)^{2/3}$ Now, follow the link: A seeker of silences am I...: 27-Sep-2007 You will see two images posted. Click on them to enlarge them. I have carried fwd the solution there. Method applied to reach till the equations 3 and 4 (given below): I used polar coordinate systems with the point A as origin. I found out radial and angular components of the accleration of C and compard them with the standard expressions with the same as given here: Polar coordinate system - Wikipedia, the free encyclopedia The following equations we finally arrive at: $\displaystyle r_{b/c}=\sqrt{r_b^2\,+\,r_c^2\,-2bc\cos \theta}\quad(1)$ $\displaystyle \frac{r_{b/c}}{\sin \theta}\,=\,\frac{r_b}{\sin \phi}\quad(2)$ These two equations define the angle $\displaystyle \phi$ and $\displaystyle r_b/c$ for us. Proceeding... $\displaystyle -\frac{k_b}{r_{b/c}^2}\,\left(1-\,\frac{r_b^2}{\sin^2 \theta r_{b/c}^2}\right)\,=\,\ddot{r_c}\,-\,r_c\dot{\theta^2}\quad(3)$ $\displaystyle -\left(\frac{k_b}{r_{b/c}^2}\frac{r_b}{\sin \theta r_{b/c}}\,+\,\frac{k_a}{r_c^2}\right) \,= \,r_c\ddot{\theta}\,+\,2\dot{r_c}\dot{\theta}\quad (4)$ $\displaystyle r_{b/c}$ is defined by first two equations. Eqations 3 and 4 then need be solved for $\displaystyle r_c$ and $\displaystyle \theta$. Plz help. Thanks. Last edited by taureau20; Sep 27th 2008 at 04:16 AM. Reason: Error in latex coding

 Tags central, dfq, forces, kinematics, mass, motion, moving, point, source, stationary

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