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Jan 25, 2021 at 13:54 comment added Cosmas Zachos For free motion, all projections of the acceleration vanish, as you illustrated. Liboff's "nvolves accelerations in the r and 𝜃 components of motion" is misleading. You now have the right idea here. You might repeat your exercise, with the same type result.
Jan 25, 2021 at 2:16 history edited Qmechanic CC BY-SA 4.0
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Jan 25, 2021 at 1:36 vote accept Lost
Jan 25, 2021 at 1:12 answer added Cosmas Zachos timeline score: 4
Jan 25, 2021 at 0:49 history edited Lost CC BY-SA 4.0
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Jan 25, 2021 at 0:47 comment added Lost @CosmasZachos Sorry. I do not understand. Where did I blow up the sign? $a_r$ comes zero. That I have calculated and it is correct.
Jan 25, 2021 at 0:43 comment added Lost @Rumplestillskin Please also check the edit I made. Also, do you mean that having a non-zero $\dot p_r$ doesn't imply a $a_r=0$ since $\dot p_r=m\ddot r$ while $a_r$ has a term of $r \dot \theta^2$ as well? I think this was the mistake I was making. But then how does Liboff conclude from non -zero expressions of $p_\theta$ that there is an azimuthal acceleration?
Jan 25, 2021 at 0:34 history edited Lost CC BY-SA 4.0
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Jan 25, 2021 at 0:19 comment added Lost @Rumplestillskin Since, $\dot p$ is non-zero, then radial force and acceleration is non-zero. This is what I reasoned. What is wrong with it?
Jan 24, 2021 at 23:53 comment added Cosmas Zachos Did you blow a sign? $ma_r= \dot{p}_r-\dot{p}_\theta^2/mr^3=0$ by your EOM.
Jan 24, 2021 at 22:25 comment added Rumplestillskin Can you explain why that implies $a_r \neq 0$? It might help to see things a bit clearer if you integrate your second equation $a_\theta$ wrt time and substitute $\dot{\theta}$ into your radial equation.
Jan 24, 2021 at 19:44 history asked Lost CC BY-SA 4.0