How can I get an equation of trajectory of any particle moving in a circular motion? I mean I want to get an equation of a circle by using the conditions given for a circular motion. We know that if a particle enters in a force field $\vec F$ with velocity $\vec v$, where force $\vec F$ is perpendicular to $\vec v$, and if the force remains perpendicular to $\vec v$ always, then the trajectory of that particle becomes a circle. We can understand it physically. $\vec F$ is always perpendicular on $\vec v$ and hence on the displacement at every position of the particle. We know the force is always directed to a fixed point from which the force is created. At any point on the trajectory, the direction of displacement is given by the tangent on that point. so we can assume the force as the radius of a circle, as it is always perpendicular to the tangent on the trajectory, which is a property of a circle. So the trajectory will be a circle. But how can I get it mathematically? Can I get an equation of a circle for the trajectory of the particle by using the conditions(like $\vec F$ is always perpendicular to $\vec v$ and hence on displacement or something like that)?
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$\begingroup$ Keep in mind that having the force perpendicular to the velocity is not the only requirement for circular motion. $\endgroup$– BioPhysicistCommented Apr 4, 2019 at 14:48
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$\begingroup$ What are the other requirements? $\endgroup$– Srijan GhoshCommented Apr 4, 2019 at 14:53
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$\begingroup$ [Here I found my answer. It is the right one, which I was searching for ](physics.stackexchange.com/questions/385363/…) $\endgroup$– Srijan GhoshCommented Apr 4, 2019 at 15:00
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1$\begingroup$ I will let you think about that :) Hint: All we can say about forces perpendicular to the velocity is that the speed of the particle will not change. Think about what additional assumption we would need to ensure circular motion, i.e. motion with a constant distance from a fixed point $\endgroup$– BioPhysicistCommented Apr 4, 2019 at 15:02
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$\begingroup$ yes, I have found it just now. It's the right one which I was searching for. @AaronStevens $\endgroup$– Srijan GhoshCommented Apr 4, 2019 at 15:02
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