A particle is moving in a circle

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A particle is moving in a circle of radius R in the xy plane. During the motion, neither the x- nor y-component of the particle's velocity exceeds v. Find the minimum possible time for the particle to complete one circle. (I.e., find the minimum possible period of revolution.)

Hint 1: I don't think it can be solved without Calculus.
Hint 2: Answer is in the form of an equation without any rounding of any numbers involved.


Considering I haven't taken Calculus yet I don't know the right way to approach this problem.
 
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Since the problem says "minimum time" you can get at least an estimate without Calculus. The particle has maximum speed [itex]\sqrt{2} v[/itex] so cannot go a distance [itex]2\pi R[/itex] is less than
[tex]\dfrac{2\pi R}{\sqrt{2} v}= \sqrt{2}\pi \dfrac{R}{v}[/tex].
 
I think it means that the max possible x-component of velocity is v. But it is worded a bit vaguely.

edit: and a similar statement for y, separately.

edit again: never mind, I am being stupid. I thought there was a problem, but there is not.

edit number 3: yeah, my mistake was thinking the question gave the max values of components of velocity. But it does not. It gives upper bounds.
 
Last edited:
HallsofIvy said:
The particle has maximum speed [itex]\sqrt{2} v[/itex]
Could you please explain your reasoning here?
 
Well how would I solve the problem with Calculus?
 
Not sure but if you consider the uniform circular motion, the conditions mentioned in the problem are satisfied.

Is it possible for you to post the answer?
 
well calculus doesn't really help to get a better answer in this case. If they told you the velocity as a function of time, then you could use calculus. But they don't tell you that.
 
Well My professor stated as a hint he doesn't think it could be solved without Calculus. Could he be wrong?
 
the way you have described the problem, I don't think you need calculus. But you should make up your own mind. (Although I guess that might be tricky if you haven't been taught what calculus is).