How Does Particle P Chase Particle Q on a Circular Path?

[Kavya Chopra]

That was not what I meant. By path, I meant the trajectory it adopts before it has attained the desired radius. After that, I had taken it for granted that the particles move in concentric circles ( in my book, there was a hint given which said that). So, you mean that it can either keep on moving in a circle, or spiral back into its centre after attaining the radius?

 

[haruspex]

I meant the trajectory it adopts before it has attained the desired radius

OK, but the plot you posted had it returning to the origin.

I had taken it for granted that the particles move in concentric circles

Without the equations, you could not be sure whether it would attain a perfectly circular path or merely spiral ever closer to one, never quite getting there. I think many would have guessed the latter (as I did), and you yourself kept mentioning a spiral.

So, you mean that it can either keep on moving in a circle, or spiral back into its centre after attaining the radius?

Both behaviours satisfy your differential equation, but only switching to the circle centred on the origin satisfies the given conditions. In an earlier post, your were recommended to rust the equations, but you can only put full trust in the equations if they encapsulate all the given information. The differential equation does not achieve that.

 

[Kavya Chopra]

The spiral I had mentioned was referring to a logarithmic spiral initially, before it attains a circular shape. More appropriately, it was a logarithmic spiral which terminated into a circle, because the radial component will keep on decreasing, and eventually it will become 0 otherwise in the case of v<u the particle would keep on moving away from Q. In the terminated circle path, at least it maintains a constant distance. The reverse spiral requires the component of the radial velocity to instantaneously become 0, and then increase again(magnitude wise). But as I said, in my book, it was mentioned that after attaining the radius, it will follow a circular path. Even though the spiral satisfies the equation What I was interested in knowing was what path it will take as it approaches the radius the first time from the centre i.e. before attaining the radius, and that path was the spiral I was referring to.

 

[Kavya Chopra]

And is there another equation that can give me the desired information apart from this differential equation, or do I have to depend only on the problem statement for the same?

 

[haruspex]

in the case of v<u the particle would keep on moving away from Q

No, it can't do that because of the "chase" condition. It can never move further from its target.

it was a logarithmic spiral

But it isn't.
I may have confused you by referring to the trajectory "you" plotted. Looking back through the posts I see it was PumpkinCougar that plotted the circular path. The circular arc of radius u/(2ω) plotted there is correct, but only until it reaches distance u/ω from the origin. Maybe you could describe this as a spiral, though certainly not a logarithmic one, but it would be more accurate to describe it as a semicircular arc.

is there another equation that can give me the desired information

Not really. You just have to be aware that the differential equation
$\dot r^2=u^2-(r\omega)^2$
has more than one solution, and make sure that you always pick the one which conforms to the "chase" condition.
In particular, watch out for $\dot r=-\sqrt{u^2-(r\omega)^2}$.

 

[Kavya Chopra]

Yes...that was what I wanted to know, because I had thought it would be a spiral but I didn't know the shape of it.Then I came across the logarithmic spiral, and it was the nearest shape that had come close to how I visualized the path. It was an intuitive guess, even though the parameters were incorrect.
Thanks a lot. It makes sense now.