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

[Kavya Chopra]

Homework Statement

Starting from the center of a circular path of radius R, a particle P chases another particle Q that is moving with a uniform speed v on the circular path. The chaser P moves with a constant speed u and always remains collinear with the centre and location of the chased Q.
a)On what path will P eventually move and how long will it take to reach on that path? Also find out the time taken by P to reach Q
Consider the cases u>v, u<v and u=v

Homework Equations

$$v=\omega r$$ (v here referring to tangential velocity)
$$\frac{d\theta}{dt}=\omega$$

The Attempt at a Solution

I can gather that at every instant P moves along a circle of new radius, growing larger and larger. Due to the collinearity of P, Q and the centre at every instant, the angular velocities of both of them will be the same. And since the speed is constant, due to the equation above the tangential component of u will be increasing and thus the component directed towards the centre will keep on decreasing. So I assume that eventually it will be small enough to be negligible and the particle P will move in a circle.
So, at last we can equate the angular velocities to get the radius of P's path.

(Also, I believe that its path of chasing Q is akin to that of a logarithmic spiral. Please correct me and give the right visualization if I'm wrong)

I could only go about this far. I had the intuition that we can find the time by calculus and used
$$\frac{d\theta}{dt}=\omega$$
But for that to give the correct answer
$$\int_{0}^{\frac{\pi}{2}} d\theta=\frac{\pi}{2}$$
And I can't figure out why these 2 have to be the limits, or even if this method is correct. As for the third part, I have no clue.
I could only make one rather obvious deduction if v>u, P can't reach Q.
Please help me with this.

 

[haruspex]

Let's just deal with v>u first. You found, correctly, that it will never reach Q, and that the path eventually approximates a circle. What radius? It remains to say how long it will take to reach a circular path. Any thoughts on that?

Next, consider v=u. Does anything change?

v<u is more problematic.

 

[PumpkinCougar95]

1. Don't think about the cases, the math would handle that for you
2. Draw a diagram and try to relate the total velocity $u$ with the radial and angular velocities
3. Make differential Equation. you can equate the radial component with $\frac {dr} {dt}$

Give the result you got.

 

[Orodruin]

I can gather that at every instant P moves along a circle of new radius, growing larger and larger.

Just to point out the obvious. If P moves along a circle, the distance to the origin cannot grow.

 

[haruspex]

Don't think about the cases, the math would handle that for

That is not so for v<u. We are told P always moves so as to get closer to Q and starts at a smaller radius. Perhaps that is not what the question intended, since it does create a bit of a problem.

 

[Kavya Chopra]

Just to point out the obvious. If P moves along a circle, the distance to the origin cannot grow.

What I mean to say is P will move along a new circle with different components of angular and radial velocity having resultant u every instant. At that instant when it's radial velocity becomes 0, it will move in a circle

 

[Kavya Chopra]

1. Don't think about the cases, the math would handle that for you
2. Draw a diagram and try to relate the total velocity $u$ with the radial and angular velocities
3. Make differential Equation. you can equate the radial component with $\frac {dr} {dt}$

Give the result you got.

As I said, by equating the angular velocities, I got the radius $$R_u=\frac{u}{v}R_v$$

And for equating the radial component with $\frac {dr} {dt}$, do we have r here as $2\pi R$ or simply as $R$
For your second observation $u^2=\omega^2+v_r^2$
Do I use the above equation in the differential Equation of $$\frac{dr}{dt}=v_r$$(radial velocity)

 

[tnich]

And for equating the radial component with $\frac {dr} {dt}$, do we have r here as $2\pi R$ or simply as $R$

I am not sure what you mean. If $\frac {dr} {dt}$ is the radial component of velocity of the chaser, what must r represent?

For your second observation $u^2=\omega^2+v_r^2$

This equation is dimensionally inconsistent. You are adding an angular velocity squared to a velocity squared. Once you fix this and substitute for $v_r$, you will have your differential equation.

 

[scottdave]

For your second observation $u^2=\omega^2+v_r^2$

The dimensions do not match on this. ω is Angle/Time , and u and v_r are Length/Time. This tells me that something should be multiplied by ω to make it (at least dimensionally) correct.

 

[Kavya Chopra]

Sorry, I made an error.
$$u^2=(\omega r)^2+v_r^2$$
So now I got a differential equation of the form
$$\frac{dr}{dt}= (u^2-(\omega r)^2)^(\frac{1}{2})$$
And I'll solve it, with limits of R from 0 to $R_u$ with $w=\frac{u}{R_u}$
Thanks a lot.
Also, for the problem of when it catches Q, maybe I can set the limits from 0 to $R_v$, and so I can get that time by the same equation. And as @PumpkinCougar95 said, the math will take care of it as the answer has a term of $(u^2-v^2)^\frac{1}{2}$, which can clearly never be negative. So, in the case of $v>u$, the time is not defined.
Thanks a lot. It's much clearer now.

 

[Kavya Chopra]

As I mentioned earlier, please confirm whether the path will be a logarithmic spiral or any other shape.

 

[PumpkinCougar95]

It is Pretty strange, i am getting a simple harmonic motion type solution.(In a frame rotating with w)

 

[Orodruin]

It is Pretty strange, i am getting a simple harmonic motion type solution.(In a frame rotating with w)

Why do you think this is strange?

 

[PumpkinCougar95]

Because by definition, P is chasing Q so why would it oscillate and move farther away?

Edit: I was plotting $r = sin( \theta )$ and I was getting a circle, but wIen i plotted $\theta = arcsin(r)$ i am getting a spiral which makes much more sense.

 

[tnich]

Because by definition, P is chasing Q so why would it oscillate and move farther away?

Edit: I was plotting $r = sin( \theta )$ and I was getting a circle, but when i plotted $\theta = arcsin(r)$ i am getting a spiral which makes much more sense.

That's odd. $\theta = arcsin(r)$ is not a spiral. Are you plotting it in polar coordinates?

 

[PumpkinCougar95]

Yes, in polar. https://www.wolframalpha.com/input/?i=plot+arcsin(r)=(theta)

Edit: Btw can someone explain why $\theta = arcsin(r)$ is a spiral while $sin(\theta) =r$ is not ?

 

[tnich]

Yes, in polar. https://www.wolframalpha.com/input/?i=plot+arcsin(r)=(theta)

Edit: Btw can someone explain why $\theta = arcsin(r)$ is a spiral while $sin(\theta) =r$ is not ?

Your spiral looks like a plot of $ρ=θ$, not $\theta = arcsin(r)$.

 

[tnich]

Your spiral looks like a plot of $ρ=θ$, not $\theta = arcsin(r)$.

I think maybe wolfram alpha is interpreting the left-hand side of $arcsin(r) = \theta$ as the dependent variable and not applying the inverse arcsin function (sin) to get r.

 

[PumpkinCougar95]

Okay. But intuitively I still think it should have been a spiral.
And this doesn't look anywhere near that.

 

[haruspex]

It is Pretty strange, i am getting a simple harmonic motion type solution.(In a frame rotating with w)

As I indicated in post #5, you need to be careful with the given condition that P "chases" Q (though there I was thinking it was only an issue for v<u). The solution obtained by blindly following the equations might not satisfy that. Your equation allows the possibility that P moves further from Q at times.

 

[Kavya Chopra]

How did you formulate the equation for the path?

 

[haruspex]

How did you formulate the equation for the path?

Your differential equation is correct, but it (and the integral that results) has another solution.
In general, given $y=\int \frac{dx}{f(x)}$, and f(a)=0, then (if allowed by other constraints on x) a solution is x=a (constant). Since that makes dx and f(x) both zero, the integrand becomes indeterminate.
Which solution applies at any point in this question is governed by the "chases" statement.

 

[Kavya Chopra]

Your differential equation is correct, but it (and the integral that results) has another solution.
In general, given $y=\int \frac{dx}{f(x)}$, and f(a)=0, then (if allowed by other constraints on x) a solution is x=a (constant). Since that makes dx and f(x) both zero, the integrand becomes indeterminate.
Which solution applies at any point in this question is governed by the "chases" statement.

I'm sorry, I didn't get what you are saying.

 

[haruspex]

I'm sorry, I didn't get what you are saying.

Go back to your differential equation. What does it tell you in the v>=u case if you plug in dr/dt=0?

 

[Kavya Chopra]

The radial component of velocity is 0, or $$u=\omega r$$

 

[haruspex]

The radial component of velocity is 0, or $$u=\omega r$$

Right. So r=u/ω, constant, is a solution to the differential equation.
Once the particle reaches radius u/ω, your equation cannot tell you what happens next. The radial velocity might be transiently zero (leading to SHM, i.e. the circular trajectory you found) or it might remain zero. Which option matches the problem statement?

 

[Kavya Chopra]

I believe it is the first one matching the trajectory of the circle( why is it SHM, though?), because eventually the particles probably move in concentric circles.

 

[haruspex]

I believe it is the first one matching the trajectory of the circle

No, that would mean it is then moving away from the particle it is supposed to be chasing.

eventually the particles probably move in concentric circles

Quite so, but the circular trajectory you plotted is not concentric with the other particle's.

 

[Kavya Chopra]

So, it does follow the path of a spiral. What I wanted to know is how can you prove it mathematically.
But I want to know what happens even if it is only transiently zero, and the mathematical aspect of it too.

 

[haruspex]

So, it does follow the path of a spiral

No, I didn't say that either.

You plotted a circular trajectory passing through the origin and having diameter u/ω. That is correct initially. It will take the particle from its starting point at the origin out to radius u/ω. But once it reaches that position your differential equation suddenly produces a second valid solution. It can satisfy that equation either by continuing around the circle back to the origin (considering r as function of time this is SHM) or by sticking at r=u/ω.
You are told that the particle is forever "chasing" (i.e. trying to get closer to) the other particle. If it continues on the SHM path back to the origin it will be moving away from its target. Only by switching to this new solution are the conditions satisfied.