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Physics olympiad problem -- struggling with polar coordinates

  1. Apr 4, 2016 #1
    1. The problem statement, all variables and given/known data

    This is a physics olympiad problem; and I am still struggling with it. I will quote it here:

    " A particle moves along a horizontal track following the trajectory $$r=r_{0}e^{-k\theta}$$, where $$\theta$$ is the angle made by the position vector with the horizontal. Recall that the velocity in polar coordinates is $$\frac{d\vec r}{dt}=\dot r \hat r+r \dot \theta \hat \theta$$. If at $$t=0, \theta=0$$; the velocity is $$v_{0}$$, Find $$\theta$$ dependence of velocity and the angle the velocity vector makes with the radial vector."

    The answer-$$v(\theta)=v_{0}$$ and independent of $$\theta$$; and $$\alpha(\theta)=tan^{-1}(+_{-}\frac{1}{k})$$ and independent of $$\theta$$.

    It makes sense, as the equation resembles a logarithmic spiral, and the results hold for it; but how do I prove it?

    2. Relevant equations


    3. The attempt at a solution
    I tried differentiating it all I could, but I am always getting an extra $e^{-k\theta}$ in the term for velocity.

    $$\vec v=\frac{d}{dt}(r_{0}e^{-k\theta}) \hat r + r_{0}e^{-k\theta}\frac{d\theta}{dt}=-kr_{0}e^{-k\theta}\dot \theta\hat r + r_{0}e^{-k\theta}\dot \theta \hat \theta=r_{0}e^{-k\theta}\dot \theta(-k\hat r+\hat \theta)--------(1)$$

    At $$\theta=0$$,
    $$\vec v= \vec v_{0}=r_{0}\dot \theta(-k\hat r+\hat \theta)------------------(2)$$


    But then how do I prove that (1) and (2) are the same; i.e. how do I get rid of $$e^{-k\theta}$$ in the first equation? Or do I have to approach it differently( but I want to stick to polar coordinates).

    Please help; I am posting this after struggling for 2 straight days; and now I am completely frustrated. What's even more even annoying that it is only a 4 mark (2 marks for each part) problem.

    Thanks in advance!
     
  2. jcsd
  3. Apr 4, 2016 #2
    To do this problem, you need to know how θ is varying with t, or if the magnitude of v is constant, or if the magnitude of dθ/dt is constant. Do they tell you any of these things?
     
  4. Apr 5, 2016 #3
    That is exactly my problem! If they mention any of the following, then it is just an equiangular logarithmic spiral and I can make a transformation $$\theta=\omega t$$ and $$k=cot \alpha$$ and get my answer; but they haven't mentioned any such thing. And I don't think this will be very complicated, as it was only a two marker.
     
  5. Apr 5, 2016 #4
    People who make up problems leave out important information all the time. It's very hard to remember to include everything when you are formulating a problem. I'm guessing that they meant to say that dθ/dt is constant.

    Chet
     
  6. Apr 5, 2016 #5

    haruspex

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    That would not yield a constant speed, but dr/dt constant would.
    @cr7einstein , are you quite sure you quoted the problem word for word? There are a few phrases that read oddly. E.g., it says the track is horizontal, then specifies theta as the angle the radius makes to the horizontal.
     
  7. Apr 5, 2016 #6
    Yes. Maybe they meant that the magnitude of the velocity vector is constant.
     
  8. Apr 6, 2016 #7
    I believe by 'horizontal track' they mean fixed in the horizontal plane. Also, I have proved that the direction is independent of $$\theta$$. However, I just need to show that the magnitude is independent of the angle, and at any instant the velocity would be the same as the velocity at t=0 and $$\theta=0$$. But for that, I would need to get rid of the exponential term. And I don't see how can I do that. Just in case, I will post the link to the question paper- http://olympiads.hbcse.tifr.res.in/uploads/ino-2015-qp-and-solutions/inpho-2015-qp-sol/at_download/file [Broken]— PDF document, 1993Kb.
    It also has a diagram (which is not very helpful though!).
     
    Last edited by a moderator: May 7, 2017
  9. Apr 6, 2016 #8
    You left out some key words of the problem statement from your OP: "of mass m slides along a frictionless track." This is not strictly a kinematics problem. It is also a dynamics problem. You need to consider a force balance on the particle; what is the tangential acceleration equal to?
     
  10. Apr 6, 2016 #9
    Sorry! My bad!. I thought it was irrelevant.
    Now usually in such cases the friction provides the centripetal force; but the track is frictionless; how can there be any force (other than the weight and the normal reaction, which cancel out) to cause any unbalanced forces? But then, what would cause the mass to move along the (almost)circular trajectory? I am even more confused now !!!! Any hints, please @Chestermiller ? I am guessing I should double differentiate the given equation of trajectory, and write down the 2nd law in polar coordinates.( But can't see how it helps me).
     
  11. Apr 6, 2016 #10

    TSny

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    Careful. You seem to be making the assumption that the normal reaction must be vertical.
     
  12. Apr 6, 2016 #11

    Orodruin

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    This is not necessary. You could simply refer to that no energy is dissipated from the system (force is orthogonal to velocity) and use energy conservation.
     
  13. Apr 6, 2016 #12
    What I was getting at was that, without friction, the tangential component of the velocity vector (i.e., the component in the direction tangent to the particle trajectory) does not change in magnitude from its initial value v0.

    @ cr7Einstein The tangential component of velocity vector is not the same as the theta component of the velocity vector (because the trajectory is not circular). Do you know how to determine the component of the velocity vector in the direction tangent to the particle trajectory? Of course, the tangential component of the velocity vector in the direction tangent to the particle trajectory is equal to the magnitude of the velocity vector, because the component normal to the particle trajectory is zero (since the particle is on a track).

    Chet
     
    Last edited: Apr 6, 2016
  14. Apr 7, 2016 #13
    @Chestermiller Sorry I really don't follow you here. The velocity vector IS tangent to the particle trajectory, isn't it? If it isn't then how will you calculate the tangential component? As far as I have done, the direction of the velocity vector is independent of theta(look at the unit vectors); but the magnitude always seems to have the irritating exponential term. So I can't really prove that the velocity at theta=0 is the same as elsewhere, can I? I too am beginning to think that this is more like a dynamics problem, but it beats me how to move on furthur.

    @TSny I don't see any reason for the normal reaction to not be vertical; unless of course I 'create' a component to provide the centripetal force; is that what you are saying?

    @Orodruin what you said actually makes sense because as there is no PE term or dissipation, the only energy term will be a constant KE throughout ( and hence the velocity is constant too). But how do I quantitatively prove that $$v=v_{0}$$?

    I solved the whole paper; but haven't been able to do this one from 4 days....and this supposed to an easy problem! totally frustrated. Sorry if I am being totally unreasonable. Appreciate the help guys, but I think that I would really need a solution now or I'm going to go crazy! Thanks in advance!
     
  15. Apr 7, 2016 #14
    Yes.
    In your original post, you wrote: $$\vec v=r_{0}e^{-k\theta}\dot \theta(-k\hat r+\hat \theta)\tag{1}$$
    The magnitude of this velocity vector is $$v=r_{0}\sqrt{1+k^2}e^{-k\theta}\dot \theta\tag{2}$$
    and the vector is, of course, tangent to the particle trajectory. Since the track is frictionless, the component of acceleration along the track is zero, and so the magnitude of the velocity vector is constant at its initial value throughout the motion. So,$$r_{0}\sqrt{1+k^2}e^{-k\theta}\dot \theta=v_0\tag{3}$$

    Chet
     
  16. Apr 7, 2016 #15

    Orodruin

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    What is the expression for the kinetic energy? Is there any potential energy?
     
  17. Apr 7, 2016 #16
    @Chestermiller but $$v=v_{0}$$ at $$\theta=0$$; the exponential term should vanish, shouldn't it?

    @Orodruin No there is no PE, so $$1/m v^2=1/2 mv_{0}^2$$ ( for any two points) gives $$v=v_0$$(which is the solution); that is understood. What I am asking is that even though $$v_{0}$$ is independent of theta, why doesn't $$v$$ become independent of $$\theta$$ after all that calculus. ( as the problem suggests to use polar coordinates, I wanted to stick to it and get a 'rigorous' proof).
     
  18. Apr 7, 2016 #17

    Orodruin

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    ##v## is independent of ##\theta##, but you can also express it as a function of ##\theta## and ##\dot\theta##. This gives you a differential equation for ##\theta(t)##.

    Edit: Hint: You can write inline LaTeX by using double hashes (##) instead of double dollars ($$). Your posts will become much more readable.
     
  19. Apr 7, 2016 #18
    Only at time t = 0. The initial condition is ##\theta = 0## at t = 0.

    You have a variable-separable first order ordinary differential equation for dθ/dt with a specified initial condition. So how do you solve it?
     
  20. Apr 7, 2016 #19

    ehild

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    vo is the speed at t=0, when theta is also 0. The problem asks the θ dependence of the speed. The speed v is the same vo all along the track, that is independent of theta. Do not mix speed v with velocity ##\vec v##.
    It has a radial component and the component normal to the radius, along the direction of ##\hat \theta## . You get the cosine of the angle the velocity makes with the radial direction by using the dot product with the radial unit vector and dividing by the magnitude of the velocity.
    To get the normal force, you can use the acceleration vector. It has a tangential and a normal component. The tangential component of the acceleration is parallel with the velocity vector. You get it if you dot-product the velocity and the acceleration and divide by the magnitude of the velocity. To get the normal component of the acceleration subtract its tangential component from it.
     
    Last edited: Apr 7, 2016
  21. Apr 8, 2016 #20
    Yes, that is understood, @Chestermiller. But the problem asks to show that the velocity ANYWHERE is independent of theta; how to prove that?
    Which ODE are you talking about? The only one I can form here is ##|v|=r_{0}e^{-k\theta}d\theta/dt \sqrt{1+k^2}##; but we don't know the speed as a function of time, do we? If you have some other equation in mind, post it please!
    @Orodruin could you please send the equation you just mentioned? I cannot find a relation between velocity and theta unless I can express velocity as a function of time; and I don't see how can I do it. (Thanks for the tip on formatting! Didn't know I could do that).
    @ehild But that is what we are supposed to "prove", isn't it? It is quite easy to prove that the velocity's DIRECTION is independent of theta( As i did in the question); but even the magnitude(speed) has an exponential term which has ##\theta##; which is not there for ##v_{0}## (i.e. at ##\theta=0##). So, how do we show that the speed is independent of theta and the same as ##v_{0}## everywhere?
    I have honestly tried my best to solve this, but I am really, really stuck. I don't know if this is appropriate, but I think it would be really nice if someone could give a full solution now. Thanks in advance!
     
    Last edited: Apr 8, 2016
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