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Non-uniform circular motion

  1. Aug 4, 2014 #1

    These equations model circular motion. Equation R is the position vector given in polar coordinates. What I've done is represent this vector onto the complex plane via equation (1). Equation (2) and (3) are the first and second time-derivatives, respectively.

    Now, the question I have is this one: It seems "magical" to me that these derivatives actually give me the radial and tangential components for velocity and acceleration. How do I demystify this? Why is this obvious, apart from the fact that e^(itheta) and ie^(itheta) model the polar unit vectors?

    In other words, why doesn't projecting vectors given in polar coordinates onto the complex plane produce contradictions? Why does it work? Why is it that the terms actually represent the tangential and radial components?

    Also, do these equations imply that the typical introductory physics problem where we have to find the centripetal force at the bottom of a ditch as simply mv^2/r wrong? Because of the term given for radial acceleration by equation (3).

    Thanks. I hope my question is clear.

    If it's useful I got this from http://farside.ph.utexas.edu/teaching/301/lectures/node89.html
  2. jcsd
  3. Aug 4, 2014 #2


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    The radial acceleration term in equation (3) is the same as the oft-taught ##v^2/r##.
  4. Aug 4, 2014 #3
    Only the negative part. The other part is referring to how my radius might be changing, which I think happens in a ditch.
  5. Aug 4, 2014 #4


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    If the radius in polar coordinates is non constant, then it isn't clear that the polar radial vector is the same direction as the "radius" in the centripetal acceleration.
  6. Aug 4, 2014 #5
    I think the fact that the term is being multiplied by e^(itheta) means it is radial, it doesn't necessarily mean that it's in the direction of the "radius" of centripetal acceleration (meaning it doesn't have to point towards the center, because it's just radial). If you consider the case r''=0 then the radial acceleration is just -r*omega^2 whose magnitude is (v^2/r) and points towards the center.
    Last edited: Aug 4, 2014
  7. Aug 4, 2014 #6


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    Well here's something that might be worth checking:

    Suppose you have uniform linear motion, with no centripetal acceleration at all, along the path ##x = 1, \, y(t) = t##.
    What does ##(\ddot{r} - r\dot{\theta}^2)## come out to be?

    (I just scribbled this out but I haven't checked the answer carefully.)
  8. Aug 4, 2014 #7
    No idea, it's making my brain hurt, what do you suggest? I think both terms (tangential and radial acceleration) would have to be zero.
  9. Aug 4, 2014 #8


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    Okay, even easier example:

    Linear motion in the outward radial direction (away from the origin) with constant acceleration ##\ddot{r} = a_0##. This implies ##\dot{\theta} = 0##.

    Is the polar radial acceleration ##(\ddot{r} - r\dot{\theta}^2) = 0?##
  10. Aug 4, 2014 #9
    I would say no. I would say polar radial acceleration is just r''
  11. Aug 4, 2014 #10
    But what's your point are you saying | (r''-romega^2) | = v^2/r and always points towards the center of the circle? Because if that's it, I don't think it's true since there is a radial velocity term in equation (2).
    Last edited: Aug 4, 2014
  12. Aug 5, 2014 #11


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    The radial velocity term in (2) is by definition radial relative to (0, 0i) in the complex/polar plane. What I'm saying is that the radial "vector" ##e^{i\theta}## in complex/polar coordinates doesn't always coincide with the "radial" vector with respect to the circular motion.

    One last example that you should be able to work out geometrically:

    Consider uniform circular motion around the point (1, 0): that is, the usual uniform motion around the unit circle, except shifted to the right by 1 unit:
    \theta(t) &= t \\
    r(t) &= 2 \cos t
    The centripetal acceleration is always directed toward (1, 0), not (0, 0), and has a fixed magnitude ##v^2/r_{(1, 0)}##, where ##r_{(1, 0)}## is the distance to (1, 0).

    Now compare the "radial" acceleration term in (3) to the (constant) centripetal acceleration at the two positions: ##\theta = 0## and ##\theta = \pi/4##.
  13. Aug 5, 2014 #12
    What is this thread really about? davidbenari, can you ask more specific questions?
  14. Aug 5, 2014 #13
    voko: I want to know why this method works; why is that just differentiating this complex number will actually give me the radial and tangential components? I want to demystify this as I said in my original question. Thanks.

    olivermsun: I see your point, maybe this is only useful if you're considering stuff relative to the origin, which I'm fine with. By the way, do you know any other mathematical ways to deal with non-uniform circular motion? Do you know of any sources I could checkout? Thanks. BTW, if I'm considering some elliptical orbit, and say the sun is in (0,0i) then the radial acceleration cannot be simply v^2/r, right? Or instead of an elliptical orbit, if I consider a non-circular ditch or hill, the same goes right?
    Last edited: Aug 5, 2014
  15. Aug 5, 2014 #14


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    If you are considering circular motion, then you can always find the center of the circle and then use the polar method you showed. Also, gravity acting on a orbiting body is always inward radial with respect to the origin (i.e., the sun) so you should be able to analyze elliptical orbits without a problem.

    One alternative way I've seen the ##v^2/r## thing derived in mechanics and vector calculus is to use a moving "TNB" (tangent-normal-binormal) frame of reference. The tangent ##\vec{T}## is in the direction of your velocity , the "normal" direction ##\vec{N}## is chosen by ##d\vec{T}/dt##, and "binormal" is just ##\vec{T} \times \vec{N}## by convention. In this construction you can quickly see how ##\vec{N}## might not line up with your ##\vec{r}## in polar coordinates.

    This is a bit more general than the polar coordinates method since it can be used for a arbitrary (non-circular) paths. A search for "TNB frame" on Wikipedia turns up Frenet–Serret formulas, which looks a little more complicated than it needs to be, but I think you can get the idea.
    Last edited: Aug 5, 2014
  16. Aug 5, 2014 #15
    I do not see any mystery. The differentiation uses the standard chain rule. Multiplication by ##i## rotates the other multiplicand by 90 degrees counter-clock wise without changing its magnitude. But you probably know all of that. What is the real difficulty here?
  17. Aug 5, 2014 #16

    I know that, but I'm actually new to complex numbers so don't assume I'm knowledgeable in this. The mystery for me is this:

    I can see how r*e^(itheta) models some position vector. Differentiating yields just another complex number, which I can rearrange such that I have some term multiplying e^(itheta) and another multiplying ie(itheta), which gives me some expression that looks as if it were a vector. Mysteriously (for me) these two terms are the actual radial and tangential components. Why is it that this is not just incidental, why was it necessarily the case that they have to be the tangential and radial components?
  18. Aug 5, 2014 #17
    If you see that ##r e^{i\theta} ## is a vector, why do you not see that the products of the differentiation are also vectors? They have the same structure, after all.
  19. Aug 5, 2014 #18
    I can see that they can be interpreted as vectors, but why does it happen to be that they actually correspond to the radial and tangential components? Why didn't something else come up as terms form my "unit vectors" ? Is this a useless question? I'm asking for why is the term for each "unit vector" the actual correct radial and tangential component.
    Last edited: Aug 5, 2014
  20. Aug 5, 2014 #19
    The radial part is easy. Every complex number is a radial vector.

    Why the derivative of ##e^{i\theta}## is tangential is more interesting. Algebraically, this is because ##e^{i\theta}## gets multiplied by ##i##, which means a 90-degree rotation. Geometrically, this is because the image of ##e^{i\theta}## is a set of unit vectors - the unit circle - so when a particular unit vector turns into one infinitesimally close to it, the direction of travel is perpendicular to the radius - in other word, tangential to the unit circle.
  21. Aug 5, 2014 #20


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    In other words, ##d\theta## is tangential (or at least normal to ##\vec{r}##) in polar coordinates.
  22. Aug 5, 2014 #21
    Yeah I can see that. (I'm sorry if I'm being frustrating). Here's my proof for why the first time derivative produces the actual radial and tangential components. I hope this clears up what the actual question I'm asking is.

    Z is the area of the square enclosed by sides e^(itheta) and r. dz is the sum of the rectangle above and the rectangle on the right. The square on the corner is ignored because its a double differential. (This is how I've seen the product rule been proven, btw).

    Now equation (1) tells me how z changes as r is kept constant and I'm just changing e^(itheta) which is equivalent to a rotation on the complex plane. Equation (2) tells me how z changes as theta or e^(itheta) is kept constant, which is equivalent to moving radially relative to (0,0i). Equation (3) gives me the total change. I'll bring out dt to be on the bottom of dz and represent the derivative (dz/dt). And there it is, one side of the addition tells me how z changes keeping 'r' constant, and the other how it changes keeping 'theta' constant. Here I can see that the actual "components" do correspond to what I was looking for.

    Now this way of doing things will be very tedious as I proceed. I want something general that tells me that the "components" will be what I'm looking for.
  23. Aug 5, 2014 #22
    I do not think that your diagram is correct. It seems to me that you treat ##e^{i\theta}## as a real number. It is not. Are you familiar with the concept "complex plane" (or the Argand plane)? Look it up. Read again what I said earlier about unit vectors.
  24. Aug 5, 2014 #23
    Yeah I'm familiar with the complex plane, that's why I feel comfortable with treating e^(itheta) and ie^(itheta) as vectors (and I know they're perpendicular). I think it's not invalid to do this type of diagram because you could work backwards by multiplying dz/dt * dt , and you would get equation (3). In any case, did this show what my actual question is? I'm completely OK with why e^(itheta) and ie^(itheta) are vectors, I just want to know why they produce the correct physical results when you differentiate 'z' . :(
  25. Aug 5, 2014 #24
    ##e^{i\theta}## and ##ie^{i\theta}## are orthogonal unit vectors, radial and transverse, respectively. So their multipliers in the first and second derivatives of ##r e^{i\theta}## are necessarily the radial and transverse components of velocity and acceleration.
  26. Aug 5, 2014 #25
    I don't agree (and I hope you can prove me wrong), because my view is that those are just numbers that have the unique property that they can be interpreted as vectors. But they are not vectors, they are just numbers.
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