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Homework Help: Change frame of reference to simplify acceleration calculation?

  1. Feb 26, 2013 #1
    I'm working problem from the upper division classical mechanics course on MIT OCW. No solutions are available. This is form the first P-set, question 4.

    1. The problem statement, all variables and given/known data

    A particle moves in a two dimensional orbit defined by

    [itex]x = A(2αt-sin(αt))[/itex]
    [itex]y = A(1-cos(αt))[/itex]

    Find the tangential acceleration at and normal acceleration an as a function of time where the tangential and normal components are taken with respect to the velocity.

    2. Relevant equations
    If you derive Newton's 2nd law in polar coordinates, we find that the acceleration on a particle is given by

    [itex]a = (\ddot{r}-r\dot{θ}^{2})\vec{e_{r}}+(r\ddot{θ}+2\dot{θ}\dot{r})\vec{e_{θ}}[/itex]

    3. The attempt at a solution

    Now I know I could use [itex] r^{2} = x^{2} + y^{2}[/itex] and [itex]θ = arctan(y/x)[/itex] and then take derivatives and so on, but this approach is very messy and time consuming. And the problem has to be solved with Newtonian mechanics. I was thinking to myself that it would be simpler if it weren't for those terms in front of the sine and cosine terms in x and y.

    Then I had an idea: why not just ignore them? And here's my hand-waving reasoning I conjured. The particle is basically moving in a circle that is shifting to the right at constant speed. If I measure the particle's acceleration in a frame of reference that is moving to the right at the same speed at the particle, it would simplify the math considerably since the radius would now be constant. In effect I'd just be finding the acceleration of a particle moving in a circle of constant radius, and the answer for the acceleration in this frame would also be the answer in the lab frame since the moving frame is not accelerating (right?). It'd be like back in freshman physics.

    But this seems too easy. I have never seen a problem done this way, and I can't find any problem like this one. Could anyone tell me if this is sound reasoning? I can't think of a reason this approach would be invalid, but it feels wrong.
    Last edited: Feb 26, 2013
  2. jcsd
  3. Feb 26, 2013 #2


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    Homework Helper

    You have to find the components of acceleration tangent to the orbit of the particle and normal to the orbit. The velocity is always tangent to the track, so you can find the tangential unit vector, and from that the outward normal vector.

    Determine the acceleration vector in Cartesian coordinates and multiply by the tangential unit vector: It is the tangential acceleration.

    You speak about cosine term but I do not see any in the equations given.

    You are right, the acceleration is the same in both frames of reference, but the tangential unit vectors are different.

  4. Feb 26, 2013 #3
    Yes, sorry the y term was incorrect, I fixed it and I'll try your suggestion.
  5. Feb 27, 2013 #4
    I ended up getting stuck again. I found the acceleration in cartesian basis

    [itex]\vec{a} = \ddot{x}\hat{x} + \ddot{y}\hat{y}[/itex]

    and dotted it with the polar unit vectors:

    [itex]\hat{r} = cos(θ)\hat{x}+sin(θ)\hat{y}[/itex]
    [itex]\hat{θ} = -sin(θ)\hat{x}+cos(θ)\hat{y}[/itex]

    But I feel like I'm going to end up with a very complicated and irreducible expression once I substitute in the functions of t for x and y. Did I make a wrong turn?

    By the way, I used

    [itex] cos\left(arctan\left(\frac{y}{x}\right)\right) = \frac{1}{\sqrt{\frac{y^{2}}{x^{2}}+1}}[/itex]
    [itex] sin\left(arctan\left(\frac{y}{x}\right)\right) = \frac{y}{x\sqrt{\frac{y^{2}}{x^{2}}+1}}[/itex]

    Here's a picture of my chicken scratch.
    http://i.imgur.com/B1CXuR3.jpg (direct link because image is big)
  6. Feb 27, 2013 #5


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    Homework Helper

    See my first post and also http://en.wikipedia.org/wiki/Acceleration for "tangential acceleration" . The particle does not move along a circle. No sense to use polar coordinates.
    Find the velocity and the tangential unit vector from it. Multiply the acceleration with the tangential unit vector. It is not the same as the polar unit vector aΦ

  7. Feb 27, 2013 #6
    echild already explained that, to get the tangential component of acceleration vector, you dot the acceleration vector with a unit vector in the velocity direction. The unit vector in the direction of the velocity vector is equal to the velocity vector divided by its own magnitude. You already wrote down the expression for the acceleration vector as [itex]\vec{a} = \ddot{x}\hat{x} + \ddot{y}\hat{y}[/itex]. Write down the velocity vector in the analogous way.

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