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A particle is moving in straight line parallel to x-axis, with uniform velocity (along y=2, lets assume). If write the equation of motion and calculate velocity in polar co-ordinates we see cos and sine dependence and hence acceleration is non-zero. But what is the physical significance of this. Non-zero acceleration means there has to be a force (Law of Inertia), but this particle is moving in straight line with constant velocity along y=2. Can someone please explain what I am missing ?

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    $\begingroup$ Can you write out the equations in their polar form in the question here? $\endgroup$
    – Kyle Kanos
    Jan 12 '17 at 12:16
  • $\begingroup$ Velocity in polar coordinates for this particle is: v = u cos θ rˆ − u sin θ θˆ (Kleppner book Example 1.15). So acceleration is non-zero. Whereas, v=xi in cartesian coordinates. Pardon me if I'm missing something. $\endgroup$
    – novice
    Jan 12 '17 at 17:10
  • $\begingroup$ How do you get from the expression for $v$ to the statement that acceleration is non-zero? The conclusion does not seem obvious to me. $\endgroup$ Jan 12 '17 at 18:21
  • $\begingroup$ It was just cosine/sine dependence of v, that made me think of it as time varying. But the answer below made me work it out and I realized what I missed. $\endgroup$
    – novice
    Jan 13 '17 at 1:48
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The unit vectors in the r and $\theta$ directions are functions of $\theta$, and $\theta$ is a function of time. What are the derivatives of the $\vec{i}_r$ and $\vec{i}_{\theta}$ with respect to $\theta$? (When I take the derivative of your velocity vector equation, I get 4 terms, and they cancel out in pairs, to give me zero acceleration)

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  • $\begingroup$ I appreciate your answer, just if you could elaborate a little, I'd be grateful. $\endgroup$
    – novice
    Jan 12 '17 at 18:09
  • $\begingroup$ I worked it out, and I could see what you are saying. Thank you so much. $\endgroup$
    – novice
    Jan 12 '17 at 18:28

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