# Can acceleration be zero in one co-ordinate system but non-zero in another system?

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 ?

• Can you write out the equations in their polar form in the question here? Commented Jan 12, 2017 at 12:16
• 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. Commented Jan 12, 2017 at 17:10
• 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. Commented Jan 12, 2017 at 18:21
• 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. Commented Jan 13, 2017 at 1:48

## 1 Answer

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)

• I appreciate your answer, just if you could elaborate a little, I'd be grateful. Commented Jan 12, 2017 at 18:09
• I worked it out, and I could see what you are saying. Thank you so much. Commented Jan 12, 2017 at 18:28