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Suppose a particle is rotating about a point at a distance $r$, then since $r$ is constant $\frac{\text dr}{\text dt}=0$ so the component of velocity along the position vector should be zero. Differentiating again, $\frac{\text d^2 r}{\text dt^2}=0$ so that component of acceleration along the position vector should be zero, which is incorrect.

So my question is that what is wrong with this reasoning, is it because I neglected that velocity and acceleration are vectors if that is so, what do $\frac{\text d r}{\text dt}$ and $\frac{\text d^2 r}{\text dt^2}$ actually represent?

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    $\begingroup$ You are mixing coordinates with constraints. "a particle is rotating about a point at a distance r" is r the distance between the center and the particle? If so dr/dt and all its derivatives are 0 or otherwise the particle would change the distance from the center of rotation, violating your first statement. $\endgroup$ – fredwhileshavin Apr 30 at 3:23
  • $\begingroup$ @fredwhileshavin I don't understand your objection. $\endgroup$ – Aaron Stevens Apr 30 at 3:31
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You have fallen into a common misconception due to familiarity with Cartesian coordinates. We are used to seeing $$F_x=m\frac{\text d^2x}{\text dt^2}$$ and $$F_y=m\frac{\text d^2y}{\text dt^2}$$ for 2D motion in Cartesian coordinates $x$ and $y$, for example.

However, it is not true in polar coordinates $r$ and $\theta$ that $$F_r=m\frac{\text d^2r}{\text dt^2}$$ and $$F_\theta=m\frac{\text d^2\theta}{\text dt^2}$$ because our unit vectors now have position dependence.

You can look more into derivations of equations of motion for planar motion, but the gist of it is that our equations actually become

$$F_r=m\left(\frac{\text d^2 r}{\text dt^2}-r\left(\frac{\text d\theta}{\text d t}\right)^2\right)$$ and $$F_\theta=m\left(r\frac{\text d^2\theta}{\text dt^2}+2\frac{\text d r}{\text d t}\frac{\text d\theta}{\text dt}\right)$$

Therefore in your uniform circular motion example, since the $r$ coordinate is not changing it must be that the radial force acting on the object is $$F_r=-mr\left(\frac{\text d\theta}{\text d t}\right)^2$$ which you should recognize is what you usually see for centripetal force (usually seen in introductory physics as $mv^2/r$, since $\frac{\text d\theta}{\text d t}=v/r$). And of course there is no $\theta$ component of the net force since $r$ is not changing and $\theta$ is not "accelerating".

This leads to other things you might be familiar with. For example notice that we have $$F_\theta=\frac{\text d}{\text dt}\left(mr^2\frac{\text d\theta}{\text dt}\right)$$ so that if $F_\theta=0$ then $$mr^2\frac{\text d\theta}{\text dt}=L$$ where $L$ is a constant which turns out to be the angular momentum (i.e. angular momentum is conserved when no net torque is present).

Also, please notice that you have to be careful with talking about, for example, acceleration and second time derivatives. We can still write Newton's second law as $$\mathbf F=m\mathbf a=m(a_r\hat r+a_\theta\hat\theta)$$ just notice that now, as I mentioned earlier $$a_r\neq\frac{\text d^2 r}{\text dt^2}$$ and $$a_\theta\neq\frac{\text d^2\theta}{\text dt^2}$$

It should also be noted that a similar thing happens with the velocity: $$v_r=\frac{\text d r}{\text dt}$$ $$v_\theta=r\frac{\text d\theta}{\text dt}$$ the radial velocity is what we would normally expect from our "Cartesian thinking", but this is not the case for the $\theta$ component. Also you can see that $$\frac{\text dv_r}{\text dt}\neq a_r$$ and $$\frac{\text dv_\theta}{\text dt}\neq a_\theta$$ This is because, as I mentioned before, the unit vectors in polar coordinates are now position dependent, so the unit vectors must also be considered when taking time derivatives. In other words, it is true that $$\frac{\text d\mathbf v}{\text dt}=\mathbf a$$ and this is where you should start when thinking about these vectors in these curvilinear coordinate systems.

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    $\begingroup$ Very nice answer. I'm a huge supporter of this being the way that Christoffel symbols be introduced in a first course in GR - I'm baffled by the tendency to hurl students into curved 4D spacetime when flat 2D space works just fine to illustrate the concept. $\endgroup$ – J. Murray Apr 30 at 3:49
  • $\begingroup$ Ah!! Now I get it, Thank you very much, I was able to derive the equations of rotational motion, just not able to find the contradiction. $\endgroup$ – Random Apr 30 at 4:06
  • $\begingroup$ @Random glad I could help $\endgroup$ – Aaron Stevens Apr 30 at 4:09
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Yes, the problem is that you neglected that velocity and acceleration are vectors and that the unit vectors themselves change from point to point in spherical coordinates. When you take derivatives you have to include the changes due to the changing unit vectors too. This is described here:

http://mathworld.wolfram.com/SphericalCoordinates.html

For uniform circular motion $r=R$, $\theta=\omega t$ and $\phi=\pi/2$, so

$$\vec r = R \hat r$$ $$\vec v = \dot{\vec r}= R\omega \hat{\theta}$$ $$\vec a = \dot{\vec v}= -R \omega^2 \hat r$$

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If r is a vector you get a different result. If it is initially r*I+0*J, with I and J being unit vectors in a plane, it can move in the J direction. More generally in a circular motion normal to I. Rigid body motions are of this sort. Also motion of planets moving under a central source of gravitational force.

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  • $\begingroup$ Looks OK to me. Reviewers, please look at the question before recommending deletion -- the question asks "doesn't $d^2r/dt^2=0$ imply acceleration is 0?" and this answer clarifies that $d^2r/dt^2$ isn't the acceleration, since $r$ is not a vector. $\endgroup$ – Abhimanyu Pallavi Sudhir Apr 30 at 5:44

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