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In cylindrical coordinates I had derived the acceleration as : the equation

for the first component of acceleration, what is the difference between radial and centripetal ?

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  • $\begingroup$ A guess: centripetal is from a point, radial is perpendicular to a line. $\endgroup$ – garyp Jan 7 at 16:54
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The radial acceleration, $\ddot r$, is responsible to change the rate of change of the radius vector, $\dot r$, which in turn is responsible to change the magnitude $r$ of the radius vector, its length.

The centripetal acceleration $-r \dot \theta^{2}$ is what makes the particle describe some curvilinear path. The "minus" signal in front of the centripetal acceleration indicates it's inward.

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If we consider only the radial component in the case when $\dot{\theta} = 0$, then it becomes clear that $\ddot{r}$ is a radial acceleration, i.e. since the angular component of velocity is $0$, the particle is only moving along a straight line through the origin, and $\ddot{r}$ is the acceleration along this line. $$\\$$ Now if we consider when $\ddot{r} = 0$, with $\dot{r} = 0$ also (incidentally, this is just the condition for circular motion around the origin), then $-r{\dot{\theta}}^2$ must be the source of what is making us turn.

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The only thing I can add to the existing answers that may be helpful is a visual representation, which can be found on Saif Rayyan's MIT course page here and is reproduced belowenter image description here

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