1
$\begingroup$

I can understand the physical existence of centripetal acceleration, the orthogonality of that with the linear (tangential) velocity - but how to physically understand jerk and snap in case of uniform circular motion?

$\endgroup$
  • $\begingroup$ 'but how to physically understand jerk and snap in case of uniform circular motion?' That's an oxymoron. In uniform circular motion, centripetal acceleration is constant. $\endgroup$ – Gert Jan 9 '17 at 14:18
  • 1
    $\begingroup$ @Gert constant in magnitude, not direction. $\endgroup$ – Rody Oldenhuis Jan 9 '17 at 14:26
  • $\begingroup$ When you say "physically understand," are you talking about an intuitive concept, or is just taking the derivatives of position, velocity, acceleration, jerk, snap... etc the kind of answer you are looking for? $\endgroup$ – Cort Ammon Jan 9 '17 at 14:42
  • $\begingroup$ I can "feel" centripetal force in uniform circular motion. What about Jerk ? I mean, jerk is change in acceleration with time. How and what can we 'feel' about that ? $\endgroup$ – novice Jan 9 '17 at 15:07
  • $\begingroup$ @novice What coordinate frame are you in when you "feel" the centripetal force? If you are in a frame that is rotating with the object, you need to allow for your own rotation. If you were moving in a circle but always facing in the same direction, you would "feel" that the centripetal acceleration was changing direction as you moved round the circle. $\endgroup$ – alephzero Jan 9 '17 at 16:47
2
$\begingroup$

For any physical system, a position can in principle be differentiated infinitely many times.

That's no different in uniform circular motion. The thing that makes it a textbook exercise though is that all components of the position are described by $A\cdot\sin(\omega t+\phi)$ and $A\cdot\cos(\omega t+\phi)$.

This means that the velocity ($d\mathbf{r}/dt$), acceleration ($d^2\mathbf{r}/dt^2$), jerk ($d^3\mathbf{r}/dt^3$) etc. all the way up to infinity, are all described by $\pm\omega^nA\cdot\sin(\omega t + \phi)$ or $\pm\omega^nA\cdot\cos(\omega t + \phi)$.

The important thing to note though is that there will be minus signs popping up all over the place. The velocity is perpendicular to the position, the acceleration perpendicular to the velocity and in the negative direction of position, the jerk will be perpendicular to the acceleration and in the negative direction of the velocity, and so on.

The takeaway lesson while studying the uniform circular motion is that although the directions are variable, the magnitude of all of them is constant, owing to the identity $\cos^2 + \sin^2 = 1$. Hence the name, uniform circular motion.

It works the same in more dimensions, it just gets more elaborate. This is when the notatational switch to vectors becomes important -- to be able to focus on the physics, not the mathematical details describing size, orientation, etc.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.