1
$\begingroup$

Suppose we have a particle moving in a circular motion governed by the equation $\vec{r} = \cos(t) \hat{i} + \sin(t) \hat{j}$, then we notice that velocity at $t=0$ is given as: $\vec{\dot{r}} = \hat{j}$, from this we can write centripetal acceleration as:

$$ \ddot{r_c} = \frac{| \hat{j}|^2}{ |\vec{r}|}$$

by the formula:

$$ \vec{\ddot{r_c} } = \frac{|\vec{\dot{r} }|^2}{|\vec{r}|^2}$$

But, suppose we a particle moving with velocity $\hat{j}$ located somewhere, then in this particles from the velocity of the particle in circular motion would be zero and hence the centripetal acceleration of the particle in circular motion is zero.

So, switches between frames seems to have cancelled out the force, this leads me to the question: In which frame do we put the velocity in for the centripetal acceleration formula?

$\endgroup$

2 Answers 2

2
$\begingroup$

from this we can write centripetal acceleration as: $$ \ddot{r_c} = \frac{| \hat{j}|^2}{ |\vec{r}|}$$ by the formula: $$ \vec{\ddot{r_c} } = \frac{|\vec{\dot{r} }|^2}{|\vec{r}|^2}$$

These formulas are not general formulas but only apply for the specific case of uniform circular motion. While $\vec{r} = \cos(t) \hat{i} + \sin(t) \hat{j}$ is uniform circular motion $\vec{R} = \cos(t) \hat{i} + (\sin(t)+t) \hat{j}$ is not. Therefore, the equations for uniform circular motion do not apply in the transformed frame since the motion is not uniform circular motion in that frame.

Note, although the motion is not uniform circular motion and therefore the acceleration is not centripetal, both frames do agree on the magnitude and the direction of the acceleration itself. So it is not that the force disappears, just that calling the force "centripetal" no longer makes sense.

$\endgroup$
2
  • $\begingroup$ Would be correct to think that in the relative frame that it's not circular motion? $\endgroup$
    – Brian
    Commented Apr 1, 2021 at 17:08
  • $\begingroup$ Yes. It is a cycloid, not a circle. $\endgroup$
    – Dale
    Commented Apr 1, 2021 at 18:12
1
$\begingroup$

Your initial equation assumes a particular coordinate system. If you move to a different system, all of your relations will change.

$\endgroup$
3
  • $\begingroup$ Yes, but could you explain what exactly is going wrong and how I can fix it a bit more elaborately? If so, I'll accept this answer and upvote . $\endgroup$
    – Brian
    Commented Apr 1, 2021 at 15:49
  • $\begingroup$ Sorry, I don't understand what you are trying to do? $\endgroup$
    – R.W. Bird
    Commented Apr 1, 2021 at 20:23
  • $\begingroup$ I added a follow up question to this see here @R.W. Bird $\endgroup$
    – Brian
    Commented Apr 1, 2021 at 20:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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