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In spherical coordinates the acceleration can be written as

$$\textbf{a} = \dot{\textbf{v}} = \ddot{r} \hat{\textbf{r}} + \dot{r} ( \dot{θ} \boldsymbol{\hat{\theta}} + \sin θ \dot{\phi} \boldsymbol{\hat{\phi}}) + \dot{r} \dot{θ} \boldsymbol{\hat{\theta}} + r \ddot{\theta} \boldsymbol{\hat{\theta}} + r \dot{θ} ( \cos θ \dot{\phi} \boldsymbol{\hat{\phi}} - \dot{θ} \hat{\textbf{r}} ) + \dot{r} \sin θ \dot{\phi} \boldsymbol{\hat{\phi}} + r \cos θ \dot{θ} \dot{\phi} \boldsymbol{\hat{\phi}} + r \sin θ \ddot{\phi} \boldsymbol{\hat{\phi}} + r \sin θ \dot{\phi} ( - \sin θ \dot{\phi} \hat{\textbf{r}} - \cos θ \dot{\phi} \boldsymbol{\hat{\theta}}) $$

and from this we have the radial component of acceleration $$\ddot{r} - r \dot{θ}^2 - r \sin^2 θ \dot{\phi}^2$$ Do we call any of the above terms as centripetal acceleration? If so why?

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2 Answers 2

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By definition, if the acceleration points towards the origin it is centripetal. So if $\ddot{r} - r \dot{θ}^2 - r \sin^2 θ \dot{\phi}^2<0$ then the radial component is centripetal.

You are most likely getting confused with introductory systems of uniform circular motion where we say "centripetal acceleration is $r\dot\theta^2$", but that is because circular motion is only ever talked about. For general planar motion you have a radial acceleration component of $\ddot r-r\dot\theta^2$, and so the latter term always gives centripetal acceleration when the motion is circular.

As you can see, in more general scenarios you can have more complicated expressions for the radial component, and there is no "centripetal term". All of it contributes to the radial component, which can be centripetal depending on the overall sign.

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  • $\begingroup$ Thank you Dear Biophysicist. So the set of terms pointing in towards origin ie towards $-r$ all compromise the ' Centripetal acceleration'. Am I correct? $\endgroup$
    – Kashmiri
    Mar 28, 2021 at 14:18
  • $\begingroup$ @Kashmiri No. The whole thing is centripetal if $a_r<0$ $\endgroup$ Mar 28, 2021 at 14:33
  • $\begingroup$ Yes that's what I meant. The whole term pointing to the centre. $\endgroup$
    – Kashmiri
    Mar 28, 2021 at 15:16
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The two negative terms combine to give a resultant, -r$ω^2$.

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    $\begingroup$ could you please elaborate on it please. What is your $\omega$? $\endgroup$
    – Kashmiri
    Mar 28, 2021 at 14:20
  • $\begingroup$ ω is the resultant angular velocity that you get after combining the two components. $\endgroup$
    – R.W. Bird
    Mar 29, 2021 at 17:57

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