Which is the centripetal term here?

Question

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?

Answer 1

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.

Answer 2

The two negative terms combine to give a resultant, -r$ω^2$.