In my textbook the definition for centripetal acceleration is given,

"The acceleration produced by virtue of change in velocity direction is centripetal acceleration"

This definition is irritating. As, the tangential acceleration can also be defined in similar way. There should be difference in definitions. And, of course the direction of the velocity keeps changing always in circular motion; so it doesn't give a good explanation.

On this I found something interesting that the tangential acceleration can be divided in two components.

  1. with direction towards center of circle and

  2. directed away from the circle or equivalently in tangent direction.

They didn't stated that whether the component

  • can be defined as the centripetal force or not. So I want to ask that are we allowed to say it centripetal force or is that what centripetal acceleration really is?
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    $\begingroup$ Yes, I think the centripetal acceleration could be usefully defined as the radial component of the acceleration. $\endgroup$ Dec 18, 2015 at 17:30
  • $\begingroup$ I forgot to ask. In what way do you think the definition from my textbook is correct or useful @Rennie? $\endgroup$
    – user66452
    Dec 18, 2015 at 17:53
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    $\begingroup$ The definition is fine if the object is moving at constant speed, because in that case there isn't any tangential component to the acceleration. It isn't a good definition when the speed varies as well as the direction. $\endgroup$ Dec 18, 2015 at 17:59
  • $\begingroup$ That definition is in the context of uniform circular motion. In that case, the speed is constant, so there is only perpendicular (to the velocity) acceleration. $\endgroup$
    – Spirko
    Dec 18, 2015 at 17:59
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    $\begingroup$ No, any motion can always be given as radial and tangential vectors. It doesn't have to be uniform. $\endgroup$ Dec 21, 2015 at 6:11

1 Answer 1


One way of looking at the different components of the acceleration is just obtaining the second time derivative of the position in polar coordinates. Lets first compare the unit vectors in rectangular coordinates and polar coordinates (2D),

  • $\widehat{u_r} = \cos\theta\widehat{i}+\sin\theta\widehat{j}$

  • $\widehat{u_{\theta}} = \perp\widehat{u_r} = -\sin\theta\widehat{i}+\cos\theta\widehat{j}$

In this basis we can write the position in the following way:

  • $\vec{r} = x\widehat{i} + y\widehat{j} = \sqrt{x^2+y^2}\widehat{u_r}=\rho\widehat{u_r}$

Now we obtain the time derivatives

  • $\dot{\vec{r}}=\dot{\rho}\widehat{u_r}+\rho\dot{\widehat{u_r}}=\dot{\rho}\widehat{u_r} + \rho\dot{\theta}\widehat{u_{\theta}}$

As we see we obtain already two components for the velocity. The first component is just the variation of $\rho$ related to a linear movement. The second component is the variation of $\theta$ related to a circular movement. Now we obtain the time derivative of the velocity

  • $\ddot{\vec{r}}= \ddot{\rho}\widehat{u_r}+2\dot{\rho}\dot{\theta}\widehat{u_{\theta}}+\rho\ddot{\theta}\widehat{u_{\theta}}+\rho\dot{\theta}\dot{\widehat{u_{\theta}}} = \ddot{\rho}\widehat{u_r}+2\dot{\rho}\dot{\theta}\widehat{u_{\theta}}+\rho\ddot{\theta}\widehat{u_{\theta}} - \rho\dot{\theta^2}\widehat{u_r}$

This expression contains all the possible components for the acceleration in two dimensions. Now coming back to your question, we see two components in the acceleration, the ones that point in the radius direction ($\widehat{u_r}$) and the other tangential to the trajectory ($\widehat{u_{\theta}}$)

  • $\ddot{\vec{r}}= (\ddot{\rho}-\rho\dot{\theta^2})\widehat{u_r}+(2\dot{\rho}\dot{\theta}+\rho\ddot{\theta})\widehat{u_{\theta}}$

In the particular case of a circular motion with no angular acceleration and contant radius ($\ddot{\theta}=0$ and $\rho = \text{constant}$) we only have

  • $\ddot{\vec{r}}= -\rho\dot{\theta^2}\widehat{u_r}$

which is the usual expression for the centrifugal acceleration.

  • $\begingroup$ I am yet unable to understand derivatives, by the way. :) $\endgroup$
    – user66452
    Dec 28, 2015 at 10:24
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    $\begingroup$ @SufyanNaeem - Ok. Then the easy explanation would be that as with any vector you can divide the acceleration in two components, perpendicular and tangential to the movement. The former only "turns" the velocity vector (centripetal acceleration) and the latter increases the velocity vector magnitude (tangential acceleration). If you have only the centripetal acceleration, the velocity vector will not change length but it will change direction (it will "turn") and you will have a circular movement. $\endgroup$ Dec 28, 2015 at 12:45

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