Tangential and centripetal acceleration in circular motion

  • #1
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I'm trying to understand the geometrical meaning of acceleration in circular motion. When I consider a particle moving in a circular path, I can clearly see that the acceleration vector can be composed of a tangential and radial component. But since the change in velocity happens over a period of time, the particle has to move a small distance along the circle (arc), so my question is this: when we say that the acceleration has a radial and tangential component, at what point on the path are we talking about? Because we could say tangential and radial to the initial point, or the final point.

Now another thing I don't understand is that if the speed (magnitude of velocity vector) is uniform, the particle's velocity vector consequently only changes in direction. This has to mean that the initial and final velocity vectors have to be of equal length, but drawing them with tails in the same origin still shows that the acceleration has a tangential component not equal to zero.
I know I'm very off track somewhere in this thinking, because it clearly makes sense in an intuitive way, but not geometrically.
 

Answers and Replies

  • #2
But since the change in velocity happens over a period of time, the particle has to move a small distance along the circle
Yes, that's true for the average. That's where differentiation comes in: the instantaneous rate of change is the limit of the change divideed by the delta time.

That way you have a unique tangential and a radial acceleration at each moment and at each point alogng the trajectory.

Very important concept, differentials
 
  • #3
Now another thing
Not really another thing. The little difference vector has a vanishing tangential component for ##\Delta t\downarrow 0## and also a vanishing component for the radial component. But the first one goes like cosine -1 and the second one like the sine of ##\Delta \theta##. So divided by ##dt## the first one gives zero and the second one ##-\omega^2 r ## in the limit ##dt \downarrow 0##.

There are more accesible explanations -- is this adequate or is differentiation new for you ?
 
  • #4
Oh yes, this actually makes sense now. You explained it very nicely, and yes I'm familiar with differentiation, just wasn't quite sure what happened in a geometrical sense. Anyways, thank you for your help :)
 
  • #5
You're welcome
 

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