Why is the centripetal acceleration direction only correct for small velocity differences?

Question

I was trying to make an image to show students how to get the direction of the centripetal acceleration from looking at the change in the velocity vectors in uniform circular motion.

However, what I discovered was that it only works if the difference in the velocity vectors are small.

Take image 1: one vector is the negative of the tangent vector at $\theta=0$ degrees, and the other at $\theta = 10$ degrees.

In this case, taking the difference between the vectors gives you a vector that points towards the circle centre.

However, in image 2: the second vector is at an angle $\theta = 30$ degrees, and you can see that when you line the vectors up and take the difference it does not point to the centre of the circle.

In both cases the size of the arrows are identical.

Why is this?

I noticed that several books/web resources show this example using vectors with large angle differences, and I can never get the difference to point towards the centre in these cases.

Am I doing something wrong or is something more subtle happening here?

My current thinking is that it's purely due to the fact that the acceleration is the derivative of the velocity with respect to time, and therefore requires infinitesimal changes in the velocity and time in order to be correct.

Your current thinking is exactly right. The direction to the center changes with the rotation, so the right direction is fixed only if the angle is infinitely small. — rodrigo, Feb 13 '16 at 23:46
As an aside there is an old hack that lets you use non-trivial parts of the circle (in fact it is usually invoked over one full turn), but it involves computing the distance covered by the position and velocity vectors in the same way (and one that might not be obvious at first blush). — dmckee --- ex-moderator kitten, Feb 14 '16 at 00:15

score 2 · Answer 1 · answered Feb 14 '16 at 00:20

2

Your vector addition is not quite right. It is $\vec v_{\text{new}} = \vec v_{\text{old}} + \Delta \vec v$

Note that the velocity vectors are tangents to the circle.

As the angle $\alpha$ gets smaller and smaller the angles $\beta$ get closer and closer to $90^\circ$.
This means that the change in velocity gets closer and closer to being at right angles to the old velocity ie directed towards the centre of the circle which is what you found by drawing your diagrams.

answered Feb 14 '16 at 00:20

Farcher

95,680

I'm not sure I understand.
I translated the vector that is tangent to the circle at $\theta=0$ (which is v_initial or v_old) to the tip of the v_new vector and then flipped its orientation to get -v_old.

Then I just added them. The only thing that should be "wrong" is the direction of the vector, which could be 180 degrees out. Either way, if I draw a line from the base (or tip) of that vector it should go through the circle centre?
– user1654183 Feb 14 '16 at 13:40
The ways that I use to draw a diagram to show the addition of two vectors is either draw them nose to tail as I have or have the tails together and complete the parallelogram. The diagonal from the the place where the tails met is the resultant vector. – Farcher Feb 14 '16 at 13:55

Why is the centripetal acceleration direction only correct for small velocity differences?

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