Uniform Circular Motion and Centripetal Acceleration

Homework Helper
In introductory physics books (or at least mine) it limits the equation $a_c=\frac{v^2}{r}$ to the sitaution where the speed around the circular path is constant. It enforces the idea that the speed is CONSTANT.

But wouldn't the equation also apply to non-constant speeds? ($a_c$ would just change from being a constant to being a function of the speed)

It would be very counter-intuitive to me if this equation did not apply to variable speeds (because why does this instant in time care about the speed of the next instant in time?)

So my question is, can you also use this equation for variable speeds?

Answers and Replies

UltrafastPED
Science Advisor
Gold Member
Some parts of your question are dealt with here: http://www.sweethaven02.com/Science/PhysicsCalc/Ch0119.pdf

The machinery required to solve for the general case of centripetal acceleration for an object constrained to travel in a circle of constant radius, but with variable speed, is discussed ... you should be able to work through to the answer on your own from this point.

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UltrafastPED
Science Advisor
Gold Member
And if you want to go further, learn the Frenet-Serret apparatus; usually taught as part of vector calculus - calc 3.

why does this instant in time care about the speed of the next instant in time?

Is not the definition of acceleration the rate of change of velocity over time?

Is it possible to have instantaneous acceleration? A classic definition of $$\overline{a} = (v_f - v_i) / t$$ would be undefined at t = 0.

jbriggs444
Science Advisor
Homework Helper
Is not the definition of acceleration the rate of change of velocity over time?

Is it possible to have instantaneous acceleration? A classic definition of $$\overline{a} = (v_f - v_i) / t$$ would be undefined at t = 0.

The mathematical definition would be the limit (if one exists) of the average rate of change (vf-vi)/(tf-ti) as tf approaches ti without actually getting there.

That is to say that acceleration is the derivative of velocity.

http://en.wikipedia.org/wiki/Derivative

AlephZero
Science Advisor
Homework Helper
Your equation does give the centripetal component of the acceleration even when the speed is changing. But if the speed is changing, there is also a tangential component of the acceleration.

You will probably meet this later on if your course deals with objects moving in a vertical circle, where the speed is greater at the bottom of the circle than at the top.

WannabeNewton
Science Advisor
...because why does this instant in time care about the speed of the next instant in time?

The following is a general remark about acceleration. Acceleration is the rate of change of velocity. You can't determine acceleration at a given instant of time by only knowing velocity at that instant of time. You need to know it in some open interval centered on that instant of time. This is part of the basic definition of a derivative.