The change in magnitude of centripetal acceleration

[Frawx]

When an object (e.g. racecar) moves around in circles with constant tangential velocity, constant centripetal acceleration is present.

What happens to the centripetal acceleration when the racecar is at rest, then increases its speed? I know that the tangential velocity increases due to the tangential acceleration, but what about the centripetal acceleration?

Since centripetal acceleration is tangential velocity squared divided by the radius, and the tangential velocity is increasing from rest, the centripetal acceleration must then be increasing as well.

How do you calculate the values for centripetal acceleration if it is changing? There doesn't seem to be a formula for it. And it seems that centripetal acceleration is changing, is there a term for the rate of change of it?

 

[andrewkirk]

If the object is moving in a circle, the centripetal acceleration must be $v^2/r$ - ie all that matters is the instantaneous linear speed and distance from the centre of the circle. If the object is not moving in a circle (eg ellipse or more complex shape) it becomes more complicated.

 

[Frawx]

If the object is moving in a circle, the centripetal acceleration must be $v^2/r$ - ie all that matters is the instantaneous linear speed and distance from the centre of the circle. If the object is not moving in a circle (eg ellipse or more complex shape) it becomes more complicated.

What if the instantaneous linear speed is increasing? How would I be able to find out the centripetal acceleration's increase as the instantaneous linear speed increases? The formula only applies to an instantaneous point in time.

Additionally, what happens when the object is not moving in a circle (e.g. ellipse)? Would you assume the ellipse to be almost equal to a circle? If it is a complex shape, do you draw many different circles?

 

[andrewkirk]

What if the instantaneous linear speed is increasing? How would I be able to find out the centripetal acceleration's increase as the instantaneous linear speed increases? The formula only applies to an instantaneous point in time.

If you know a formula for the linear speed as a function of time then you automatically have a formula for the centripetal accel as a function of time. Just square the first formula and divide by $r$. That's all there is to it.

Perhaps what you are wondering about is what would happen with an object that is constrained to move on a circular track, to which a constant circumferential acceleration is applied?

 

[CWatters]

The "v" in v^2/r is the instantaneous tangential velocity.

So if v was changing with some complicated function of time like v= 3t^3 + t^2 then you just substitite it so..

A = v^2/r
Becomes
A =(3t^3 + t^2)^2/r

Note that A is just the centripetal acceleration NOT the net acceleration. If you want to calculate the net acceleration you must do the vector sum of the tangential acceleration and the centripetal acceleration. That's not difficult because they are at 90 degrees to each other so Pythagoras can help.

 

[A.T.]

If it is a complex shape, do you draw many different circles?

Conceptually yes. You use the local radius & center of curvature to determine the magnitude and direction of the centripetal acceleration.

https://en.wikipedia.org/wiki/Curvature

 

[Frawx]

Thanks for the responses! Turns out that the answer to this question was simpler than I had thought.

The formula for centripetal acceleration at an instantaneous point in time is a_c=v²/r.

The formula for centripetal acceleration for a changing tangential velocity over time is a_c=(v(t))²/r.

 

[A.T.]

Thanks for the responses! Turns out that the answer to this question was simpler than I had thought.

The formula for centripetal acceleration at an instantaneous point in time is a_c=v²/r.

The formula for centripetal acceleration for a changing tangential velocity over time is a_c=(v(t))²/r.

Well, it's the same formula, regardless if the speed changes or not.