Tangential acceleration in circular motion

1. Jun 24, 2014

We know that the magnitude of tangential component of acceleration is,
atangential = dv/dt (where v is speed)
So clearly atan = 0 for uniform circular motion (as v is constant)

But what about non-uniform circular motion?
I can see atan = 0 only when v = constant. But in non uniform circular motion v is not constant.
Does it mean that atan can never be zero if the particle is moving in non-uniform circular motion?
(It never seems to be the case.)

Last edited: Jun 25, 2014
2. Jun 25, 2014

CWatters

Tangential acceleration is the rate of change in the magnitude of the velocity vector (eg the rate of change of speed). The normal acceleration is the rate of change of the direction of the velocity vector.

If something is in non-uniform circular motion the speed must be changing.

3. Jun 25, 2014

My confusion arises from this diagram. Why atangential = 0 at the top and at the bottom point? Isn't the magnitude of velocity changing at these two points? Why not?

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4. Jun 25, 2014

Tanya Sharma

Is there any tangential component of force at the top and at the bottom ?

5. Jun 25, 2014

Well, now I think I was a bit silly in asking so.
Let me rephrase my question.
In the diagram (post #3), tangential acceleration is zero at the top and the bottom. Is this because velocity is minimum and maximum at those two points?
If so, can I say that tangential acceleration is zero when the particle is at it's maximum or minimum speed?

6. Jun 25, 2014

Thank you. I missed that point. Actually I was thinking in terms of velocity.
Still I wonder, if no tangential acceleration (hence tangential force) is acting at the top, and the particle has zero speed, why the particle simply doesn't fall down in a straight line from the top?

7. Jun 25, 2014

Tanya Sharma

In case of roller coasters there is some mechanism ( eg seat belts ) to hold the person .

But if you consider a mass attached to a string undergoing vertical circular motion ,then you are right ,the mass would simply fall down.

Edit :The mass would fall down if the speed is zero at the top.

Last edited: Jun 25, 2014
8. Jun 25, 2014

Did you mean fall down in a straight line?

9. Jun 25, 2014

Tanya Sharma

Yes.

10. Jun 25, 2014

But that doesn't happen really. Does it?
When I attach a mass with string and create a vertical circular motion by holding that string, the mass never falls down in straight line when it reaches the top, rather it keeps moving in circular path.

11. Jun 25, 2014

Tanya Sharma

I was responding to post#7 where you were discussing the case when the speed is zero at the top .

12. Jun 25, 2014

Oh sorry.
So in this diagram, speed is not zero at the top ? (Thanks for your patience!)

13. Jun 25, 2014

Tanya Sharma

Yes . Speed is not zero .

14. Jun 25, 2014

Then speed is constant there right? (Because atangential = dv/dt is zero there.)
But just before the particle reached the top, it was slowing down and right after it crosses the top, it speeds up... So what does it mean to say speed is constant at the top point?

15. Jun 25, 2014

ehild

You are right. If the mass keeps on moving along the circle up to the top it will continue its circular motion.
To move along the circle, the string must be taut: the tension must be greater than zero. If the kinetic energy at the bottom of the circle is the same as the potential energy at the top , that is, the speed would be zero there, the string becomes slack earlier and then the mass will not follow the circular path.

So the speed can be zero at the top of the circle if you hold the mass there and release it.
When the mass is attached to the end of a rigid rod, it is possible to reach the top with zero speed, as the rod can exert both pulling and pushing force. The string is able to pull only.

ehild

16. Jun 25, 2014

ehild

The speed is not constant anywhere. It has a minimum at the top.

ehild

17. Jun 25, 2014

If speed is not constant anywhere (hence not at the top too) how can atangential = dv/dt = 0 at the top?
(By v I mean speed)

18. Jun 25, 2014

Bandersnatch

Yes, you can say that. These points are the local extrema of the function Vtan(t). Meaning, the first derivative(i.e., tangential acceleration) is zero at those points.

http://en.wikipedia.org/wiki/Maxima_and_minima
http://en.wikipedia.org/wiki/Critical_point_(mathematics)
http://en.wikipedia.org/wiki/First_derivative_test

Whenever a continuous function changes from increasing to decreasing, or vice versa, it will have a critical point(local maximum or minimum) there.

For example, look at a ballistic trajectory of a rock thrown into the air. Draw a function of height vs time(i.e., y(t) ). It'll be a familiar parabola. As the stone reaches the topmost point, you can see that to the left of that point(i.e., earlier) the height always increases, while to the right it always decreases. This means that the first derivative(velocity, Vy(t) ) is zero exactly at that point.

Remember that when you talk about the value of any function at one particular point, you can't say whether it is constant or not. It makes no sense to say that, as you need an interval to talk about changes. At one point, any function will have just one value.

The stone momentarily stops at the top of its trajectory, and it has 0 velocity in the y direction there, even though the velocity is never constant.

So, back to your circular motion. At the extreme points, the velocity is maximum/minimum, but to tell whether it changes or not you need to look at the whole thing.

19. Jun 25, 2014

Thanks

20. Jun 25, 2014

dean barry

α = rotational acceleration in ( rad / sec ) / sec

Tangential acceleration ( ( m / s ) /s ) = α * r