Tangential acceleration in circular motion

[Adjoint]

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

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

 

[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.

 

[Adjoint]

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

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

 

[Tanya Sharma]

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

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

 

[Adjoint]

I can see a_tan = 0 only when v = constant. But in non uniform circular motion v is not constant.
Does it mean that a_tan can never be zero if the particle is moving in non-uniform circular motion?

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?

 

[Adjoint]

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

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?

 

[Tanya Sharma]

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?

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.

 

[Adjoint]

Did you mean fall down in a straight line?

 

[Tanya Sharma]

Did you mean fall down in a straight line?

Yes.

 

[Adjoint]

If you consider a mass attached to a string undergoing vertical circular motion, the mass would simply fall down (in a straight line).

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.

 

[Tanya Sharma]

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.

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

 

[Adjoint]

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

 

[Tanya Sharma]

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

Yes . Speed is not zero .

 

[Adjoint]

Then speed is constant there right? (Because a_tangential = 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?

 

[ehild]

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.

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

 

[ehild]

Then speed is constant there right? (Because a_tangential = 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?

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

ehild

 

[Adjoint]

The speed is not constant anywhere.

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

 

[Bandersnatch]

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?

No .

Yes, you can say that. These points are the local extrema of the function V_tan(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, V_y(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.

 

[Adjoint]

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.

Thanks

 

[dean barry]

α = rotational acceleration in ( rad / sec ) / sec
r = radius in metres

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