# Circular motion and acceleration

1. Feb 8, 2016

### Prannu

If you are twirling a rock around your head with a rope (assuming it is uniform circular motion), then the only acceleration that is acting is radial. So if you take a look at the rock at any given instant, its velocity is perpendicular to its acceleration. My question is, why does the rock not gain any speed as a result of its acceleration? Why do textbooks say that the speed is constant, despite the fact that there is an acceleration (a net force) on the rock?

2. Feb 8, 2016

### mfig

Remember that acceleration means a change in velocity. Velocity is a vector quantity characterized by speed (magnitude) and direction. Since the rock is continuously changing direction, it is accelerating even though the speed stays the same.

3. Feb 8, 2016

### Prannu

But why should the rock's velocity change in direction only? As long as there is acceleration, over a given period of time, dv = a*dt, indicating both a directional and magnitude change in velocity.

4. Feb 8, 2016

### A.T.

You could just as well ask why it doesn't loose any speed. If the angle between acceleration and velocity is less than 90° speed increases. If its more than 90° speed decreases.

The acceleration isn't constant over any given period of time

5. Feb 8, 2016

### mfig

It is hard to know how to answer when you ask why a rock's velocity should change in direction only. All I can tell you is that does change in direction only.

Think about it! You have a string and a rock. You are spinning it at a constant rate, so you know for a fact that the speed (magnitude) is not changing. You can measure this by timing how long it takes the rock to complete once circle and verifying that it always takes the same amount of time. But the rock is constantly changing direction. So the facts of the case are these:

No change in magnitude.
Change in direction.

As I said before, velocity is a vector with a magnitude and a direction. A change in velocity is called acceleration. The change can be in:

1. The magnitude of the velocity or
2. The direction of travel or
3. Both the magnitude and the direction.

There are examples of all three cases in nature. You have found a case where there is only a change in direction. But either way, any of those three cases is a change in velocity and is therefore an acceleration, by definition.

6. Feb 8, 2016

### Prannu

Yes, I completely agree with you in that the speed of the rock is fixed in uniform circular motion. I also agree with you in that acceleration is only causing a change in the direction only of the velocity. However, I still do not understand why, at any instant, the acceleration should cause only a change in direction. Please take a look at the attached image. In it, vf = sqrt(vi^2 + (delta-v)^2). As delta-t approaches 0 (but is still not 0), vf approaches vi (delta-v = a*delta-t), but is still not exactly equal to vi. This difference may be small, but over a enormous period of time, shouldn't the difference between vf and vi add up, and become significant enough to be observed?

*vi = initial speed
vf = final speed
delta-t = change in time
delta-v = change in speed

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7. Feb 8, 2016

### A.T.

An "instant" has zero duration.

8. Feb 8, 2016

### Staff: Mentor

It should change in direction only because you are not providing a force in the direction tangent to the circular path. You are only providing a force in the direction perpendicular to the circular path. So it can't have a component of acceleration along the path. It can only have a component of acceleration perpendicular to the path.

9. Feb 8, 2016

### Tom.G

Ambiguous usage of the word "acceleration" seems to have crept in here. The word Acceleration is often (improperly) used to indicate a force. Actually, it is defined as "a change in speed or direction", not the force causing the change.

re post 6: "...I still do not understand why, at any instant, the acceleration should cause only a change in direction."

Try turning it around. Look at it as the change in direction is causing the acceleration. The rock is attempting to move in a straight line. The rope is providing a force to prevent straight line motion.

re post 1: "So if you take a look at the rock at any given instant, its velocity is perpendicular to its acceleration."

Replacing "acceleration" in that sentence with "centrifigal force" may clarify things a bit.

10. Feb 9, 2016

### FactChecker

Because the acceleration is always exactly at right angles to the velocity vector, it never adds or subtracts from the magnitude of the velocity. It only changes the direction.

11. Feb 11, 2016

### drvrm

i think you will make an error if you take centrifugal force which is outward and its not acting - on the stone only centripetal force is acting towards the center.

to clarfy the concept you can draw a vector diagram of the stone say at time t and a later time t+ dt-
say its at point A initially and after lapse of time dt it has moved to B -both A and B are on the circular path and draw now two tangents at A and B designating the velocity vectors at t and t+dt
let it be v1 and v2 naturally their directions will be making an angle theta with each other -the same angle made by the two radii at A and B.
Now you have to find change in velocity- so draw the two vectors at angle theta separately and the tip of v1 and v2 will be joined depicting the change in velocity in time interval dt say dv -you can apply triangle rule for the three vectors v1 ,v2 and dv - v1+dv=v2 so dv is along AB
now dv/dt is the acceleration and taking limit to dt tending to zero will give you acceleration vector. the magnitude of the third vector dv is radius times theta- but the direction of the acceleration will move towards the centre as dt tends to zero-
so you can see that the centripetal acceleration drives the stone on a circular arc but its speed does not change if the force is constant -the acceleration will come to v^2/radius.

12. Feb 12, 2016

### jbriggs444

A slightly different take on the error... The diagram as shown is biased. It is not time-symmetric. It takes the current direction of the acceleration as a given and extrapolates into the future on the assumption that the acceleration stays constant at its starting value. But it would be equally reasonable to extrapolate into the future based on the assumption that the acceleration stays constant at its ending value.

An approximation that uses the starting acceleration vector over a finite interval predicts in a small increase in velocity. An approximation that uses the ending acceleration vector over a finite interval predicts in a small decrease in velocity. The truth lies somewhere in the middle.

The smaller you make the intervals, the smaller the gap within which the truth can hide. As the intervals shrink toward zero width, the only place that the truth can hide is exactly at zero.

13. Feb 13, 2016

### drvrm

yes thank you for pointing out the error pranu is making -its a conceptual error of 'rate of change' depiction by differential -one has to go to the limits of dt tending to zero.

14. Feb 14, 2016

### Prannu

Oh, I see now. I was making a mistake in approximating the time interval: it was too large, which is why the speed changed also. Because dt approaches an infinitesimally small value, the delta-v has 0 contribution towards the speed, and therefore, the speed is constant. However, this dv (or dv/dt) is enough for the particle's velocity to change in direction only.

Thank you all for helping me clarify this problem!
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