# Angle of a perfect circle

1. Sep 5, 2005

### golith

Hi,
Just a quick question
I have a circle with radius of 5 metres, what is the angle of the circle?
or is there a simple formula to work this out.

Thanx

2. Sep 5, 2005

### LeonhardEuler

What is meant by "angle of the circle"?

3. Sep 5, 2005

### golith

the angle of the circle would be the constant angle (? angular velocity?) you would be on as you went around the horizontal ride. If its radius is 5 metres and is travelling at 1 m/s with a mass of 100kg in a circle, i believe i will need to know its angle to find out the force applied at any given time in its coarse.
Did this clear it up a bit? :)
thanx

4. Sep 5, 2005

### LeonhardEuler

What do you mean by "the constant angle you would be on"? If you are traveling at 1m/s around a circle and the radius is 5 meters, then your angular displacement depends on how much time has passed.

5. Sep 5, 2005

### golith

Hay there, just re-reading some other information and have another question?
If your travelling in a straight line and your velocity is 1 m/s and you want to achieve a g-force of 5G would you therefore have to accelerate at (9.8m/s*5) or 48 m/s to affect 5 g's Thanx

6. Sep 5, 2005

### golith

Yes this would be the vectoring im at now, but would time now be in reference to your velocity ?

7. Sep 5, 2005

### LeonhardEuler

Yes, that would do it. (except its 9.8m/s^2*5=49m/s^2)
Time would simply be the amount of time that has passed since you started moving.

8. Sep 5, 2005

### golith

Now thats very good but what if that straight line is now a circle as perfectly round as humanly possible. would this still apply or has angular displacement now componded the g force and therefore not have to accelerate at 49 m/s^2 but less ?

Thanx

9. Sep 5, 2005

### golith

Last edited by a moderator: Apr 21, 2017
10. Sep 5, 2005

### LeonhardEuler

If a particle is changing direction, it is accelerating even if it is not speeding up or slowing down. If the radius of the circle is r and the particle's speed is v, then it's acceleration without changing speed is equal to $\frac{v^2}{r}$. So, just by moving quickly enough in a small enough circle, you could have an acceleration of 5 g. For example, moving at 16 m/s in a circle of radius 5 gives an acceleration of a6^2/5=51.2m/s^2, which is about 5.2 g. If the particle also speeds up while moving in a circle, its acceleration will be the vector sum of the linear acceleration and the centripital acceleration.

11. Sep 5, 2005

### golith

Ah ha, Have U been told you are a legend :)
This has helped no end.
The last part of my query then is the last part being the vector sum of the linear acceleration and the centipital acceleration.
Could I express this by stating a(centripital) is V(2/t)/R
Linear acceleration as Vf-Vi/t = * m/s/s
If so then how do you do the vector sum of these two quantities? :)
That would be it and the computer would be able to do the rest to apply whats required upn the object at any given time.
This would then negate having to travel at high speeds to fool the particle into thinking it is travelling faster than it is, just deacceleration is the bugga now then in how to affect a rapid deacceleration when your not actually travelling that fast. Anyhow i will get there.
I added you to my buddy list and hope to be able to employ your good talent in the future. I'm also trying my hand at thermodynamics and quantum physics in general to be able to grasp concepts required in the future.
Hope to talk to you again.

Sorry for the waffle
Long live Formula one

12. Sep 6, 2005

### LeonhardEuler

The linear and centripital accelerations are perpendicular vectors, so they can be added by the pythagorean theorem. If $a_c$ is the centripital acceleration and $a_l$ is the linear acceleration, then the total acceleration is:
$$\sqrt{a_c^2 + a_l^2}$$

13. Sep 9, 2005

### golith

Yet again another so simple answer in this world of complexities.
I tip my hat to you sir.