Calculating the Angle of a Perfect Circle: A Simple Formula

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SUMMARY

The discussion centers on calculating the angle and forces involved in circular motion, specifically for a circle with a radius of 5 meters. Participants clarify that the angle of a circle is related to angular displacement, which depends on time and velocity. They discuss the relationship between linear and centripetal acceleration, emphasizing that the total acceleration can be calculated using the Pythagorean theorem. The conversation also touches on achieving specific g-forces through acceleration in circular motion.

PREREQUISITES
  • Understanding of circular motion and angular displacement
  • Knowledge of centripetal and linear acceleration
  • Familiarity with basic physics equations, including acceleration formulas
  • Ability to apply the Pythagorean theorem in physics contexts
NEXT STEPS
  • Study the principles of circular motion in physics
  • Learn how to calculate angular velocity and acceleration
  • Explore the relationship between speed, radius, and g-forces in circular motion
  • Investigate advanced topics in dynamics, such as vector addition of forces
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Physics students, engineers, and anyone interested in understanding the dynamics of circular motion and forces involved in high-speed scenarios.

golith
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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
 
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golith said:
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
What is meant by "angle of the circle"?
 
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 traveling 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
 
golith said:
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 traveling 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
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.
 
Hay there, just re-reading some other information and have another question?
If your traveling 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
 
LeonhardEuler said:
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.
Yes this would be the vectoring I am at now, but would time now be in reference to your velocity ?
 
golith said:
Hay there, just re-reading some other information and have another question?
If your traveling 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
Yes, that would do it. (except its 9.8m/s^2*5=49m/s^2)
Yes this would be the vectoring I am at now, but would time now be in reference to your velocity ?
Time would simply be the amount of time that has passed since you started moving.
 
Now that's 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
 
http://www.sixflags.com/parks/overgeorgia/pdf/HighSchoolWorkbook.PDF

looking at this to inspire me and in particular the horizontal accelerometer.
 
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  • #10
golith said:
Now that's 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
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 centripetal acceleration.
 
  • #11
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(centripetal) 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 what's 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 traveling faster than it is, just deacceleration is the bugga now then in how to affect a rapid deacceleration when your not actually traveling 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
The linear and centripetal accelerations are perpendicular vectors, so they can be added by the pythagorean theorem. If a_c is the centripetal acceleration and a_l is the linear acceleration, then the total acceleration is:
\sqrt{a_c^2 + a_l^2}
 
  • #13
Yet again another so simple answer in this world of complexities.
I tip my hat to you sir.
 

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