- #1

golith

- 17

- 0

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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- Thread starter golith
- Start date

- #1

golith

- 17

- 0

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

LeonhardEuler

Gold Member

- 860

- 1

What is meant by "angle of the circle"?golith said:

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

- #3

golith

- 17

- 0

Did this clear it up a bit? :)

thanx

- #4

LeonhardEuler

Gold Member

- 860

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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.golith said:

Did this clear it up a bit? :)

thanx

- #5

golith

- 17

- 0

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

- #6

golith

- 17

- 0

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

- #7

LeonhardEuler

Gold Member

- 860

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Yes, that would do it. (except its 9.8m/s^2*5=49m/s^2)golith said:

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

Time would simply be the amount of time that has passed since you started moving.Yes this would be the vectoring I am at now, but would time now be in reference to your velocity ?

- #8

golith

- 17

- 0

Thanx

- #9

golith

- 17

- 0

http://www.sixflags.com/parks/overgeorgia/pdf/HighSchoolWorkbook.PDF

looking at this to inspire me and in particular the horizontal accelerometer.

looking at this to inspire me and in particular the horizontal accelerometer.

Last edited by a moderator:

- #10

LeonhardEuler

Gold Member

- 860

- 1

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 [itex]\frac{v^2}{r}[/itex]. 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.golith said:

Thanx

- #11

golith

- 17

- 0

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 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

LeonhardEuler

Gold Member

- 860

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[tex]\sqrt{a_c^2 + a_l^2}[/tex]

- #13

golith

- 17

- 0

Yet again another so simple answer in this world of complexities.

I tip my hat to you sir.

I tip my hat to you sir.

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