Calculating Centrifugal Force for 300KG Object in 5m Radius Circle

[golith]

Hi,
I'm attempting to workout what is the G-force applied to an object that is traveling at 1 m/s/s in a perfect circle with a radius of 5m. The circle is vertical not horizontal. The total weight being propelled is 300KG and is running on the inside of a track. There is no air resistance or friction to worry about here either. I can not work out all the required data to computate the answer as i want to be able to manipulate the applied G-Force affected on the object at any given time, but 1 m/s/s is as simple as they come for this purpose. Yes this has already been achieved.

Thanx

 

[Doc Al]

...an object that is traveling at 1 m/s/s .

Not sure what you mean by this. Did you mean to specify the object's speed?

 

[golith]

HI,

Well the objects speed is at 1 m/s/s and is traveling around a circle with a radius of 5m. i am playing unsuccessfully with
centripetal acceleration.
Angular velocity is also not working.
I need a formula that will allow adjustment to the speed at anytime to allow G- Force to be apllied to the object going round and around the circle.

 

[Manchot]

Ok, speed cannot be in units of m/s/s. Did you mean that the speed is 1 m/s?

Anyway, to find centripetal acceleration, remember that for an object traveling in a circle, the magnitude of the acceleration is always \frac{v^2}{r}. That should really be all you need.

 

[golith]

Aha so the constant speed is 1 m/s say and its radius is 5m
therefore i need to substitute v2 =1m/s and r=5m, if this gives the desired result then where to then to find the applied G-force ont he onject. or a(centripetal) give this?

(Of all the things I've ever missed i miss my mind the most :( )

 

[Doc Al]

By "applied G-force" I assume you mean the normal force that the surface exerts on the object. To find that, use Newton's second law at the point in the motion that you want to analyze. (The normal force varies as the object goes around the circle.) You already have the centripetal acceleration (or can compute it).

Or if you mean what "centrifugal force" that the object experiences, then that is an outward force equal (in magnitude) to the centripetal force. But realize that its not the only force acting on the object.

 

[golith]

Hmm so this is now getting more complex than first thought.O.k then here it is in a nutshell. I want to simulate real time driving experiences of a formula one car and have a cockpit going round a vertical circle (can be horizontal thou) the occupant inside will feel the g- force the same as if they were in a car of that acceleration/deceleration which is expressed as G-forces. i.e 300Kph down to 140 kph in 3 seconds applies a G-Force of 5G's for approx 3 seconds at its maximum.
I want to translate a formula that will allow the computer to push/pull the cockpit around and simulate this affect. But what other pitfalls am i not seeing then.

Thanx for the time by the way

 

[HallsofIvy]

Since the wheel is vertical, at the bottom, both the weight and "centrifugal" force will be downward and will add. At the top, the weight will be downward but the "centrifugal" force will be upward!

 

[Doc Al]

Use a horizontal circle and you avoid the complication of the person's weight. The outward centrifugal force on a person going in a horizontal circle equals m v^2/r. (This "force" is due to being in an accelerating frame.)

 

[golith]

Ok. so m=mass v=velocity and r=radius. The accelerating frame i understand and yes your right about keeping it horizontal,a lthough i have been able to keep the model in my head independent of mass.
So, is there something I am missing in the whole G-force thing as I'm still trying to discover the data required to actually express the answer in units of g-force or is this actually not the correct way of expressing this. if not then the formulas that I am playing with will allow the computer to effectively increase decrease acceleration/velocity to affect the g-force change

was that confusing or what, I've been teaching myself the physics 101 thing and love it, although now applying it to real life is the fun challenge. I used all my chalk on the drive way with lots of diffrent circles and equations to figure it out and found that the circle being perfectly round the angle is always the same therefore the forsce applied is going to be a constant in the equation. therefore the outward and inward force would be equal, or in otherowrds by discovering the inward angle then the outward angle is a simple reflection, or vice versa, this was one of the inputs i was working with to give the g-force

sorry for the waffle :)

 

[Doc Al]

The centrifugal acceleration is just v^2/r; to express that in terms of g, just divide by g: v^2/(rg). (g, of course, is the acceleration due to gravity, equal to about 9.8 m/s^2.)