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Calculating G-Force

 
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Jun3-12, 02:38 PM   #1
 

Calculating G-Force


How do you calculate G-Force? I know that one g is the force gravity exerts on a particular mass (I think). But how exactly do you figure it out, particularly with objects spinning in a circle, like a merry-go-round?
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Jun3-12, 02:39 PM   #2
 
V^2/R

Velocity in m/s with that number squared divided by the radius in meters.
Jun3-12, 02:43 PM   #3
 
Quote by Ferraridude View Post
V^2/R

Velocity in m/s with that number squared divided by the radius in meters.
So, G-Force is the same as centripetal acceleration?
Jun3-12, 09:21 PM   #4
 

Calculating G-Force


Quote by SignSeeker7 View Post
So, G-Force is the same as centripetal acceleration?
"G force" is just the ratio of a given force to the force due to gravity on the same object. Think of it as a unit of measure, it's not a special kind of force. Typically, "G forces" are referring to pseudo forces experienced in an accelerating reference frame.

Now, if you wanted to know the gravitational attraction between two objects, then you'd use Newton's law of gravitation.
Jun3-12, 10:29 PM   #5
 
Quote by SignSeeker7 View Post
So, G-Force is the same as centripetal acceleration?
Yes, if you are going in a circle, or even changing direction. You can also get g force from linear acceleration. If you were in a rocket that could accelerate at 4 g's straight up, or 39.2 m/sec^2 you would be subject to 5 g's. Four from the rocket and one from the Earth.
Jun4-12, 07:47 PM   #6
 
If you were going around a merry go round with a radius of 15 meters at a velocity of 8 m/s, this is how you would calculate the G-forces.

8^2=64
64/15=approximately 2.67.

So now, you have your m/s^2 for the acceleration.

2.67/9.8 m/s^2 (9.8 m/s^2 is the acceleration of the Earth's gravity)=.2724

So, we can now see that the centripetal force on a merry go round rider with that velocity around a circle with that radius, we are feeling .2724 G's of acceleration, or 27.24% of Earth's gravitational pull.
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