# Calculating g force for flywheel

1. Nov 3, 2015

I need to determine the RPM just that the flywheel (that has an offset shaft) will travel faster than gravity (g force). Meaning my offset shaft will be falling faster than a falling object per se.

So flywheel has a diameter of 0.1524m.

h=1/2g*t^2
t=0.144s (assuming we did 1.5x(9.8m/s) with h=0.1524m)

angular velocity = 3.14/t = 3.14/0.144s = 21.82 rad/s
Therefore the RPM is 208 RPM (1 rad/s = 9.55 RPM)

Is this the right approach? Thank you.

2. Nov 3, 2015

### Staff: Mentor

Where does that formula come from and what does it represent?
Why pi/t?

It is not the right approach.

There is a simple formula for centrifugal/centripetal forces, you can directly use it.

3. Nov 3, 2015

I am simulating a ball dropping at the top of a circle and the time it takes to reach the bottom, hence using pi. The ball is under the g force.

But ive seen this
http://www.calctool.org/CALC/phys/newtonian/centrifugal

This calculates the centrifugal force but i need the vertical component to be greater than g (downward gravity force)

4. Nov 3, 2015

### Staff: Mentor

Ah, then I misunderstood your first post. I thought you were interested in the instantaneous acceleration.

Okay, then your approach is right.

5. Nov 3, 2015

So the offset shaft will fall (or in reality, rotate about the center) at 1.5gs if I have 208 RPM?

The centrifugal approach from the link I posted has it higher, 600+ RPM

6. Nov 3, 2015

### Staff: Mentor

I checked the numbers. With your diameter, I get 0.176 seconds of free-fall time. That corresponds to 17.823/s or 170.2 rpm.

Using those numbers, the tool gives 2.47 g acceleration. That looks reasonable. Don't forget to convert diameter to radius.