Calculating g force for flywheel

[wakadarc]

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.

 

[mfb]

h=1/2g*t^2

Where does that formula come from and what does it represent?

angular velocity = 3.14/t = 3.14/0.144s

Why pi/t?

It is not the right approach.

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

 

[wakadarc]

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 I've 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)

 

[mfb]

I am simulating a ball dropping at the top of a circle and the time it takes to reach the bottom

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

Okay, then your approach is right.

 

[wakadarc]

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

 

[mfb]

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.