# Vertical Centripetal Force to push an object

1. Oct 5, 2012

### Vonnesy

I have this device, It has 3 masses 30 grams each (Red ones) the other masses are not considered. The masses are spinning at 600 rpm at that velocity this will reach maximun radius. The maximun radius the masses and reach due to the centripetal force is 15cm. Can this device push a 3kg mass at 600 rpm and if not what speed should the masses be spinning in order to push 3kg or how much must the masses be in order for them to push 3kg at 600 rpm

I tried calculating the centripetal force by taking the 3 masses as one so just considered 90 grams object.

I am kind of lost and need some guidance.

I attatched an image of what the device might look like. This is an idea I am working on and I want to find out if it could work as expected I need to use centripetar force to push and object.

File size:
15.9 KB
Views:
135
2. Oct 5, 2012

### Staff: Mentor

Can you elaborate about how you are trying to convert a spinning circular motion into a pushing linear force?

3. Oct 21, 2012

### Vonnesy

Well basically the red objects have some weight and as they spin they will be pushed upwards and pushing upwards will make the bell crank turn and the green objects will push the shaft

4. Oct 21, 2012

### Vonnesy

This is what i mean.

Radial forces are an important component of the process that keeps any object traveling in a circular orbit. When a known mass (such as a ball) spins around in a circle a set distance (the radius) from the center point at a constant speed, the radial force pushes the mass out, away from the center. "http://www.wisegeek.com/what-is-radial-force.htm"

5. Oct 21, 2012

It's kind of hard to understand what you're after. Do you have a desired radial velocity and want to see if the motor you are using is efficient for this speed?

6. Oct 21, 2012

### Vonnesy

Not really I am trying to push a turbine's tail to overall to "control" a yaw system. The axial force come as a result of the centripetal force on the bell crank. I have to calculate the axial force exerted at 600 rpm and see if that force can overcome the wind load on the tail

7. Oct 21, 2012

### AlephZero

Sorry, but I can't figure out what your mechanism is supposed to do.

Can you draw two pictures to show it stationary and rotating, so we can see what moves where?

8. Oct 21, 2012

### Vonnesy

Ok so this is how it looks...

#### Attached Files:

File size:
15.5 KB
Views:
102
• ###### rotating.JPG
File size:
15.4 KB
Views:
129
9. Oct 22, 2012

### AlephZero

OK. It's like a Watt centrifugal steam engine governor, turned on its side, and presumably with some sort of "return spring" force pushing the rod to the left when it slows down again to move the weights back in.

You need to take moments about the pivot of the L-shaped bracket.
When it has rotated out through some angle $\theta$ between 0 and 90 degrees, the masses will be at a radius $r_0 + L\sin\theta$, so you will have a radial force $m(r_0 + L\sin\theta)\omega^2$ at an axial offset $L\cos\theta$ from the pivot. This is balanced by the pushing force F at a radial offset of $L \sin\theta$.

You will find the maximum push force when $\theta$ is somewhere between 0 and 90 degrees, depending on the relative lengths of $r_0$ (the radius to the pivot points) and $L$ (the length of the arms of the L-shaped piece).

At 90 degrees, the pushing force will be be zero.

The above ignores any friction between the arms and the end of the rod, as the arms pivot.