Calculate distance of launched particles from a rotating disk At my education we are busy with creating a small device that can spread thaw grains.
Our plan is as follows:
The particles should be spread by launching them from a fast rotating disk.
The particles are dosed and dropped on the disk.
Now we have to do calculations on this.
The folowing information is given:


*

*The amount of particles is 1 cm3

*Mass of 1 cm3 of particles is 0.0027 kg.

*Radius of the disk is 2.5 cm.

*Height of disk above ground is 0.1 m.


What we want to know:
How fast should the disk rotate to launch the particles 0.3 meter away (horizontally).
What I tried:


*

*Each particle can have its own radius to the centerpoint of the disk. But of course I can take an average.

*I tried to calculate the amount of kinetic energy needed to move the 1 cm3 of particles 0.3 meter away. At a starting height of 0.1 m. But I think I’m wrong.

 A: The below results ignore air resistance, calculating the air resistance on such small particles might be more trouble than it is worth so hopefully this will give a ball-park figure and you can adjust for air resistance through trial and error.
Below is an image of how I imagined your system.

The particles fall from the funnel and drift outwards on the rotating disk. They leave the disk at the edge. 
We know a particle falling straight down will be accelerated by gravity so the air time we have is given by:
$$ T_{air} = \sqrt{\frac{2S}{a}} = \sqrt{\frac{2 * 0.1}{9.81}} $$
The particles leave the disk at the same velocity (horizontal) as the disk's rim.
$$ v_{x} = \frac{2 \pi r}{T_{d}}$$
Where $T_{d}$ is the period of the disk and r is the radius of the disk. We want this velocity to be able to make the particle travel $0.3m - 0.025m$ (the minus because we already travel to the edge of the disk) in our time $T_{air}$ so we have:
$$ \frac{0.3-0.025}{T_{air}} = v_{x} = \frac{2 \pi r}{T_{d}} $$
Rearranging to find the period of rotation (what I think will be most useful to you) we get:
$$ T_{d} = \frac{2\pi r}{0.3-0.025} \sqrt{\frac{2 * 0.1}{g}} = 0.8155...s $$ 
So you want to try to get your disk rotation once ever 0.8 seconds. How accurately you can do this depends on your equipment I suppose.
Assumptions made:


*

*Particles move at the same velocity as the disk at it's edge (they may not due to lack of friction).

*No air resistance.

*Particles don't reach terminal velocity whilst falling down (could increase air time and make them go further).
