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.
  • $\begingroup$ Are the particles randomly scattered on the disk? $\endgroup$
    – TheFool
    Feb 1, 2017 at 10:43

1 Answer 1


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. Particle distribution

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).


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