# Calculation for Motor Selection

I need help with the selection of a motor and a motor driver

I need to spin a 700g (65mm radius) round ball at 35,000 RPM. I need to bring it up to speed slowly, so I also need a motor driver/speed controller and something that tells me what the current rpm is. I have attached an image for reference.

How should I know what torque or power the motor should have?

• Is the ball situated on the motor axis or is it located on an arm extending from the axis? Jul 27 at 12:53
• @fred_dot_u It is located on the motor axis. Jul 27 at 13:04
• your specification is insufficient. What is needed is the acceleration you want to give the device (how long until top speed), and then we need to know what the steady state drag on the system is (what is the power to keep it going). While we're at it, the hard thing here is building a device that will rotate at such a high rate without shaking itself apart. Jul 28 at 4:09
• As Tiger guy said, the hard part will be balancing this properly. If the mass is offset by only 0.1mm at 35,000 RPM, the unbalance forces will be about 1kN. You also need to check the shaft whirling speeds, otherwise your first run will end with a bang, and a 700g sphere flying across the room looking for something it can break (and that something might be you, if you are in the way). Jul 28 at 21:37
• You also need to realize that the maximum linear speed of the surface of the sphere will be about Mach 0.8. The aerodynamic drag will be large and the device will be the center of a small whirlwind while it is running, unless you plan to operate the device in a vacuum chamber. Jul 28 at 21:45

The text in your image could be improved by making it lighter than the background, but that's not important to the solution.

By using "torque required for rotating mass" as the search terms, I found many resources to assist your goal. Unfortunately, most of them use Imperial measurements and it appears that you're using the metric system. Not to worry, Torque for Dummies (paraphrased) has the calculations in metric.

The solution from the linked page is aimed to calculate torque for a spinning disk, not a sphere, but I suspect the mass portion of the calculation and the diameter/radius are the critical inputs.

Say that a DVD has a mass of 30 grams and a diameter of 12 centimeters. It starts at 700 revolutions per second when you first hit play and winds down to about 200 revolutions per second at the end of the DVD 50 minutes later. What’s the average torque needed to create this acceleration? You start with the torque equation:

A DVD is a disk shape rotating around its center, which means that its moment of inertia is

Thanks to NMech for the correction for the rotational inertial of a sphere, maintaining the original link information above.

$$\frac{2}{5} m r^2$$

The diameter of the DVD is 12 centimeters, so the radius is 6.0 centimeters. Putting in the numbers gives you the moment of inertia:

Here’s the angular equivalent of the equation for linear acceleration:

But because the angular velocity always stays along the same axis, you can consider just the components of the angular velocity and angular acceleration along this axis. They are then related by

First, you need to express angular velocity in radians per second, not revolutions per second. You know that the initial angular velocity is 700 revolutions per second, so in terms of radians per second, you get

Similarly, you can get the final angular velocity this way:

Now you can plug the angular velocities and time into the angular acceleration formula:

The angular acceleration is negative because the angular velocity of the disk decreased. The negative acceleration then leads to a reduction in this angular velocity.

You’ve found the moment of inertia and the angular acceleration, so now you can plug those values into the torque equation:

Obviously, you'll have to replace the example values with your own. This is left as an exercise to the reader.

Note: all quoted text and formulae images directly imported from linked site.

• minor improvement The moment of inertia for a solid sphere is $\frac{2}{5} m R^2$ . Jul 27 at 14:06
• the other thing I am worried is that the rpms are really high, so you might have additional friction, which might need additional torque to overcome. Jul 27 at 14:08
• @NMech, I don't know how to incorporate your comment formula into the post. If you have edit capabilities, please add that in. If not, tell me now to place it in the body. Jul 27 at 15:06
• One must hope that the OP recognizes the difference between the quoted formulae and the added correction and incorporates it into the calculations. Thank you for the corrective edits. Jul 27 at 16:58
• This ignores the aerodynamic drag on the sphere and the fact that the sphere will act like a centrifugal air pump drawing air in from the "poles" and throwing it out at the "equator". The aerodynamic effects will completely swamp the torque required to accelerate the mass and inertia of the sphere. Jul 28 at 21:49