T = (1/2)*M*r

^{2}*(n/60)*2*p/t [Nm]

Where

n = speed (rpm)

T = Torque (Nm)

I = moment of inertia

a = angular acceleration (rad/s

^{2})

M = Mass

r = disk radius (m)

t = time to top speed (s)

For instance: If your wheel had an MI of 0.2875kg-m

^{2}and you wanted to spin it up to 16000 rpm in 20 seconds, you would need a motor capable of delivering approximately 24N-m of torque.

Note that although the formula implies that you can use a really small motor if you’re willing to wait a long time to get up to maximum speed, this is not necessarily the case if you the mismatch between the motor and load is too large. That’s because motors will stall if hooked up to a load too heavy to start –static friction in the bearing is one contributor to such a stalling torque which prevents the motor from being able to start up the rotation.

You can also use the above formula to calculate the amount of time it will take for your motor to spin up a load of a specific inertia to a specific speed.

The formula above is derived from the simple equation

T = I*a

We substitute

a = w/ t

I = (1/2)*M*r

^{2}for a cylindrical disk

w = (n/60)*2p

Please be careful to substitute the right formula for the moment of inertia to match the object being spun up, for instance if you were spinning up a sphere, the moment of inertia would have to be changed to (2/5)* M*r

^{2}. For a ring, with most of the mass on the outer edge, the moment of inertia would be M*r

^{2}.

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