sizing DC motor needs

Have a 40 lb wheel that I need to spin at around 3 - 5 rpms. About 20" in diameter. Its for a display. What size/type DC motor would I need? I can figure out the drive system. Thanks

• Go metric, man, go metric. You need to calculate the moment of inertia and then specify the required angular acceleration. This is what's going to determine the power requirements. Friction estimation would also be useful. – Transistor Apr 12 '18 at 17:36

If all the motor needs to do is spin the wheel around 3- 5 RPM then I suggest finding a simple Brushed DC motor with an integrated gear box that spins at that RPM when the rated voltage is applied. When looking for a motor you should make sure the torque the motor outputs can defeat the friction of your drive train and have enough torque left over to accelerate the wheel up to speed in a reasonable amount of time.

Working through your problem: Lets say we want to spin up to 5 RPM in five seconds.

We know that rotational acceleration can be described by

$$\mathrm d \omega = {T \over J}$$

Where:

• $\mathrm d \omega$ is rotational acceleration,
• $T$ is torque applied, and
• $J$ is inertia of load (Your wheel in this case, we can ignore the motor's inertia for it will be very small compared to the load)

We know dw because we want to reach 5 RPM (0.514 Rad/s) from 0 RPM in at most 5 seconds.

$$\mathrm d \omega = {0.514 \over 5} = 0.105\:\mathrm{Rad/(s^2)} = 1\:\mathrm{RPM/s}$$

We can approximate the wheel as a disk of uniform density:

$$J = {Mr^2 \over 2}$$

Where $M$ = Mass (Kg) and $r$ = radius (meters)

$$J = 0.585\:\mathrm{kg\:m^2}$$

The needed torque to accelerate the load then can be calculated to be:

$$T = J \mathrm d \omega = 0.0613\:\mathrm{Nm} = 6.13\:\mathrm{Ncm}$$

But this is only one part of the load the motor will experience. The other part is friction. The friction torque is something you can determine experimentally by seeing how much torque it takes to move the wheel at standstill. Add the frictional torque to the torque needed to accelerate your load and you have the minimum torque your motor needs to be rated for.

Hope this helps.

• Thanks for the edit Phil looks much better, I see now the format help for math is in the LaTex section. – Clipboard_Waving_Enginerd Apr 12 '18 at 18:53

There are insufficient data to answer your question.

To be able to spin the wheel in steady state, the motor must supply only enough power to overcome friction. This is independent of the wheel's weight and size. You have not specified friction.

The time required to accelerate the wheel up to speed will depend on the size of the motor. A larger motor can get the wheel up to speed faster. But the energy required to accelerate the wheel to your target speed depends on the geometry of the wheel, which is also not fully specified. A 40 pound, 20" wheel with the mass concentrated at the edges will require much more energy to accelerate than one with the mass concentrated at the center.

However, given the low speed involved, it's likely the motor you'll require is "small", and energy requirements will be dominated by friction in turning the wheel, and in the drive system. Were I designing such a system, I'd be inclined to whip up a prototype with whatever cheap motor I had lying around, and determine the energy requirements empirically. And I'd add a healthy margin considering a display system is probably running continuously, and you'll want it to be robust even as the bearings wear out, dust accumulates in the moving parts, and so on.