Need to choose DC motor with correct specs

AI Thread Summary
To select a suitable DC motor for rotating a 15kg aluminum cylinder at a constant speed of 3000 RPM, it's essential to consider both torque and voltage specifications. The torque required can be calculated using the moment of inertia of the cylinder, while the back EMF constant (Ke) will determine the relationship between applied voltage and speed. Applying a voltage of around 90V is advisable to minimize power losses, but the motor must also provide sufficient torque at this speed. Understanding the torque-speed curve from the motor's data sheet is crucial, as it indicates whether the motor can maintain the necessary torque at 3000 RPM. Properly matching these specifications will ensure the motor can effectively drive the cylinder.
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Hi everyone,

As part of a project, I need to rotate a horizontal solid aluminium cylinder of approximately 15kg mass, 15cm radius and 8cm height such that its surface speed is a constant 160km/h. I need a D.C motor capable of achieving this. The acceleration of the wheel does not matter much - around 1-2m/s^2 is okay.

I know I need to convert the speed into RPM, and that I've done: approximately 3000RPM, but what reading should I look for to know whether the DC Motor will be able to supply this output of 3000RPM? I don't think it is as simple as getting one that reads 3000RPM, because there is the torque rating, the voltage, the current, etc. If possible, I want to use around 90V because it will reduce power losses in the motor. But what's the link between HP and speed of a motor? It's quite confusing. As you are aware by now, I'm quite a beginner at mechanics.

Kindly give me some guidelines or suggested specifications, or tell me if I need some more starting information.

I just need to know how I can select a motor suitable to drive this cylinder around at 3000RPM.

Any help would be greatly appreciated.
 
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Welcome to PF:

Here are some rules of thumb for a DC motor that I have learned:

The motor torque τ is proportional to the current I in the windings: τ = KtI, where Kt is the torque constant, measured in Nm/A or oz-in/A. The torque constant is usually given in the motor data sheet.

So, if you figure out the moment of inertia of your cylinder, then you know what angular acceleration a given torque (and hence motor current) will provide. This matters if you want to reach the 3000 rpm in specific amount of time.

But there's more to it than that. The second rule of thumb is the that back EMF VBEMF of the motor is proportional to the speed ω of the rotor: VBEMF = Keω. The constant Ke is called the back EMF constant, it is usually measured in V/krpm (volts per kilo-rpm) and is also usually given in the motor data sheet. The back EMF is an induced voltage in the winding that opposes the applied voltage.

So, let's say you apply a voltage of V0 across the motor windings (90 V in your case). Then Ohm's law says that the voltage across the windings is equal to the current times the resistance of the windings (the winding resistance is also usually given in the data sheet):

V0 - VBEMF = IR

V0 - Keω = IR

From this equation, you can see that the final speed that the motor will spin up to is entirely determined by what voltage you apply across the windings. This is because, when Keω = V0, the back EMF will be equal to the applied voltage, and hence the total voltage across the windings will be 0. If the voltage across the windings is 0, then the current will be 0. If the current is 0, then there will be no more torque, and hence the motor cannot speed up any further. So, the higher you make V0, the faster the motor will spin when it reaches its final speed. You need to select V0 such that you can reach a speed of 3000 rpm given the back EMF constant of your motor. Keep in mind also that the final speed reached by the motor will be slightly slower than V0/Ke, because there always has to be *some* current in the windings in order to overcome whatever frictional torques may be present in the system.

You can also see from these equations that you would expect a linear torque-speed curve at any given applied voltage, since I = V0/R - Keω/R, and τ is proportional to I. In other words, the amount of torque that you can get out of the motor decreases linearly as the rotation speed increases.

In reality, the torque speed curve may have a more complicated shape than this, because the rules I have given above may only be approximations to the true behaviour of the motor. A motor data sheet usually has a graph of the torque vs. speed curve in it. This will at least give you a sense of whether the motor can give the required performance. If the torque falls off to 0 at a speed way less than 3000 rpm, then you probably won't be able to accelerate up to 3000 rpm!
 
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