Torque Required for Rotating a Cylinder

In summary: Welcome to the PF.By "plinth", I take it that the cylinder is vertical, and you want it to rotate about its vertical axis?In any case, the torque required to keep it rotating at a constant angular velocity is zero, except for the torque required to overcome losses like friction.
  • #1
Alvaro Vargas
4
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Hello everyone, I am new to this forum, and here is my first question.

I need some help in a particular case regarding torques. Most of the examples given anywhere on the web and books have rotations which are affected directly by gravity like pulleys and such, but, what if I want to calculate the torque needed for a servomotor that's under a cylinder with a certain mass and radius which makes it to rotate at constant RPM about its axis, not horizontally, but vertically, kind of like a plinth.

Would (T.of.motor)= (mass.of.cylinder* gravity * radius.of.cylinder) in a practical way?
 
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  • #2
Alvaro Vargas said:
Hello everyone, I am new to this forum, and here is my first question.

I need some help in a particular case regarding torques. Most of the examples given anywhere on the web and books have rotations which are affected directly by gravity like pulleys and such, but, what if I want to calculate the torque needed for a servomotor that's under a cylinder with a certain mass and radius which makes it to rotate at constant RPM about its axis, not horizontally, but vertically, kind of like a plinth.

Would (T.of.motor)= (mass.of.cylinder* gravity * radius.of.cylinder) in a practical way?

Welcome to the PF.

By "plinth", I take it that the cylinder is vertical, and you want it to rotate about its vertical axis?

In any case, the torque required to keep it rotating at a constant angular velocity is zero, except for the torque required to overcome losses like friction. The torque that is required to spin it up to speed is the same torque to overcome the losses, plus τ = I * α. Where I is the Moment of Inertia, and α is the angular acceleration. Are you familiar with Moments of Inertia? http://en.wikipedia.org/wiki/Moment_of_inertia
 
  • #3
Exactly, the cylinder is vertical and I want it to rotate about its axis.

I am somewhat familiar, I=(m*r^2)/2 for a solid cylinder, but that's where my dilemma comes in, a servomotor has a pre established maximum torque, the one that I'm using is about 5.1 kg-cm, so if I have that maximum torque and the moment of inertia from the cylinder, does it mean the cylinder will just rotate with an angular acceleration= t/I?

It doesn't make too much sense to me, that would mean any cylinder could rotate regardless of the moment of inertia, just with very low angular accelerations if I is too large.
 
  • #4
Alvaro Vargas said:
Exactly, the cylinder is vertical and I want it to rotate about its axis.

I am somewhat familiar, I=(m*r^2)/2 for a solid cylinder, but that's where my dilemma comes in, a servomotor has a pre established maximum torque, the one that I'm using is about 5.1 kg-cm, so if I have that maximum torque and the moment of inertia from the cylinder, does it mean the cylinder will just rotate with an angular acceleration= t/I?

It doesn't make too much sense to me, that would mean any cylinder could rotate regardless of the moment of inertia, just with very low angular accelerations if I is too large.

That sounds like a pretty small motor. How big/heavy is the cylinder?

If there is no friction or other losses, then a small motor can spin up a heavy cylinder, just pretty slowly. But for real systems, you will have bearing friction, which is usually maximum when the cylinder is not moving (similar to how the static coefficient of friction is generally larger than the dynamic one). And it will be the stall torque of the motor that will have to overcome that bearing friction. Is gearing down the motor an option?
 
  • #5
The weight of the cylinder is about 30 Kg and has a 12 cm radius, I'am also neglecting friction since the drum is over a bushing, maybe that could be interchanged with a safety percentage (I don't know what percentage would be adequate for it), so, if I'm not wrong, and I hope you could redirect me, by the amount of weight I assume I would have to gear down the motor, probably to about a 3:1 ratio to ensure movement. So if I gear it down, the torque would be 5*3 Kg-cm and its maximum rated speed would be 54 RPM/3, and after it I would have to recalculate angular acceleration which would mean a little increase on how fast it starts to move?

That seems more like I'm finding how fast it is going to start moving, not if the motor is strong enough to start moving the drum, am I wrong?
 
  • #6
Alvaro Vargas said:
The weight of the cylinder is about 30 Kg and has a 12 cm radius, I'am also neglecting friction since the drum is over a bushing, maybe that could be interchanged with a safety percentage (I don't know what percentage would be adequate for it), so, if I'm not wrong, and I hope you could redirect me, by the amount of weight I assume I would have to gear down the motor, probably to about a 3:1 ratio to ensure movement. So if I gear it down, the torque would be 5*3 Kg-cm and its maximum rated speed would be 54 RPM/3, and after it I would have to recalculate angular acceleration which would mean a little increase on how fast it starts to move?

That seems more like I'm finding how fast it is going to start moving, not if the motor is strong enough to start moving the drum, am I wrong?

Do you have a motor in mind? Can you post a link to its datasheet? What is its rated stall torque?

Do you have the cylinder and bearings already? You can measure the torque it takes to overcome its static friction, and compare that to the motor's datasheet numbers...
 
  • #7
The datasheet is this one, it doesn't say much besides what I have described before though:

https://www.pololu.com/product/1248/specs

And I'm not using bearings, just a base that makes barely any contact with the bottom part of the drum and it will be well lubricated.
 

FAQ: Torque Required for Rotating a Cylinder

What is torque?

Torque is a measure of the force that causes an object to rotate around an axis. It is often represented by the symbol "τ" and is measured in units of Newton-meters (Nm) in the SI system.

How is torque related to rotating a cylinder?

In the context of rotating a cylinder, torque is the force that is applied to the cylinder to make it rotate around its axis. It is necessary to overcome the inertia and friction of the cylinder in order to achieve rotation.

What factors affect the torque required to rotate a cylinder?

The torque required to rotate a cylinder depends on several factors, including the mass and shape of the cylinder, the speed at which it is rotating, and the type and amount of friction present. The distance between the axis of rotation and the point where the force is applied (known as the lever arm) also plays a role in determining the required torque.

How is torque calculated?

Torque can be calculated by multiplying the force applied to the cylinder by the distance from the axis of rotation to the point where the force is applied. This is known as the lever arm. In equation form, it can be represented as τ = F x r, where τ is torque, F is force, and r is the lever arm.

What are some real-world applications of calculating torque for rotating a cylinder?

The concept of torque and its calculation is important in many fields, including engineering, physics, and mechanics. Some specific examples of real-world applications include calculating the torque required for a motor to rotate a cylinder, determining the torque needed to open a jar lid, and designing gears and pulleys for machinery.

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