Massive rotating drum on bearings with friction

In summary, the conversation discusses determining the coefficient of friction between the bearings and the drum in a setup where a mass m' is suspended from a massless rope wound around a drum of mass m. The solution includes adding a force couple to account for the rotation motion and the resulting force exerted by the drum on the bearing. The question arises whether this approach is correct and if it takes into account the additional torque produced by friction. Clarification is provided that the friction exists between the inner hole of the drum and the bearings. Suggestions are made to draw the friction force in a specific direction and to solve for variables such as angular acceleration and force on the shaft.
  • #1
erore
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I came across the following example. I have a mass [itex]m'[/itex] that is suspended from a massless rope which is wound around a drum of mass [itex]m[/itex]. The radius of the drum is [itex]R[/itex]. The drum is fixed on a shaft with bearings of radius [itex]r[/itex]. I am supposed to determine coefficient of friction between the bearings and the drum but for now I am stuck with the forces diagram.
I created something like that:
attachment.php?attachmentid=51334&stc=1&d=1349007090.gif


In my solution I add a pair of forces of the same magnitude and opposite direction [itex]F^{T''}[/itex] and [itex]F^{T'}[/itex]. [itex]F^{T''}[/itex] and [itex]F^{T}[/itex] make a force couple which generates some momentum about the axis of the drum (which contributes to rotation motion). However, [itex]F^{T'}[/itex] acts on the center of mass of the drum and adds up with gravitation force. Thus, the resulting force exerted by the drum on the bearing is bigger than if there was no [itex]m'[/itex]. Consequently the friction is bigger too.

My question is whether this is correct. The solution I have only calculates the torque due to [itex]m'[/itex] and does not introduce the force couple therefore it does not have the additional force on the center of mass and consequentially no additional item in the normal force and friction respectively.

I am also curious, the friction itself produces a torque, if my reasoning above is correct, it would create another force couple, giving rise yet to another force acting on the center of mass of the drum which could go on and on. This is confusing for me at the moment.
 

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  • #2
Friction between what and what? Is the drum attached to something or in contact with something? That will somehow balance the force FT (otherwise the whole setup accelerates downwards).
 
  • #3
I am sorry, the first paragraph was nonsense. I corrected. Basically, there is a hole in the middle of the drum of radius r and a shaft with bearings is inserted into that hole. The friction I mention is between the inner hole of the drum and the bearings.
 
  • #4
Ah. I think I would draw the friction force at the upper side and in the other direction.
Afterwards, I would add variables for the angular acceleration and the force on the shaft and solve for those.
 
  • #5


I can confirm that your reasoning is correct. The additional force couple generated by F^{T''} and F^{T'} does indeed contribute to the overall torque on the drum and can affect the normal force and friction between the bearings and the drum. This is because the force couple creates a moment arm, which can increase the torque and therefore the friction force.

Additionally, the friction force itself can indeed create another force couple and contribute to the overall torque. This can create a feedback loop where the friction force increases the torque, which in turn increases the friction force, and so on. This is a common phenomenon in rotating systems and is known as self-excited vibration.

In order to accurately determine the coefficient of friction between the bearings and the drum, it is important to consider all of these forces and their effects on the system. Your approach of including the force couple in your force diagram is a valid and necessary step in understanding the dynamics of this system.
 

FAQ: Massive rotating drum on bearings with friction

1. What is the purpose of a massive rotating drum on bearings with friction?

The purpose of a massive rotating drum on bearings with friction is to store and transport large amounts of material. The rotation of the drum helps to mix and evenly distribute the material, while the bearings and friction reduce the amount of energy needed to rotate the drum.

2. How does the friction in the bearings affect the rotation of the drum?

The friction in the bearings helps to slow down the rotation of the drum, which is important for controlling the speed and stability of the drum. It also helps to evenly distribute the weight of the material inside the drum, preventing any imbalances that could affect the rotation.

3. What are the benefits of using bearings in a massive rotating drum?

Using bearings in a massive rotating drum allows for smoother and more efficient rotation. The bearings reduce the amount of energy needed to rotate the drum, making it easier to operate and saving on energy costs. They also help to evenly distribute the weight of the material, preventing any damage to the drum or its components.

4. How is the speed of rotation controlled in a massive rotating drum?

The speed of rotation in a massive rotating drum can be controlled in several ways. One method is by adjusting the torque or force applied to the drum. Another method is by using gears or pulleys to change the speed of rotation. Additionally, the friction in the bearings can also be adjusted to control the speed of rotation.

5. What are the potential drawbacks of using a massive rotating drum on bearings with friction?

The main potential drawback of using a massive rotating drum on bearings with friction is the wear and tear on the bearings. Over time, the friction and pressure can cause the bearings to wear out, leading to increased maintenance and replacement costs. Additionally, if the bearings are not properly maintained, it can affect the stability and efficiency of the drum's rotation.

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