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Required torque to rotate a solid cylinder around its axis.

  1. Nov 12, 2013 #1
    I'm a mechanical engineer and I am more specialized in structural calculations than dynamic calculations and now I'm faced with a basic dynamic problem and would need some help.

    I have a solid cylinder (shaft) that I want to rotate around its axis. The cylinder is supported by two bearings at its ends. No other loads than the weight are applied to the cylinder. I already know how to calculate the torque required to start rotating the cylinder up to the desired speed within the required time. What I want to know is how to calculate the torque required to maintain the cylinder rotating at the desired speed once it reaches it. From what I understood, it is dependent on the friction at the bearings only.

    The formula I have found is T=μ*m*g*R where :
    T : the required torque
    μ : the friction coefficient
    m : the weight of the cylinder
    g : earth gravity
    R : radius at contact bearings/cylinder

    1. I would like to know if the torque calculated using this formula is for only one bearing or is it independent of the number of bearings, in other words, the calculated torque is the final torque or should I multiply it by the number of bearings ?

    2. In the case this formula is independent of the number of bearings, what happens if the bearings are different, shall I consider the bearing with the highest friction coefficient or is there another formula ?

    3. Apart from the frictional torque, is there any other torque I should consider in my calculations of the required torque to maintain the cylinder at the required speed ?

    4. If there are no other torques involved than the frictional torque, once the cylinder has reached its desired speed and if we assume there is no friction at all, does that mean that the cylinder will continue to rotate forever ?

    Now back to the torque required to accelerate the system, the formula I know for accelerating the cylinder up to the desired speed is T=Jα where :

    T : the required torque
    J : moment of inertia of the cylinder
    α : acceleration of the cylinder

    5. During my studies I have only learned to use this formula since we always neglected the friction. Now that I'm faced with a real life case where the friction is very important to be neglected, I'm wondering if friction doesn't intervene in the calculation of the required torque to speed up the cylinder too ? Logically, as friction increases, the torque required to speed the cylinder up to the desired speed increases too, am I right ? If I am, how to calculate this torque ?

    6. Once all torques calculated, what would be the final torque (power) I should consider for the choice of the motor ? Should it be the maximum of the torque required to speed up the cylinder up to the desired speed and the torque required to maintain it at that speed or should it be the sum of the two torques ?

    7. An additional question that is not necessary for my case but would like to have an answer to if possible : what happens if there is a radial load applied on the cylinder in the following cases :
    a. The load is fixed and doesn't rotate with the cylinder ?
    b. The load is rotating with the cylinder ?

    I would appreciate any answer or partial answer to my 7 questions. Also, any links for demonstrating these equations or explaining the phenomenons involved would be appreciated.
     
  2. jcsd
  3. Nov 12, 2013 #2

    Baluncore

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    Science Advisor

    Apart from bearing friction there will be a couple of other things (your point 3) that influence power needed to continue rotation. The first is windage or if the shaft runs through a liquid there will be viscous drag losses. Secondly, the shaft must be doing something apart from rotating, so if for example drops of water or material fall onto it, they will be accelerated before they fall off. That will consume energy.

    Some idea of shaft diameter, length, RPM and application would help to identify energy flow.

    Your point 4. “does that mean that the cylinder will continue to rotate forever ?” is outside the forum rules, they do not permit discussion of perpetual motion machines.
     
  4. Nov 13, 2013 #3
    Hi Baluncore,

    Thank you for the partial answer, the shaft was actually designed and manufactured badly, there are just cloth wires moving over it (over it and not around it) but it is oversized which makes it very heavy and therefore the weight of the tapes very negligible and there aren't actually real bearings, the shaft is supported at its ends in direct contact with the rest of the machine without the use of intermediate bearings which makes the friction coefficient extremely high at the support locations compared to the friction between it and the cloth wires.

    I cannot make any change to the machine, all I need to do is select the right motor but my questions aren't just related to this problem, I want to understand the methodology for general cases, whenever I need to start and maintain a shaft in rotation around its axis because I will have a lot of work to do on that in the future.

    I hope this explains better the case. I'm not interested in actual values, this is why I'm looking for global answers and general formulas.

    Again, I thank you for your partial answers, so in general, I should account for any other kind of friction (drag, fluid friction...)

    You mentioned that if something is falling on the shaft, it will consume energy, are there global formulas to use in that case or shall one rely on simulation software ? Because in my knowledge this case belongs to rapid dynamics (impact) and it is hard if not impossible to solve analytically. Also, how is the energy is consumed ? Is it due to the impact or the friction or both and are there other phenomenons to take into consideration ?

    Finally my point 4 is just a global question, I'm not trying to discuss the perpetual motion of machines, I could have asked the question differently and said, assuming a cylinder is rotating around is axis, what phenomenons prevent it from continuing to turn forever ? But anyway, I googled "perpetual motion machines" and now I have my answer so thank you again.

    What I really need an answer to now is if the number of bearings matter : whether the cylinder is supported at each end by 1 bearing or 2 bearings or more change something to the required torque calculated through the equation :

    T=μ*m*g*R where :

    T : the required torque
    μ : the friction coefficient
    m : the weight of the cylinder
    g : earth gravity
    R : radius at contact bearings/cylinder

    In other words, shall I calculate this formula once for one bearing and use the result directly as the required torque, or should I multiply or divide the result by the number of bearings ?

    Thank you for your help again.
     
  5. Nov 13, 2013 #4
    If you've got it supported at two ends, then each bearing sees half the weight. That's the same, in that formula you've posted, as one bearing seeing the full weight, assuming both bearings have the same coefficient of friction.
    (2 * m/2) = (1 * m)

    Torque will be the summation of the various torques required to move the thing. Static friction initially, then kinetic friction + torque to accelerate and any other misc. bits of friction/resistance. You said:
    This is somewhat, semantically, incorrect. In fact, the torque required to accelerate the mass (and overcome inertia) is the same, but the total torque required is higher because you must constantly overcome friction.
     
  6. Nov 14, 2013 #5
    Hi Travis,

    Thank your for the clarification about the formula, it totally makes sense and it is actually obvious, I should have thought of it.

    Thank you for the explanation.

    I don't quite get it, you say total torque and torque required to accelerate the mass, I thought that at the beginning, there is only one torque which is the torque to accelerate the mass, and once the system has reached the design speed, this torque is no longer required and a new one is needed (the one that needs to overcome friction).
     
  7. Nov 14, 2013 #6
    The system will always have inertia and friction. When you are starting it up, you must overcome the friction that is working against movement, and the object's inertia. When it gets up to speed, that friction will still be present, you'll be working against that in order to maintain the shaft's rotational energy. There's always friction, when you start the thing from a stop it will be a higher component as the static friction coefficient in most systems is typically much higher than the coefficient of kinetic friction while it's moving.
     
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