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Homework Help: Tension of rod connecting two cylinders

  1. Jul 27, 2016 #1
    1. The problem statement, all variables and given/known data

    Two cyllinders of equal size and of mass M (one hollow, one solid) are connected along their common axis by a light rod and placed on an inclined plane with an incline of α (both cylinders are on the same height initially). The system of cylinders then begins to roll down the inclined plane without skidding. Find the tension in the rod.

    Given variables: M, α, g (gravitational acceleration)

    2. The attempt at a solution and the problems I'm facing

    My biggest problem is imagining what causes the tension in the first place. My guess is, due to the cylinders having equal mass, but different moments of inertia, their rate of descent would be differ if they weren't connected by the rod (the hollow one would roll slower because of a larger moment of inertia). So since they are connected, the difference of their altitudes creates the tension, which makes the system roll down in a curved trajectory. Is that where tension arises from? Cause otherwise (if the cylinders were identical) I would think that the tension would be equal to 0.
  2. jcsd
  3. Jul 27, 2016 #2
    Can you paint image?
  4. Jul 27, 2016 #3
  5. Jul 27, 2016 #4
    You needn't to find where tension arises from. You just care when tension arises, the rod is always strained. And the acceleration of 2 points connected with 2 cylinder along the rod will be equal.
  6. Jul 27, 2016 #5
    The goal you prescribed is indeed more useful to me than the one i prescribed :D But the question remains very similar - does tension arise from the difference in altitude of the cylinders?
  7. Jul 27, 2016 #6
    Yes, I think the tension isnt constant.
  8. Jul 27, 2016 #7
  9. Jul 27, 2016 #8
    But the model that the system would roll down the inclined plane in a curved trajectory (as shown in my drawing) is correct?
  10. Jul 27, 2016 #9
    Well, kind of. Since tension arises from the stretching of the rod, you should calculate how much the rod has changed its length, which isn't equal to calculating what is the difference in the heights of the cylinders. Check the image:
    The change in length is simply c(t)-a=(a^2+b^2)^1/2 -a. You can also choose to work with vectors, so the length difference will be the magnitude of the vector(c-a)=b(t) which as you can see is not necessarily the difference in heights because the difference in heights is just one of the components of the vector b(t).
    But all this assumes that the two cylinder's trajectories don't affect each other..

    So, what troubles me is that I would expect that since there is a force between the two cylinders(the tension), wouldn't that mean that their trajectories shouldn't necessarily be straight lines? I think that the tension between them should result in curved trajectories.

    Please, somebody correct me if I am wrong somewhere!

    Attached Files:

  11. Jul 27, 2016 #10
    I'm quite sure we are dealing with a model where the rod that does not expand
  12. Jul 27, 2016 #11
    Oh, OK. In that case, yes, the trajectory will be the one in your drawing but since there is no friction, why would the left body(in your drawing) move in a trajectory that implies that the centripetal force is in the same direction as the centripetal force acting on the right cylinder? The centripetal force acting on both cylinders is equal in magnitude but opposite in direction(it's the tension). So, I think that the drawing is not exact. What do you say?
    Also, your explanation at the start of this discussion is correct, but I only disagree in the part that it is the height difference that causes the force. Why do you say that it is exactly the height and not other distance measured on the plane?
  13. Jul 27, 2016 #12
    First of all, there is friction - static friction. It does not do any work but it exists nonetheless. To answer the question about the centripetal force - it's a good point, but how else would the cylinders roll without skidding if they didnt follow the same trajectory?
    I now think there's a better way to explain the cause of the tension. It's not a specific distance, but rather the fact that the two cylinders would move differently if they were not connected - there must exist a force then that balances the motion and makes the angular velocities of both the cylinders equal.
  14. Jul 27, 2016 #13
    Oh, I thought there was no friction. So, in this case, you are indeed correct and it's the friction that causes the trajectories with the same angular momentum. Also, about the distance, that was what confused me. You talked about distances so I thought that you managed a partial solution that involved distances.

    You said "there must exist a force then that balances the motion and makes the angular velocities of both the cylinders equal.". Well, there you have it. Since there is static friction, then you already know the trajectory. So, you know that their angular momentum is the same, so now you just have to calculate the tension through the equations of motion of the two bodies since you already know the component of the force of gravity along the plane for each body. I mean, since your intuition is right, use the trajectory as a given(equal angular momentums). Did I get it wrong somewhere?
  15. Jul 27, 2016 #14
    What you said all sounds nice and straight-forward, but I only know the masses and the angle of incline. I do not know the trajectory, I only drew a rough sketch of it. And for clarity, could you add in the equations that arise from your claims? Without the equations, what you are saying does not seem to help the solving process. Thanks!
  16. Jul 27, 2016 #15
    Well, I am just trying to help. In your question you said "My biggest problem is imagining what causes the tension in the first place". This implies that you need help with it conceptually. Typically, whatever help is provided to somebody, it should be appreciated! Thanks!
  17. Jul 27, 2016 #16
    Indeed it should, and while I did sound quite unappreciative of your help, I am very thankful for your input. It's just that your conceptual advice hasn't quite made me understand this problem to the point where I could see a way to describe everything mathematically, and I hoped you could help with that :) Sorry if I came off as rude or unappreciative
  18. Jul 27, 2016 #17


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    If there is no skidding at all, the trajectory will not be curved. Moreover, even if the trajectory were curved, I agree that (a) there is no tension in the rod initially and (b) it would thereafter change with time, so does not permit a single answer.

    I strongly suspect the question was supposed to ask for the torsion in the rod, not the tension.
  19. Jul 28, 2016 #18
    In the language that the problem was written in, it strictly says to find a force.

    However, if we imply that the rod with torsion creates a torque on the cylinders (one along the direction of rotation, the other against the direction of rotation), then I think we can find a quantity T / R (ratio of torque over the radius of the cylinder).

    Newtons second law of translational and rotational motion for both the bodies:

    Hollow cylinder:

    Mg sin α - Ffr1 = Ma
    Ffr1R + T = Ihollowβ=MaR

    Solid cylinder:

    Mg sinα - Ffr2 = Ma
    Ffr2R - T = Isolidβ = ½MaR,

    where Ffr1 and Ffr2 are the friction forces.

    From the 4 equations I obtained that T / R = 3/2 * mg sinα

    Are the equations correct? Is this quantity even useful?
  20. Jul 28, 2016 #19


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    I agree with the preceding equations but get a different answer.
  21. Jul 28, 2016 #20
    zemaitistrys, I agree with you that what causes a [I am going to use the word 'force'] to be exerted on the rod is due to the different moments of inertia. I also agree where haruspex stated, "if there is no skidding at all, the trajectory will not be curved". The statement in the problem about there being no tension initially, I believe that statement is there to indicate that there is no initial frictional force acting on the two cylinders from the inclined plane surface BEFORE it is released. I would assume that there is no stretching of the rod. So once this dumbbell is released, I believe the rod will be applying a constant torque to each of the cylinders - a torque on the hollow one to cause it to 'speed up', and on the solid one to cause it to 'slow down'.
  22. Jul 28, 2016 #21


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    That would appear to be the basis of @zemaitistrys' equations, but I get a different result from them.
  23. Jul 28, 2016 #22
    I started from scratch and, using zemaitistrys notation, I got T/R = (1/7)mg sinα
    Knowing how prone I am to making silly mistakes, I don't have much confidence in that answer.
  24. Jul 28, 2016 #23


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    That's what I got by a slightly different route. I took moments about the points of contact with the plane so that I did not care about the frictional forces.
  25. Jul 29, 2016 #24
    Okay, haruspex, here's my confession: Like I said, I am prone to making silly mistakes - and I did so in my solution to this problem. I realized it after I was done when I went back and looked at zemaitistrys' equations. Somehow, without it being a conscious decision, I used the same friction force (the same variable) for both disks. However, because of the symmetry of the mass, it turned out that the frictional forces are indeed equal. It sure would be nice if I could say it was intuitively obvious to me that the friction forces were equal and that's why I did it, but that's not what happened. :(
  26. Jul 29, 2016 #25


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    @zemaitistrys deduced they were the same from the requirement that the linear accelerations are the same.
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