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Period of Oscillation of a Cylinder Attached to Two Springs

  1. Aug 13, 2017 #1
    1. The problem statement, all variables and given/known data
    A solid cylinder of mass, M, is connected to two springs of total stiffness, k. The springs are connected tangentially (on top) to the cylinder. The other ends of the springs are attached to walls. What is the period of oscillation of the cylinder assuming that it does not slide on the floor?

    2. Relevant equations
    T= 2π/ω

    3. The attempt at a solution

    So the springs are going to provide a tangential force on the cylinder, which will cause it to rotate. I'm struggling a little bit with the concept of two springs, but I did some googling and it just seems like they act as one spring with a combined spring constant, which is known in this problem. So force of the springs is going to apply a torque:


    Torque can also be expressed in terms of moment of inertia and angular acceleration.


    I can set these two guys equal to each other:


    Moving them all to one side to get a first-order differential equation:


    Angular acceleration is a=Rα, so I can substitute and use x'' for a:

    Ix''/R + kRx = 0

    Then to have x'' on its own, I multiply by R and divide by I:

    x'' + [kR2/I]*x = 0

    Using the moment of inertia of a cylinder: I=0.5MR2:

    x'' + [2kR/MR2]x=x''+ [2k/M]x=0

    This would make my angular frequency, ω, (2k/M)0.5, and I can use T= 2π/ω to get


    I'm concerned that the differential equation should be different because there are two springs involved, but I'm not really sure how. The way I solved the problem resembles how I would solve an oscillation problem with one spring too much :P
  2. jcsd
  3. Aug 13, 2017 #2


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    I'm not sure of the setup. Does the cylinder lie on a surface so that the cylinder can roll on the surface? Something like this?
  4. Aug 13, 2017 #3
    Yes like that!
  5. Aug 13, 2017 #4


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    OK. You will need to consider how to include the constraint of rolling without slipping in your analysis.

    What prevents slipping?

    Start with a carefully drawn free-body diagram for the cylinder.
  6. Aug 13, 2017 #5
    There's probably friction that prevents slipping.

    I thought the condition a=Rα was a condition that was only true in the case of no slipping. Is there another one/is this not the one I need?

    The forces in the x-direction I have in my free-body diagram now are:

    -Force of the Springs (they always act in the same direction)
    -Rolling Friction (always opposite of force of the springs)

    I believe those are the only two forces in play here. They are both tangential forces, so they both apply torque to the cylinder, and they apply torque in the same direction. So would my torque equation be?


    This seems a little odd as friction appears to be helping the motion, but I understand its there and necessary for rolling.

    Do I set this equal to τ=Iα?

    Thanks for replying!
  7. Aug 13, 2017 #6


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    Yes, definitely.

    Yes, that's a correct condition. Of course you need to be clear on the meaning of the acceleration, a, here. In particular, which point of the cylinder has this acceleration, a?
    Rolling without slipping implies a certain relation between the angle of rotation of the cylinder and the amount, x, by which the springs stretch or compress. You seem to have assumed that x'' = a where a is the acceleration in the relation a=Rα. Is that actually the case?

    Yes, if you mean that when one spring exerts a force to the right then the other spring also exerts a force to the right. Of course, sometimes the springs exerts a force to the right and sometimes to the left.
    For me, it's not obvious which way the friction force will act. But, you can assume a direction for the friction in your free-body diagram and then work it out based on this assumption. When you then work it out based on your assumption, you can look at the sign of the answer for the friction force to see what direction it actually points. So, you may assume that the friction acts oppositely to the spring force. Your answer will then tell you if your assumption is correct.

    Yes, the torques would act in the same direction if the forces act in opposite directions.
    Where are the "lever arms"?

    If the spring and friction forces oppose each other, then the toques will help each other.

    Yes (once you get the correct expression for the torque)
  8. Aug 13, 2017 #7
    I would interpret "two springs of total stiffness" in the same way; as the effective stiffness. If, instead, the problem gave two stiffnesses, one for each spring, then it will just be a small additional problem to find the effective stiffness in terms of the two.

    We know that x=0 is an equilibrium position, or else oscillation will not occur about that point. Further, this problem becomes highly complicated for large displacements. So this means somewhere in solving the problem, we will need to eventually make some approximation which takes only the lowest order effect.

    I might be overlooking something, but the problem seems rather simple if you take the ground-contact point to be the origin of rotation. Then only the springs cause torque. This gives me an angular frequency slightly higher than what you got in the OP.

    If you want to take rotations about the center, you will have to account for first order variations in the location of the center of the ball. I just skimmed your OP, so I'm not sure, but I would guess that's the most likely reason our answers differ. (Based on what @TSny just posted (#6) that seems to be what you've overlooked.)
  9. Aug 13, 2017 #8
    Linear acceleration, a, is the acceleration of the center mass.

    This makes a lot of sense actually. If friction acts directly on the axis of rotation, it won't cause any torque or rolling, so I don't have to worry about it. I'll use this approach for this problem.

    Can't a=Rα be used to find the linear acceleration of the center mass from an angular acceleration? If so, the acceleration of the movement mass relates to its movement and displacement (2nd derivative), and the displacement of the cylinder compresses/stretches the springs. Other than this, I'm not sure how to relate α and a.

    Using the point where the cylinder touches the ground as the axis of rotation:

    Finding Inertia using Parallel-Axis Theorem-




    Getting the First-Order Differential:




    T= 2π / (4K/3M)0.5

    Well it seems like my new angular frequency is smaller than my angular frequency in my original post, so I seem to have gone wrong somewhere. Is it my assumption a=αR?

    Thanks guys!
  10. Aug 13, 2017 #9
    Yes that's a nice advantage. There's actually a more important (and probably more subtle) reason for choosing that point. It is the 'instantaneous axis of rotation' for the cylinder. If a ball is rolling without slipping, that is the point which is at rest (in the ground's eyes). This means, for small x, the contact point is pretty much stationary. (A bit more "precisely," the motion of the contact point depends squarely, not linearly, on x. So in taking only linear effects, the motion is ignorable.)

    This is what went wrong with your solution; it's not because you messed up handling friction, but because you ignored the fact that your origin of rotation is itself moving! So when you rotate the cylinder a bit, the center moves closer towards one side, which gives an additional contribution to x, which you left out. (You might convince yourself qualitatively that leaving that effect out should give a smaller frequency.) If we choose the ground, then it is ok that we ignore this contribution, because it's not a "first order effect," it's just a "higher order effect."

    I hope I am not confusing you with this loose explanation!

    You did more in this step than just factor out 1.5MR.... you also input α=x''/R ... was this on purpose? If so what is your reasoning? If not what is the proper relationship?
  11. Aug 13, 2017 #10
    I used the relationship a=αR rearranged; a/R=α. Linear acceleration is the second derivative of position, so I substituted x'' for a to get the First-Order Differential Equation format.

    As far as I know that's the proper relationship 0.o

    So if I wanted to use the center of the cylinder as the axis of rotation, would I need to express the motion of the cylinder with two equations, a translational and rotational one?

    Thanks! I feel like I'm getting close.
  12. Aug 13, 2017 #11
    It is not your relation between α and a which TSny is unhappy with, it is your relation between a (or α) and x'' (and I am unhappy with it too!)

    You must remember x'' represents the displacement of the springs.
    I think it will help to imagine two different methods of "producing x" (that is, two different ways to displace the springs).
    One way is to only rotate the cylinder without moving it's center.
    Another way is to only move the center without rotating the cylinder.
    The case of rolling without slipping "produces x" due to both these effects. By saying x'' = a you are only accounting for one of these effects.
  13. Aug 13, 2017 #12
    We must keep in mind the pictures of what we are doing! If we are taking torques about the ground, then we must also take the angles about that same point... so α represents the the acceleration of the angle with the ground...
  14. Aug 13, 2017 #13
    So currently am I only accounting for the translational or rotational "method" of "producing x?" I'm inclined to say I'm accounting for the rotational movement only right now because I started with τ=Iα, but I'm not positive.

    Would "translational" method be Newton's 2nd Law, F=ma? I could find acceleration from this. Would the "total x-acceleration" be the sum of these two, and this is what is x''?

    So the angular acceleration in the expression τ=Iα is the acceleration of the angle with the ground. That means that the linear acceleration of a point that is distance R away from our axis of rotation is moving with a linear velocity of αR I think? Or does the expression a=αR just not matter here?

    I'm confused at what I've got right and wrong at this point xD
  15. Aug 13, 2017 #14
    Okay, but are the ends of the springs R from the axis of rotation?
    Understandable, since we've been mixing discussions about two different approaches. Sleep on it, reread things if you still aren't clear.
  16. Aug 13, 2017 #15


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    OK. Let's review it. One approach to the problem is to use the general law Στc = Ic α. Here, the torques are about the center of mass, c, of the cylinder. It appeared to me that this was your original approach. There were a couple of mistakes in your first attempt at using this approach.

    One mistake was that you left out the torque that friction produces about the center of mass. As you have suggested, you will need another equation to deal with the additional unknown friction force. You are correct, that the additional equation is ΣFx = mac, where ac is the acceleration of the center of mass.

    Another mistake was writing x'' = a, where the acceleration on the right side is apparently representing ac (the acceleration of the center of the cylinder). But x represents the amount of stretch of the spring. So, x is the displacement of the point of attachment of the spring, not the displacement of the center of the cylinder. It might help to compare the velocity of the top of the cylinder with the velocity of the center:
    For rolling without slipping, how does vA compare to vc ?


    Another approach to the problem was suggested by @Hiero . For rolling without slipping, you can choose the point of contact of the cylinder with the surface, point P, as the origin and set up ΣτP = IP α. (But please keep in mind that this equation would not be valid if the cylinder were slipping; whereas, Στc = Ic α could still be used!) This simplifies things greatly since you don't need to worry about friction in ΣτP = IP α (why?). So you won't need ΣFx = mac. But you will still need to relate x and θ correctly.


    So, you can try either approach.
  17. Aug 13, 2017 #16
    Alright I took a short break from physics and just tried the problem and I'm cautiously optimistic. For now, I'm going to ignore my initial approach and go with Hiero's because it seems simpler, though I'll probably attempt my original way tomorrow :P

    I think my biggest misunderstanding was my concern and focus around the acceleration of the center mass rather than where the spring is in contact with the cylinder.

    The linear velocity at point A should be greater than at point C. Angular acceleration is the same throughout a rotating object, and α is related to a through αr=a; the further away the point from the axis of rotation, the greater the linear acceleration (I'm using small r in the general equation as distance so I don't get it confused with big R, the radius in this problem). This is why me using αR=a is incorrect, because it gets me acceleration of the center mass, but that isn't relevant in this problem. The acceleration and position of the top of the cylinder is what's related to the acceleration and compression/stretching of the springs (x).

    I'm confident I have the correct moment of inertia: IP=0.5MR2+MR2=1.5MR2

    Using τ=Iα:


    Using τ=Fr; in this case the force is applied a distance 2R away from the axis of rotation:


    Set the two torque equations equal to each other and move to one side:

    1.5MR2α + 2kRx

    Now the linear acceleration of the cylinder at the point where its in contact with the spring is α(2R)=a, so a/2R=α

    1.5MR2(a/2R) + 2kRx = 0

    Now I think I can say "a" equals to x'' and isolate x'':

    x'' + (8/3)(k/M)x = 0


    T= 2π / (8k/3M)0.5

    I think I understand this problem now- do you guys see any issues?

    Thanks for sticking with me!
  18. Aug 13, 2017 #17


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    Great. That all looks good to me. (The k in the answer is the total effective spring constant due to both springs.)
  19. Aug 14, 2017 #18
    One of the easiest ways to formulate this problem is to start with the system kinetic energy. This will easily account for both translation and rotation. Then you can add in the springs with a potential energy term. The energy approach is the often overlooked much simply way in most mechanics problems.
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