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A Rope with a mass

  1. Nov 15, 2012 #1
    1. The problem statement, all variables and given/known data

    Given a rope with a constant length L, and a mass M.
    The rope is connected in one side, and moving with angular velocity on a frictionless surface.
    What is the tension along the rope?

    Hint:
    Look at a small mass element of the rope and consider the forces acting on it, and therefore arrive at a differential equation.

    2. Relevant equations



    3. The attempt at a solution
    Im a bit new to this way of looking at problems by looking at them in very small parts, but it does look beautiful, so any help with getting me acquainted with it will be much appreciated! I do have the math needed, but not the intuition of how to break a problem to its infi parts.

    So, since an angular velocity is mentioned, I obviously thought about a circular movement. The rope is moving along the circumference of a circle. Which means a force F must be applied to its connected end, and the force pointing to the center of the circle.
    On the other hand, when I look at a very small element of rope, its a particle moving in circular motion. But what are the forces acting on it? F would act only at the connected end, wouldnt it? So the force acting on the particle is T1 towards one side tangential to the circle, and T2 towards the other tangential side? How does that explain it's circular motion? (he must have some force towards the center, not on the tangential axis)

    What force causes the circular motion when looking at a very small mass particle?

    Any help would be much appreciated!
     
  2. jcsd
  3. Nov 16, 2012 #2

    Simon Bridge

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    Welcome to PF;

    You have done free-body diagrams before?
    Model the rope as lots of little masses separated by massless ideal ropes just like in those diagrams only where you had only two or three masses, here you have lots of them.

    The force F is only applied to the mass nearest the center - yes.
    But that first mass has a tension force the other way.
     
  4. Nov 16, 2012 #3
    Hi!
    Thanks for the response!
    I did do free body diagrams before, but those were very simple. I never did one where you need to consider infinitely many small elements. My my issue is with the specifics of breaking this problem to the little masses you mentioned. I know its the way to go, I just dont know how to do it.

    I added a picture from which I dont know how to continue.
     

    Attached Files:

  5. Nov 16, 2012 #4

    haruspex

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    The statement of the problem is unclear to me. What does it mean "is connected in one side"? Does it form a circle? If so, there's no dT. The tension will be constant. Instead, consider dθ and dm. You have a mass dm = ρrdθ, two instances of T in almost opposite directions (difference being dθ), and a resultant centripetal force.
     
  6. Nov 16, 2012 #5
    Hi! Thanks so much!
    Would you mind explaining a bit more about the equation you wrote, dm = ρrdθ?
    How did you arrive at it? What does p,r represent?
     
  7. Nov 16, 2012 #6

    haruspex

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    Sorry, I hoped the choice of letters was sufficiently standard as to be self-explanatory.
    ρ is the density (mass per unit length) of the rope.
    In polar co-ordinates centred on the centre of the circle, (r, θ) is the location of an element of the rope subtending dθ at the origin. Its length is therefore rdθ and its mass, dm, is ρrdθ.
    The centripetal force will bisect dθ, and will be provided by the radial component of the tension (each side of the element).
    Can you obtain an F=ma equation from that?
     
  8. Nov 17, 2012 #7
    Thank you! So, if I have dm = ρrdθ I have F=ρ*dθ* r^2 * w^2, right? And there are 2 tensions causing the force F, which is T*d0/2 (I added a picture of finding F, do you mind telling me if Im on the right track?).
    So 2*ρ*r^2*w^2=T?
    Where is the expected differential equation?...

    And also, theoretically speaking, how do we know that the tension is constant in the rope?
     

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    Last edited: Nov 17, 2012
  9. Nov 17, 2012 #8
    Is it possible that the question is refering to this scenario (in the added picture),

    and therefore:

    dm/dL = p
    T-(T+dT)=p*dL*r*w^2
    -dT=p*dL*r*w^2
    and then somehow replacing dL with dr to get T as a function of r?
     

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    Last edited: Nov 17, 2012
  10. Nov 17, 2012 #9

    haruspex

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    Yes, I think that may be the intent. OTOH, it means the tension is not constant but is a function of distance from the attached end. "What is the tension along the rope?" makes it sound as though it is one value.
    Yes - except that I would have written dr everywhere you've written dL. L is the total length. Whether it's + or - depends which end you're measuring from. I would measure from the fixed end, so tension should increase with distance.
     
  11. Nov 17, 2012 #10
    Edit: Uh, sorry. Misread the response :D
     
  12. Nov 17, 2012 #11

    haruspex

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    As I said, I would not have written dL in the first place. If r is the distance of an element from the fixed end, the length of the element is dr, so it's dm = p dr.
     
  13. Nov 17, 2012 #12
    Thank you! So, is this correct?
    dm = p*dr
    T-(T+dT)=dm * r * w^2
    T-(T+dT)=p*dr* r * w^2
    dT/dr=-p*r*w^2

    T=-0.5r^2 * p * w^2 + C

    Is this correct? And... how do I determine C?
     
    Last edited: Nov 17, 2012
  14. Nov 17, 2012 #13

    haruspex

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    Looks good. To determine C you need a boundary condition. Is there some value of r where you know what the tension must be?
     
  15. Nov 17, 2012 #14
    Nope, there is no such boundary condition. However, perhaps I can know something about the tension in the farthest point of the rope? The length is L, so I can write an f=mLw^2 equation there, if only I knew what m to apply. Its not the entire mass of the rope, right?
     
  16. Nov 17, 2012 #15

    ehild

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    What forces act to the small piece of rope at the farthest point of the rope? Is it something pulling it outward?

    ehild
     
  17. Nov 18, 2012 #16
    I think that on the outer point there is only T, the max T, inward. Causing the circular motion.
     
  18. Nov 18, 2012 #17

    ehild

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    At the end of the rope, only the inward tension acts on a small piece of the rope. So dmRω2=T(R). dm can be arbitrary small. What about T(R)?

    ehild
     

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  19. Nov 18, 2012 #18
    Well, according to the obtained equation, T(L)=-0.5L^2 * p * w^2 + C.
    And as you say, considering the forces, -dmLw^2=T(L). And p is M/L, and dm=p*dL
    So:
    dmLw^2 = 0.5L^2 * p * w^2 + C
    dmLw^2 = 0.5L * M * w^2 + C
    C=dmLw^2-0.5L * M * w^2
    C=Lw^2(dm-0.5M)=-0.5MLw^2

    ?
     
  20. Nov 18, 2012 #19

    haruspex

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    Dropped a minus sign.
     
  21. Nov 18, 2012 #20
    I did that intentionally, since both Ts should be in the same direction. Is that wrong?
     
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