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Conceptual doubt in a rotating ring

  1. Oct 29, 2014 #1
    1. The problem statement, all variables and given/known data

    A metal ring of mass m and radius R is placed on a smooth horizontal table and is set rotating about its own axis with a constant angular speed ω. What is the tension in the ring ?


    2. Relevant equations


    3. The attempt at a solution

    Consider a small element ds=rdθ .Tension T acts at the two ends .If I resolve tension in horizontal and vertical components Tcosdθ/2 and Tsindθ/2 respectively , the horizontal components cancel and the vertical components add up . The net vertical component F = 2Tsindθ/2 .

    Now here I am little unsure how this F is acting as the component of force providing the required centripetal acceleration to this element ds .It is the direction of this force which is troubling me.

    How is F acting along the line joining the mid-point of this small segment ds to the center of the ring (radial direction)?

    My thinking is that it is because the Center of Mass of the element ds lies somewhere along this line joining the mid point of ds to the center of the ring .

    Should I consider F to act at the mid point of the element ds OR to act at the COM of the element OR something else ?

    Could somebody help me understand this .

    Many thanks .
     
    Last edited: Oct 29, 2014
  2. jcsd
  3. Oct 29, 2014 #2

    haruspex

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    I guess you mean tangential and radial components.
    I don't understand your puzzlement. Can you post your diagram?
    Since ds is small, you can treat it as a particle at radius r.
     
  4. Oct 29, 2014 #3
    I have attached a picture .

    The tension T is acting at the two ends of the arc AB (element ds) . The arc subtends an angle θ at the center .This makes the two tensions separately to be acting at an angle θ/2 with the horizontal .

    The horizontal components cancel whereas the vertical components add up to give a net force F.

    How is the net force F= 2Tsinθ/2 acting along the line joining the mid-point of this arc M to the center of the ring C ?

    Are we considering the two tensions to be acting at the COM of the arc (which lies on the line MC) ? OR should we consider the two T's to be acting at the mid point M of the arc ?
     

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  5. Oct 29, 2014 #4

    haruspex

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    It seems obvious to me, but I'll try to analyse it somewhat:
    Three things completely determine a force: magnitude, direction (its two vectorial attributes) and line of action.
    In vector terms, clearly the two individual forces add up to the one force of double the magnitude through M. If you consider torques about some point, the torques of the two individual forces add up to the torque of the one force of double the magnitude through M. There are no other considerations.
    Once you have reduced it to the single force along MC, it does not matter at which point along MC you consider it to act.
     
  6. Oct 29, 2014 #5
    Is it wrong to consider the two T's to act at the COM of the arc (which lies somewhere along the line MC) and then resolve them in components ?
     
  7. Oct 30, 2014 #6

    haruspex

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    Yes and no. They don't literally act at the COM of the arc since they act tangentially at the ends of the arc, but it is clear that their resultant will act along lime MC, so you can treat them as acting at the COM.
    Sorry, but I am struggling to understand what your difficulty is. In the OP you correctly observed that the tangential components of the two forces cancel, leaving only the radial components, which you can add together.
     
  8. Oct 30, 2014 #7
    Keeping everything else same (same two T's acting at same angles at the end points),suppose the arc under consideration is not circular (symmetric) and is of any arbitrary shape ,how would you determine the line of action of force in this case?

    Would you still say that the line of action of net force F is along MC(M may no longer be the mid point of arc) ? MC represents the direction as in the previous post .
     
  9. Oct 30, 2014 #8

    haruspex

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    If two forces have equal magnitude then their resultant bisects the angle between them, and passes through the point where the lines of action of the two forces intersect.
     
  10. Oct 30, 2014 #9
    Thanks haruspex .

    So finding the resultant of the two forces has nothing to do with the COM of the circular arc .Right ?

    And it is just a coincidence that in the OP ,the line of action passes through COM . The net torque produced by the two T's would be zero .

    But if we have any other arbitrary shape ,then the line of action may not pass through the COM .There may be a net torque about the COM .

    Does this make any sense to you ?
     
  11. Oct 30, 2014 #10
    Another thing I would like to clarify is that you mentioned about tangential and radial directions . What is a tangential direction in case of a circular arc ? Tangent at which point ? Radial at which point ?

    Is it that since we are dealing with a differential element ,there is a unique tangential and radial direction ?

    Again ,thanks for your patience .
     
  12. Oct 30, 2014 #11

    haruspex

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    If you mean some kind of distorted ring under tension, it must still be true that the resultant of the two tension forces at the ends of an element have no net torque about the COM of the element. If they did, the element would spin, accelerating. The two forces need not be equal in magnitude in this case. E.g. consider a square of wire, and an element part the way along one side.
     
  13. Oct 30, 2014 #12
    Ok.

    Are the above two statements of post#9 correct ?

    Please reply to post#10 .
     
  14. Oct 30, 2014 #13

    ehild

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    The acceleration of the COM of a system of mass points is determined by the resultant of all external forces and the total mass of the system, as if the total mass was concentrated in the COM. You also know that the force acting on a rigid body can be shifted along the line of application. It is irrelevant where these lines intersect.
    The acceleration of the COM of the line segment multiplied by its mass is equal 2Tsin(dθ/2) as you derived. The line segment moves along a circle of radius R with angular speed ω, so the force above is equal to the centripetal force dmRω2, where dm is the mass of the line segment,dm=Mdθ/(2pi)
    You can also apply the approximate formula sin(dθ/2) =dθ/2.
     
  15. Oct 30, 2014 #14
    Thanks ehild

    I know how to solve the problem . But I feel am struggling with some calculus and geometry concepts .Please help me in understanding the basics of calculus as applied in this problem.

    If I consider a differential element (curve) then the tensions are acting at an angle to the arc ,but at the same time I can treat this element as a point particle .

    So how is it that this element has a curve shape but at the same time it may be treated as a point particle ? Should I look at it as a point particle or a collection of particles ?

    Also ,what are the tangential and radial directions of this differential element(arc) ?
     
    Last edited: Oct 30, 2014
  16. Oct 30, 2014 #15

    ehild

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    You know that the forces acting on a rigid body are equivalent with a resultant force (applied at the CM) and a couple.
    In this case there is no torque, only a resultant force. See picture. The parts of the ring neighbour to the line element both exert force, tangential to the ring at the ends of the sector. You can shift these forces and draw their sum. In this case, the line of the resultant force goes through the middle of the orf the small arc and points at the direction of the centre of the ring.


    ringtension.jpg

    In general case of a system of point masses, ##m_i \ddot{\vec {r_i}}=\vec {F_i}##. Adding all : ##\sum {m_i \ddot{\vec {r_i}}}=\sum {\vec {F_i}}##

    On the right side, the internal forces cancel each other, and the sum of external forces remains.
    The left side is ##M \sum \frac{m_i \ddot{\vec {r_i}}}{\sum {m_i}}= M\ddot{\vec r_{COM}}##.

    So you have the equation for a system of point masses, that ##M\ddot{\vec r}_{COM}=\sum \vec F(external)##.
     
  17. Oct 30, 2014 #16
    Thanks ehild and haruspex .
     
  18. Oct 31, 2014 #17
    Is it possible to define tangential and radial directions for this differential element (arc) ds ?
     
  19. Oct 31, 2014 #18

    ehild

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    ds is small. You can consider the element a single point. Any resultant force applied at it has a radial component, pointing to the centre of the circle and a tangential component, perpendicular to it. This tangential component is zero now.
     
  20. Jun 27, 2015 #19
    My doubt is that why we are considering tension on element of ring as T and not dT?how force on element can be force on ring?
     
  21. Jun 27, 2015 #20

    haruspex

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    Tension is not the same as force. Rather, it is a pair of equal and opposite forces. If you conceptually cut the ring at any point, there is a force magnitude T pulling on each side of the cut.
    The net force on the element comes from a force of magnitude T pulling at each end, at slightly different angles.
     
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