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Pendulum with circular motion

  1. Apr 8, 2012 #1
    1. The problem statement, all variables and given/known data



    2. Relevant equations

    Ac= V^2/R

    physics questions.jpg Theta is made by the upwards component of tension and tension itself.
    3. The attempt at a solution

    I know Sum of all forces = MA. Though for circular motion you always subtract the bigger number by the smaller one (bigger one being the force inwards so circular motion can happen), though for this I dont know what to subtract the tension by.. Gravity is pointing downwards, its not on the same axis of the inwards force, I have no idea what to do.

    Note: I have the answer though it makes 0 sense to me. It is T-mgcos(theta)=Mv^2/R
     
  2. jcsd
  3. Apr 8, 2012 #2

    OldEngr63

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    Using the figure that you have, try writing the sum of forces in the horizontal and vertical directions, setting each equal to the appropriate m a factor. That should get you started.
     
  4. Apr 8, 2012 #3
    well... sum of the forces X = Tsintheta
    and for y it is Tcostheta - mg
     
  5. Apr 8, 2012 #4
    Though Iv never seen a circular motion question like this, I dont know how to set it up.
     
  6. Apr 8, 2012 #5
    There was a question like this though, where the pendulum is at the 9'oclock, and the equation was T=Ac(MV^2/r), It didnt use gravity anywhere. Though the answer to this one does have gravity as T-mgcos(theta), i dont understand how it works.
     
  7. Apr 9, 2012 #6

    rock.freak667

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    Right, so if you rotate the axes such the y-axis points in the same direction of the tension T and the x such it is tangential you can easily see the equation form.


    You know that the centripetal force is given by Fc= mv2/r and it is the resultant force that points in the direction towards the center of the circle.

    Along the y-axis we have T (pointing towards the center) and mgcosθ (pointing away from the center). So the resultant force pointing towards the center will be T-mgcosθ i.e. T-mgcosθ=mv2/r


    At the 9 and 3 o'clock positions, the weight mg will have no component which points towards the center thus the tension T will just be equal to the value of the centripetal force mv2/r.
     
  8. Apr 9, 2012 #7

    OldEngr63

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    rockfreak667, I doubt that the OP knows any such thing of the sort. Let him work though the problem and discover that for himself.

    OP, do you know about using spaces and parentheses to make clear the mathematics you are writing? It would help if you would use them.

    You have the force sums, now, what can you say about the acceleration terms? Remember that this is simply motion on a circle.
     
  9. Apr 9, 2012 #8
    I dont understand what mgcosθ comes from. I dont see it ?
     
  10. Apr 9, 2012 #9
    you say the 9 and 3 o'clock positions have no gravity component which points towards the center, I dont see the angle one having any gravity components pointing to the centre either. The gravity is straight down how can you have any components for it such as mgcosθ?
     
  11. Apr 9, 2012 #10
    I only see one component pointed inwards and thats tension, I dont see anything point oppositely to it.
     
  12. Apr 9, 2012 #11

    OldEngr63

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    So, now look at the kinematics of the point, moving on a vertical circle with uniform motion. Write the equation for its position and start differentiating until you have expressions for its acceleration components. The you can write the right side of the force equations.
     
  13. Apr 9, 2012 #12
    I know the right side of the question, thats the easy part Ac = MV^2/R.

    What I dont understand is where T-mgcosθ comes froms. I understand why T is there because it is the greater force that allows circular motion to occur, though I dont see where the mgcosθ is. I see no force opposing the tension.
     
  14. Apr 9, 2012 #13

    OldEngr63

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    And is MV^2/R the acceleration in the horizontal direction or the vertical direction (since that is the easy part)?

    I think you are way to fixated on T - m g cos(theta) to listen to what I am telling you. Please re-read what I have asked you to do, and then give it a try.
     
  15. Apr 9, 2012 #14
    it is in the vertical directs because the gravity is pointing down?
     
  16. Apr 9, 2012 #15

    OldEngr63

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    No. Try again.
     
  17. Apr 9, 2012 #16
    so the acceleration is horizontal, though i dont understand why ?
     
  18. Apr 10, 2012 #17

    OldEngr63

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    In post #3, you wrote ƩFx = .... and ƩFy = .... (you did not actually write that, but you wrote equivalent expressions). Now that indicates that there is both a horizontal and a vertical acceleration. What I have been trying to get you to do is go back now, and work though the kinematic relations, starting with

    X(t) = R sin(θ)
    Vx = R ω cos (θ)
    Ax = - R ω2 sin (θ)

    Y(t) = R cos(θ)
    :

    After you generate the acceleration terms, then complete the right side of both F = m a equations and see what you can do from there.
     
    Last edited: Apr 10, 2012
  19. Apr 11, 2012 #18
    What you just said I dont understand, I think I need to go back to 3 o'clock position. Because I still dont understand how that one works out.

    The tension on that one is horizontal, and the gravity is vertical. I dont know how T=MV^2/R, gravity is not used in that equation why? For that one its just sum of the x forces = MA. Why isnt gravity used in that one?
     
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