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Various locations during a vertical circular motion

  1. Nov 26, 2015 #1
    For a vertical circular motion, its standard to solve for tension at the 4 common points on the circle (North, South, East, West)

    My question is, is it possible to solve for tension at other locations?
    Example :https://www.dropbox.com/s/c4q1sprphol2tzb/20151127_020345.jpg?dl=0

    Can i resolve vertically and horizontally,
    I.e. vertically : Tcosθ =mg,
    Horizontally : Tsinθ=[mv^2/r]sinθ
    Doesn't seem quite right though..

    Can advise/explain on how to approach/solve this questions?

    Thanks a million!!!
     
  2. jcsd
  3. Nov 26, 2015 #2

    TSny

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    Does the object have only centripetal acceleration, or does it also have some tangential acceleration?

    You stated that Tcosθ = mg. Wouldn't this imply that the vertical component of total acceleration is zero? Why would that be true?

    When you wrote Tsinθ=[mv^2/r]sinθ you have not accounted for tangential acceleration.

    I would recommend that you take components along the string.
     
  4. Nov 26, 2015 #3


    OHHhh
    Vertically, Tcosθ=mg + rα
    Where α is angular acceleration, rα is tangential acceleration

    Horizontally ,Tsinθ=[mv^2/r]sinθ + rα?

    I have learn from my lecture notes at that the east and west positions just horizontally T=[mv^2/r] will do
    As for top position its tension = weight - mv^2/r

    And for bottom position tension = W + mv^2/r
    Is it correct? ??

    When do we have to consider tangential acceleration?
     
  5. Nov 26, 2015 #4

    TSny

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    rα is the tangential acceleration, but it's direction is not vertical. To make the above equation correct, you would need to take the vertical component of rα.

    Here you would need to use the horizontal component of rα.

    Yes. You should be able to derive these results.

    If you are looking for the tension, you do not need to consider the tangential acceleration if you consider the application of F = ma along the direction of the string.
     
  6. Nov 27, 2015 #5

    How to apply newton's 2nd law in the direction of the string?
    ∑F=ma,
    T=mv^2/r?
    Please help):
     
  7. Nov 27, 2015 #6

    TSny

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    You know that there are two forces acting on the pendulum bob. Can you find the component of each of these forces along the string? Note that this is the same as finding the "centripetal" component of each force. One of the forces happens to act in the centripetal direction, so the centripetal component of that force is the entire force. However, the other force does not act in the centripetal direction. You will need to use a little trig to find the centripetal component of that force.
     
  8. Nov 27, 2015 #7
    Only tension is acting in the direction of centripetal force, weight is vertically down?
     
  9. Nov 27, 2015 #8

    TSny

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    Right, the weight acts vertically down. But the weight does have a nonzero component along the string. You need to find this component of the weight.
     
  10. Nov 27, 2015 #9
    Weightcosθ?
     
  11. Nov 27, 2015 #10

    TSny

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    Yes. Now set up the second law along the centripetal direction:

    ΣFc = mac

    where the "c" subscript refers to components of force or acceleration in the centripetal direction.
     
  12. Nov 28, 2015 #11
    ΣF=ma
    In the direction of centripetal acceleration,
    T-Wcosθ=mv^2/r?
     
  13. Nov 28, 2015 #12

    TSny

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    Yes, that looks good.
     
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