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Why is tension greatest at bottom in circular motion?

  1. Nov 17, 2015 #1
    Hi. I am having some difficulty with circular motion.


    An object is spun vertically on a rope a) when would the string be most likely to break?

    The object would most likely to break when it has the most tension, so at the bottom of the circle it is Ft-Fg=Mv^2/r which is (Ft=Mv^2/r + Fg) while the top is Ft=Mv^2/r - Fg. But I dont understand the concept of this, because if both Tenstion and Gravity are going in the same direction how can it be possible? Thanks.
     
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  3. Nov 17, 2015 #2

    HallsofIvy

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    Tension and gravity are NOT "in the same direction". The tension in the rope is always directed toward the middle of the circle. At the bottom, that means tension is UPWARD while gravitational force is directed DOWNWARD. The two are in opposite directions, not the same direction.
     
  4. Nov 17, 2015 #3
    But if bottom one goes up other goes down how can bottom have the greatest tension?? I thought it would be top because in that case both the force tension and force gravity would go down
     
  5. Nov 17, 2015 #4

    NascentOxygen

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    If you swing a bunch of keys on a keychain around so they trace out a circle in the vertical plane, you'll quickly discover that it is impossible to have the keychain go slack near the bottom of the circle, but it's easy to cause it to go slack near the top of the circle. When the chain goes slack, this indicates there is no nett force acting through the chain.

    How to account for the chain going slack at the top of this circular orbit? That's what you need to address.
     
  6. Nov 17, 2015 #5
    The tension force will always be directed along the string, which will always be pointing to the center.

    Gravity always points downwards.

    When the object is at the bottom, tension points up and gravity points down. When balancing the forces, [itex]T - W = mv^2/r[/itex], and [itex]T = mv^2/r + W[/itex].

    When the object is at the bottom, both the tension and gravity point downwards. [itex]T + W = mv^2/r[/itex] and [itex]T = mv^2/r - W[/itex].

    Clearly, T at the bottom is greater than T at the top by twice the weight.

    I encourage you to draw the Free-Body Diagrams and balance the forces yourself if what I wrote didn't make sense. You can also try this when the string is at some angle to the horizontal, you should find that the tension will range from [itex]T = mv^2/r - W[/itex] to [itex]T = mv^2/r + W[/itex].
     
  7. Nov 20, 2015 #6
    Got it ty so much for help :)
     
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