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Homework Help: Uniform circular motion question

  1. Oct 14, 2011 #1
    Please look at the picture....
    It is of a blue object in four locations that is swung around with a rope ( red line ) in a circle with constant velocity.....

    I wanted to know if my diagrams are correct for each seperate location... and I was hoping someone could check my work below to tell me if it is correct...

    1. ƩFx = 0; ƩFy = -T - W = (m)(a) and T = W

    2. ƩFx = -T - W = (m)(a) ƩFy = 0 and T> W

    3. ƩFx = 0; ƩFy = +T - W = (m)(a) and T>W

    4. ƩFx = T - W = (m)(a) ƩFy = 0 and T>W

    Attached Files:

  2. jcsd
  3. Oct 14, 2011 #2
    And for an object somewhere other than the first four points... would the objects have both nonzero acceleration for ƩFx and ƩFy ???

    Attached Files:

    • 22.jpg
      File size:
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  4. Oct 14, 2011 #3
    Is the angular velocity constant? If so, then there will be no acceleration "around" the loop. The idea is that the object has centripital(sp?) acceleration towards the center. As the thing moves around the circle, the reference frame changes along with it. If acceleration is always pointing in the positive y (or whatever you called the center), then it can't have separate components can it?
  5. Oct 14, 2011 #4
    I never said acceleration was around the circle... if you noticed I always said acceleration points to the sum of the vectors which is in the direction of Tension force.....
    I added another picture with a coordinate axis... Im not sure if it will make a diff,,,?

    And velocity is constant, I mentioned uniform circular motion...

    Attached Files:

    Last edited: Oct 14, 2011
  6. Oct 14, 2011 #5
    I was referring to this:

    Which the answer would have been no if you were using the coordinate system where the center point is positive y regardless of where the object is around the circle. (A polar type grid).

    If you use the grid in your image then yes. Acceleration will have component vectors for the times when the object is not at 0, pi/2, pi, etc. which would have to be summed in your forces.
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