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Motion in a circle

  1. Jul 30, 2005 #1
    Two wires with 2.0kg sphere is tied to a pole with 1m in between the wires. each wire makes an outward angle with the pole of 60 degress and an inward angle with the pole of 30 degrees.

    for what speed is the tension the same in both wires and what is that tension.

    well i figured you needed to find the tension in the vertical direction and then subsitute that into the tension in the horizontal and find the radius. but im not sure exactly how to go about doing this. so any help would be nice =] thanks
  2. jcsd
  3. Jul 30, 2005 #2
    I'm not sure whether I understood your problem but I think you need these:

    [tex] T_{vertical} = G [/tex] and the horizontal component is causing the rotational movement, so [tex] T_{horizontal} = m\frac{v^2}{R} [/tex]
  4. Jul 30, 2005 #3
    I would need to see a picture or have a better physical description as to whats going on here, but for the general idea of the problem. You are correct, in the fact that the sum of forces in the radial (horizontal) direction must equal the centripetal force. This is 1 equation. You are also correct that the sum of forces in the vertical direction must equal zero. Between these two equations you should have 3 unknowns, both the tensions and the velocity of the object. Your third equation is simple, its one of the constraints in the problem... t1 = t2. So you have three equations and three unknowns, the problem is solvable. I actually would find it slightly more general to write the magnitude of centripetal force as m*r*omega^2 as opposed to mv^2/r because that velocity only corresponds to a single radius. Granted we assume the radius doesnt change due to any of the deformable characteristics of the wire, I still think its more general to determine the angular velocity which is omega, having units of rad/sec.
  5. Jul 30, 2005 #4
    oh no i forgot to attach the inital picture.. ok basically its two strings attached to a vertical pole with 1m of space between them, it looks like a triangle attached to a pole with the interior angles 30,30,120 the 30 degree angles are with the pole. so if u break that up its two triangles (30,60,90).

    this is what i got so far since the interor angles 30 and 30 are the same both sides should have the same length, so i used the law of sines and found the length of the rope and since they both have the same angle they have the same length. after finding this i was able to find the radius = 0.289 by trig.

    so my equations i set up were..

    T(1)cos60+ T(2)cos60=mv^2/r (im using v because the answer needs to be in m/s)

    T(1)sin60-T(2)sin60-mg=ma=0 T(1)sin60-T(2)sin(60)=19.62

    and then they said that T(1) = T(2) and that zeros out the equation so it doesnt work?? i dunno
  6. Jul 30, 2005 #5
    One can directly conclude that this is incorrect since the centripetal motion is two dimensional, ie the r-vector has both horizontal and vertical components. Since you are writing down two equations of motion in vertical and horizontal direction, you must be consequent. The left hand side of this equation contains components of forces in ONE single direction, while the right hand side is two dimensional. You see my point ?


    Besides, could you add a figure ?
    Last edited: Jul 30, 2005
  7. Jul 31, 2005 #6
    the picture of the problem

    i finally figured out how to put the picture on

    Attached Files:

  8. Jul 31, 2005 #7
    o yeaa its not drawn to scalee
  9. Jul 31, 2005 #8

    Doc Al

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    Staff: Mentor

    The only radial forces on the sphere are the horizontal components of the tension in the wires. So those forces create the centripetal acceleration. (Use some geometry to find the radius of the circular motion.)
  10. Jul 31, 2005 #9
    for my radius i got 0.289m

    i understand that the x components are the radial components and u set that equal to centrip. accel, but what i dont get is how to find the tension
  11. Jul 31, 2005 #10

    Doc Al

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    Staff: Mentor

    You find the tension by writing equations for both horizontal and vertical forces on the sphere. (The vertical forces are in equilibrium.) With those two equations, you'll have all the info you need to find the speed.
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