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Tip to tail vertor diagram using pythagorian theorem?

  1. Jan 24, 2005 #1
    If a stone is spun around in a vertical circle, is the tension in the string the centripetal force and the force of gravity?

    If the stone is at the 6 o'clock position, is the tension in the string the centripetal force plus the gravitaional force?

    At the twelve o'clock, is it the centripetal force minus the gravitational force?

    And at the 3 o'clock position is the tension calculated by a tip to tail vertor diagram using pythagorian theorem?

  2. jcsd
  3. Jan 25, 2005 #2


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    The vector sum of the tension and gravity is the net force on the stone. The component of gravity tangent to the circle applies a torque which changes angular velocity. The sum of the component of gravity parellel to the string and the tension (which is also parellel) must add to the centripetal force (for the instantaneous angular velocity, which, don't forget, is changing) if the stone is to remain in circular motion. So you're right for 6 and 12 oclock but I don't think your 3 oclock method is correct.
  4. Jan 25, 2005 #3
    Ok, but at the 3 o'clock postion is the sum of the tension the centripetal force acting along the horizontal plane and the gravitational force acting along the vertical plane?

    So then would this be a tip to tail vector diagram using the pythagoras rule?

    Sorry if i don't seem o get this to quickly, but circular motion and centripetal force is a section that i find quite difficult not to mention if tension is involved.
  5. Jan 25, 2005 #4
    let's get vector-free for a minute.... :smile:

    let's get vector-free for just a moment.......
    are you assuming no air friction? yes? good!
    can you assume no other losses, like friction in the string where it's connected to something? yes? good!

    that simplifies things.

    next, you didn't specify the velocity (tangential) of the stone.

    if the stone's velocity is too low, it won't complete a circle in a vertical plane around the center of the circle. if the velocity is high enough, (again, assuming no air friction to slow things down...) the stone will continue to circle the "hub" with maximum velocity at the 6:00 position and minimum V at the 12:00 position.

    at some speed, wouldn't the tension in the string at the 12:00 position be zero? (this is the "balance point" between too slow to complete the circle and fast enough to keep spinning).

    +af :confused:
  6. Jan 25, 2005 #5


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    This is a 2 dimensional problem, everything is in the vertical plane. As I said, the component of gravity tangent to the circle (which at 3 oclock is all of it) contributes to angular acceleration. The component along the string (none of it at 3 oclock) plus the tension equals the centripetal force. So at any given time:

    T+mg cos(theta) = mv2/r

    where theta is the angle between the string and 12 oclock.
  7. Jan 26, 2005 #6
    Thank-you, that sounds great.

    The angle between the 3 o'clock(the string) and the 12 o'clock positon is 90 degrees, but when you punch cos 90 into your calculator, it comes up zero.

    What do i do now?
  8. Jan 27, 2005 #7


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    I've already given you the exact equation you need, which was probably more than I should have done. If you can't figure it out from here, this class is probably too advanced for you.
    Last edited: Jan 27, 2005
  9. Jan 27, 2005 #8
    But if cos 90 equals zero, then the tension would just equal the centripetal force, so then the upward force of the stone is then equal in magnitude to the gravitational force, right?
    Last edited: Jan 27, 2005
  10. Jan 27, 2005 #9


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    What upward force of the stone? The net force on the stone doesn't have to be zero.
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