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Rock on string

  1. Jul 15, 2005 #1
    as shown: in the picture, look at figure d) (attachment) don't worry about figure a) --- given:

    720 gm rock

    each string has a tension of 29 N

    the rock is held by two of the 45 cm strings with ends 56 cm apart and whirled in a circle between them. Neglect gravity.
    What is the radius of the circle of motion?

    what is the maximum speed the rock can have before the string breaks?

    Attached Files:

  2. jcsd
  3. Jul 15, 2005 #2


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    Hmm, this is a curvilinear trajectory dealing with only normal acceleration component (radial), apply Newton's 2nd Law. Now that i think about it you can calculate the radius throught trigonometry :smile:. If you have any more questions i will come back later today, after a good night sleep.
    Last edited: Jul 15, 2005
  4. Jul 15, 2005 #3
    well, i found the radius to be 35.22 cm. and i also found the respective angle between direction of acceleration and the string tensions. from each string to the r as shown, angle is 38.48 degrees. but i don't know where to go from here
  5. Jul 15, 2005 #4


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    Yes now you need to apply Newton's 2nd Law

    Now with [itex] \theta [/itex]

    [tex] T \cos \theta + T \cos \theta = m \frac{v^2}{R} [/tex]

    solve for v.
  6. Jul 15, 2005 #5
    didn't work cyclovenom,

    but how did you come up with that equation?
  7. Jul 15, 2005 #6


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    It should have worked, well the Tension have 2 components, one of them is acting radially, and because there are two tensions both of them hace 1 component acting as the centripetal force. Rememeber to convert the mass of the rock to kilograms!.
    Last edited: Jul 15, 2005
  8. Jul 15, 2005 #7


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    Horizontal components of T:
    [tex]T_x = Tcos\theta[/tex], where [tex]\theta = sin^{-1}(\frac{d}{2s})[/tex], where d = 0,56m and s = 0,45m

    So, the equation solves into:
    [tex]v = \sqrt{\frac{2Tcos(sin^{-1}(\frac{d}{2s}))\sqrt{s^2-(\frac{d}{2})^2}}{m}}[/tex]

    In my markings [tex]sin^{-1} = arcsin[/tex]
    Last edited: Jul 15, 2005
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