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Circular Motion and speed

  1. Oct 1, 2008 #1
    A 930 g rock is whirled in a horizontal circle at the end of a 1.3 m-long string. If the breaking strength of the string is 110 N, what is the maximum allowable speed of the rock?

    OK the main problem that i am having with this equation is the fact that no angle is given. Is there a way to figure out the angle?

    since the length is given: the radius is r = L cos (theta)
    you need theta to calculate the velocity with the equation V= sq root ((g L cos 2 theta)/ sin theta)
  2. jcsd
  3. Oct 2, 2008 #2


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    This is the first time I have witnessed these kinds of equations, but may I just ask, why are you using the equation [tex]r = L cos \vartheta[/tex] ? The question states that the circle is horizontal, therefore the angle is 0o and hence, [tex]cos \vartheta = 1[/tex] which means the radius = length of string.
  4. Oct 2, 2008 #3


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    Oh sorry, after re-reading your problem I see where you had the trouble. The angle was not specifically given, it was instead discretely provided. In the information where it said that the rock was spun in a horizontal circle, this gives you all the information you need for the angle :smile:
  5. Oct 2, 2008 #4
    Well length of string is constant 1.3 m regardless of the angle.
    The two forces acting here are gravity (9.81 ms^-2) and the centripetal force.
    The tension in the string can be a maximum of 110N as you stated, this implies that you have a right angle triangle with (9.81 * 0.930)N pointing down and 110N along the hypotenuse. With this information it is possible to calculate the angle for breaking point. However that is unnecessary, you can just use pythagoras' theorem to find the horizontal component of that force.

    Re-arrange [tex] F = \frac{mv^2}{r} [/tex]

    [tex] v = \sqrt{ \frac{Fr}{m}} [/tex]

    That should be all that you need.
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