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Circular motion

  1. Oct 19, 2004 #1
    the sloping side is frictionless, it is spun at constant speed by rotating the wedge. show that when a mass rises up the wedge a distance L, the speed of the mass is v=sqr(gLsin0)
    I got Fx=m(v^2/r) = mgsin0
    can I say r=L ????
    I got stuck, could you give me some hints thank you

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    Last edited: Oct 19, 2004
  2. jcsd
  3. Oct 19, 2004 #2


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    Hint: Gravity is the centirpetal force because it is pulling the block towards the center.
  4. Oct 19, 2004 #3
    What is your specific problem with this? Tried resolving normal reaction in the right direction? Do you know what the radius of the circular path is?
  5. Oct 19, 2004 #4


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    Yes r will be L.

    [tex] mgsin\theta = m \frac{v^2}{l} [/tex]

    [tex] \sqrt{lgsin\theta} = v [/tex]
  6. Oct 20, 2004 #5
    thank you, however, i do not quite understand why r=l. Could you please explain
  7. Oct 20, 2004 #6
    Where you get the equation m(v^2/r)=mgsin0. I have search for many reference book but can't find this equation .Can you tell me ?
  8. Oct 20, 2004 #7
    In my analysis, I resolve the normal reaction in the horizontal and vertical direction. This gives,

    [tex]N\cos\theta = mg[/tex]
    [tex]N\sin\theta = \frac{mv^2}{l\cos\theta}[/tex]

    I divide equation 2 by equation 1 to get

    [tex]v^2 = gl\sin\theta[/tex]

  9. Oct 20, 2004 #8
    mg sin theta is the component of the weight acting down the incline (resolve mg in two directions one parallel and the other perpendicular to the plane). Drawing a well labeled freebody diagram might help.
  10. Oct 20, 2004 #9


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    Yes Maverick, that's a correct analysys for the way your coordinate system was put, although i prefer your way than mine, because it is using the vertical coordinate system (most used in Circular Motion) rather than the inclined one.

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