Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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

    Attached Files:

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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook