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Rigid body rail question

  1. Jan 20, 2009 #1
    here is the question
    and how i tried to solve it

    [​IMG]
     
  2. jcsd
  3. Jan 20, 2009 #2
    This is a frame of reference type of problem. Yes, the marble will release from the track when the normal force is zero. In order for the marble to travel in a circular path a centripetal acceleration must be present. If the observer is the marble, it experiences an inertial force called the centrifugal force. This is the reactionary force caused by the track. So, in this reference frame the centrifugal force is the same magnitude as the centripetal force (v2/R) but in the opposite direction. Therefore, when the gravitational force equals the centrifugal force this is the point where the marble will leave the track.
     
  4. Jan 24, 2009 #3
    i described the equation
    can you continue them?
     
  5. Jan 25, 2009 #4
    I'm sorry. The web page with your image will not load at this time so I cannot view your equations.
     
  6. Jan 26, 2009 #5
  7. Jan 26, 2009 #6
    Yes, I can see your image. Thank you. Now, express omega in terms of v (omega = v/R). This will give you an expression for v as a function of alpha. Use the expression for centripetal force in terms of v and equate to the radial component of the graviational force acting on the marble.
     
  8. Jan 28, 2009 #7
    ok what now
    the total energy is
    [itex]


    & {\rm{mg2R = mgR(1 + cos}}\alpha {\rm{) + }}{{mv^2 } \over 2} + {{I({v \over r})^2 } \over 2} \cr
    & \cr}
    [/itex]
     
  9. Jan 28, 2009 #8
    Your equation can be solved for v2. Now,

    [tex]F_{centripetal}=m\frac{v^2}{R}[/tex]

    Find the radial component of the gravitational force as a function of alpha and set it equal to the centripetal force.
     
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