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Work & energy - friction

  1. Dec 3, 2009 #1
    1. The problem statement, all variables and given/known data
    A ball is released from height h. The friction coefficient between the straight part and the ball is 'u'. I need to find the smallest h so that the ball doesn't fall off the track.
    The angle alpha=45 degrees.


    2. Relevant equations
    Work of non-conservative forces = Change in mechanical energy


    3. The attempt at a solution
    I calculated the work done by the friction force while the ball is going down the straight part.
    Wf=-u*mg*cos(alpha)*X

    X=the length of the straight part.

    We get that:
    Wf=-u*mg*h

    since cos(alpha)/sin(alpha)=1

    Now I want to say that Wf=change in mechanical energy, so:

    -u*mg*h=mg(0-h)+0.5*m(v^2-0)

    when v=the velocity of the ball when it reaches the end of the straight part. We get:

    v^2=g*h(1-u)

    Now to the second part, the frictionless rail. Since all the forces are conservative now:

    0.5*m*V1^2=0.5*m*V2^2+mg2R

    When V1^2=g*h(1-u) -----> (the velocity we found before)
    V2=the velocity at the top of the loop

    So I want that V2>0, so after some work we get:

    h>(2R)/(1-u)

    but when I put in a numerical answer I'm told that I'm wrong. Is there a mistake in my solution?
     

    Attached Files:

  2. jcsd
  3. Dec 3, 2009 #2
    The condition v>0 at the top is not sufficient for the ball to stay on the track.
    You need to have the reaction force acting on the ball at the top to be >=0.
     
  4. Dec 3, 2009 #3
    yeah you're right I forgot, thanks..
     
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