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Rotational and potential energy problem - Mastering physics

  1. Nov 9, 2008 #1
    1. The problem statement, all variables and given/known data

    "In a lab experiment you let a uniform ball roll down a curved track. The ball starts from rest and rolls without slipping. While on the track, the ball descends a vertical distance h. The lower end of the track is horizontal and extends over the edge of the lab table; the ball leaves the track traveling horizontally. While free falling after leaving the track, the ball moves a horizontal distance x and a vertical distance y."

    "Calculate x in terms of h and y, ignoring the work done by friction."

    h - height of ramp ball rolls down
    y- height of track above ground
    x - horizontal distance ball travels while falling
    g - acceleration due to gravity
    m - mass of ball
    r - radius of ball

    2. Relevant equations

    m*g*h = 1/2*m*v^2+1/2*I*(omega)^2 (potential energy equal to kinetic energy and rotational energy)
    I= 2/5*m*r^2 (moment of inertia for a sphere)
    (omega)=v/r
    x = x_0+v_0*t+1/2*a*t^2


    3. The attempt at a solution

    I solved for the velocity of the ball at the bottom of the ramp, mass and radius cancels out and got gh=6/10*v^2, so v=sqrt(10/6*gh). Then I tried to find the time it would take for the ball to fall y and got y=1/2gt^2 as there was no initial vertical velocity. So, i got t=sqrt((2y)/g). I then multiplied the velocity by the time because there should be no horizontal acceleration, and I got x = sqrt(6/10*gh)*sqrt((2y)/g) and mastering physics said "Your answer either contains an incorrect numerical multiplier or is missing one." so, I'm not sure what I should do, where I went wrong or whatever, so if anyone has any help, it would be appreciated, thanks.
     
  2. jcsd
  3. Nov 9, 2008 #2
    [tex]2E = 2m\vec{g}\vec{h} = m\vec{v}^2 + I\vec{\omega}^2 \left(=\frac{I\vec{v}^2}{\vec{r}^2}\right) = m\vec{v}^2 + I\vec{\omega}^2 + 2mgy[/tex]. Check this out. :)
     
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