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Rotational Energy and Linear Momentum Problem

  1. Nov 16, 2013 #1
    1. The problem statement, all variables and given/known data

    In the figure here, a small, solid, uniform ball is to be shot from point P so that it rolls smoothly along a horizontal path, up along a ramp, and onto a plateau. Then it leaves the plateau horizontally to land on a game board, at a horizontal distance d from the right edge of the plateau. The vertical heights are h1 = 4.5 cm and h2 = 1.00 cm. With what speed must the ball be shot at point P for it to land at d = 3.5 cm?

    fLI4qPI.jpg

    2. Relevant equations

    E'=E
    Translational KE = mv^2/2
    Rotational KE = Iω^2/2
    y = y0 + v0T + aT^2/2

    3. The attempt at a solution

    (0.5)mv0^2 = (0.5)mv'^2 + mgh1
    Which simplifies to v0 = sqrt(2gh1 + v'^2) (1)

    To find v': d = v'T => v' = d/T
    To find T: h2 = (0.5)gT^2 => T = sqrt(2h2/2)

    So, v' = d/sqrt(2h2/2) which is approximately 0.77 m/s.

    Plugging this into (1) v0 = 1.217 m/s.

    However, this is not the right answer.

    So, I then attempted it again this time taking into account that the moment of inertia of a solid sphere is (2/5)mR^2.

    (0.5)(2/5)(mR^2)ω0^2 = (0.5)(2/5)(mR^2)ω'^2 + mgh1

    v = rω
    (1/5)(mR^2)v0^2/R^2 = (1/5)(mR^2)v'^2/R^2 + mgh1
    v0^2 = v'^2 + 5gh1
    v0 = sqrt(v'^2 + 5gh1)

    Then v' is the same in this attempt as the previous, so v0 = 1.67 m/s.
    However, this is also not the right answer.

    It appears there is not enough information to take into account energy lost to friction so I'm assuming that's negligible. So, I'm at a loss.
     
  2. jcsd
  3. Nov 16, 2013 #2
    You also have to take into account the linear energy at P.
    [tex]K_P=\frac{1}{2}mv_i^2 + \frac{1}{2}I\omega_i^2 [/tex]
    Also at the top, consider the energies for gravity, movement, and rotation.
    [tex]mgh_1+\frac{1}{2}mv_2^2+\frac{1}{2}I\omega_2^2[/tex]
    When the ball then falls, It must have a final energy greater than [itex]mgh_2[/itex] (so it could be rolling along and make it a distance d away, not just survive the fall.) You have already shown that you know I for a sphere to be [itex]\frac{2}{5}mr^2[/itex], and [itex]\omega[/itex] can be calculated as [itex]v/r[/itex].
    When you put the equations together you will be able to calculate the correct answer.
     
  4. Nov 16, 2013 #3
    Oh shoot, that's obvious. I don't know why when I remembered to take into account rotational kinetic energy I dropped off translational kinetic energy.

    Thank you.
     
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