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A small ball rotating and along a curve and landing at a certain distance

  1. Oct 28, 2009 #1
    1. The problem statement, all variables and given/known data

    In Figure 11-38, 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 = 5.9 cm and h2 = 2.5 cm. With what speed must the ball be shot at point P for it to land at d = 9.7 cm?

    http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c11/qu_11.1.14.gif

    2. Relevant equations


    3. The attempt at a solution

    I have a fairly solid idea how to calculate the kinetic energy required to make it up the ramp using the rotational kinetic energy, but I'm not quite sure about landing at distance d, once on top of the plateau. I have two ideas

    1) Use equations for projectile motion to determine the point of impact, by somehow finding the acceleration that that point

    2) Use the equations for kinetic energy to compute the work done by travelling over that distance and factoring that into my original equations to compute the total kinetic energy..

    I did try to work on this problem earlier in the week and I can't remember what I tried, but I'll probably look for them later today.

    Will either of these ideas work? Maybe both? Neither?

    Thanks in advance..
     
  2. jcsd
  3. Oct 28, 2009 #2
    you can assume that the acceleration downward is soley due to gravity.

    I'd say go for projectile motion to find the speed.
     
  4. Nov 16, 2009 #3
    We are given information,
    h1 = 5.9cm = 0.059m ,
    h2 = 2.5cm = 0.016m
    d = 9.7cm = 0.097m

    Apply conservation of energy at the point P and the at the height h1 we get

    (1/2)mU^2 = mgh1+ (1/2)mV^2
    U^2 = 2gh1+ V^2 --------(1)

    If the ball is moving with the inital velocity V it coveres the distance of d then

    d = V (2h2/g)^1/2
    V = d / (2h2/g)^1/2
    = ----------- m/s
    substitude the value of V in equation (1) we get the value of intial velocity U = --------- m/s
     
    Last edited: Nov 16, 2009
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