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Sphere Rolling Down Hill into Projectile Motion

  1. Dec 8, 2015 #1
    1. The problem statement, all variables and given/known data
    Mountaineers are simulating a massive avalanche snowball by using a big Styrofoam ball (m = 75 kg; r = 1.50 m).They give it an initial impulse of 1500 kg m/sto start it rolling down the hill without slipping (no snow sticks to the ball, and ignore air resistance). Near the bottom, the hill first levels off, and then rises at a 15 degree angle to the horizontal to then abruptly end at the edge of an absolutely vertical 100 m high cliff, which launches the ball into the air. The height between the top of the hill and where it is launched into the air is 500 m. At the cliff bottom:

    a) With what speed and at what angle, with respect to the horizontal, will the Styrofoam ball land?

    b) How far from the cliff will the Styrofoam ball land?

    c) How many revolutions will the Styrofoam ball make while in the air?

    2. Relevant equations
    Kinetic Energy of Snowball = 7/10 mv^2, Range = v_x *t, y-y_0 = v_y +1/2a*t^2
    w = v/r
    3. The attempt at a solution
    a) Applying Conservation of Energy to the motion down the hill, mgh = 7/10mv^2, so v^2 = 10/7 gh or v = 83.7 m/s. For the subsequent projectile motion v_x = 83.7 cos(15) = 80.8, t = 7.23 s so v_y = -49.2. Therefore v_f = 94.6 m/s at an angle of arctan(49.2/80.8) = 31.3 degrees below the horizontal.
    b) Since Range = v_x *t = 80.8*7.23 = 584.2 m.
    c) Angular velocity won't change once the snowball loses contact with the hill so w= v/r = 83.7/1.5 = 55.8 rad/s. The angular displacement during the fall is w*t = 403.4 rad or 64 revolutions.
     
  2. jcsd
  3. Dec 8, 2015 #2

    haruspex

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    What about the initial velocity?
    It will be much easier to follow your working, spot any errors in it, and tell you what and where those errors are if you work entirely symbolically, only plugging in numbers right at the end. This will involve inventing symbolic names for all the given data. It's an excellent habit to get into. (It also minimises rounding errors.)
     
  4. Dec 8, 2015 #3
    I forgot to include the initial velocity due to the impulse.
    a) Using the impulse-momentum theorem: I = Δp = mv ⇒v_0 = 20 m/s and ω_0 = 13.33 rad/s.
    For the motion down the hill to the cliff mechanical energy is conserved so 1/2 m*v_0^2 + 1/2 I*ω_0^2 + mg*h = 1/2m*v_1^2 + 1/2 I*ω_1^2.
    Since ω =v/r and I = 2/5m*r^2, the above energy conservation becomes 7/10m*v_0^2 + mgh = 7/10m*v_1^2. Therefore v_1^2 = v_0^2 + 10/7*gh = 7400. Therefore v_1 = 86 m/s and ω_1 = 57.35 rad/s.
    For the motion of falling off the cliff 1/2m*v_1^2 + mg(100) = 1/2m*v_f^2 ⇒v_f = √(v_1^2 + 2g(100)) = 96.72 m/s. v_x is a constant at 83 m/s so v_y = √(v_f^2 -v_x^2) = 49.65 m/s.
    The angle below the horizontal is arctan(v_y/v_x) = 30.9 °.
    b) v_y = 86*sin(15) - gt ⇒ -49.65 = 22.6 -9.8t ⇒ t = 7.4 s. Since the range is v_x*t the snowball lands 83*7.4 = 636.4 m from the cliff.
    c) Angular velocity doesn't change after the ball leaves the hill so Δθ =ωt = 57.35*7.4 = 424.39. Revolutions is Δθ/2π = 67.5 revolutions.
     
  5. Dec 9, 2015 #4

    haruspex

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    Haven't checked all the numbers in detail, but that all looks about right.
     
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