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Rolling a ball down a hill

  1. Aug 8, 2009 #1
    newbie question, say i'm rolling a ball down a smooth , differentiable, and frictionless 1-Dimensional hill V(x) from a point x_i

    find the location x(t) and the velocity dx/dt(t) of this ball at some arbitrary time t.

    what would be the general approach towards such a problem?

    note that for x_i such that dV/dx(x_i) = 0, dx/dt=0
     
    Last edited: Aug 8, 2009
  2. jcsd
  3. Aug 8, 2009 #2

    russ_watters

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    Staff: Mentor

    You can't roll a ball down a frictionless hill, you can only slide it down a frictionless hill.

    You can use the angle of the hill and geometry to find the acceleration. You need no other information besides the angle of the hill and the value of g.
     
  4. Aug 9, 2009 #3
    I disagree with russ watters. Don't you need Radius and moment of Inertia, Or that you are telling about sliding case. At that case you are right.
     
  5. Aug 9, 2009 #4
    In the rolling case for a hard solid ball of uniform density with out friction;

    m*g*h = rotational K.E + linear K.E

    mgh = (1/2)*I*w^2 + (1/2)*m*v^2
    I = moment of inertia
    w = angular velocity of ball
    v = linear speed

    now a rolling ball of diameter r turning at frequency has a linear speed v of;
    v = 2*pi*r*f = w*r

    so w = v/r

    Therefore:
    mgh = (1/2)*I*w^2 + (1/2)*m*v^2
    becomes:
    mgh = (1/2)*I*(v/r)^2 + (1/2)*m*v^2
    = (1/2)*((2/5)*m*r^2)*(v/r)^2 + (1/2)*m*v^2
    g*h = (1/2)*((2/5)*r^2)*(v/r)^2 + (1/2)*v^2
    = (7/10)*v^2
    v= sqrt((10/7)*g*h

    So if h = 1m, d = 1m
    v = 3.72 m/s
     
    Last edited: Aug 10, 2009
  6. Aug 9, 2009 #5

    rcgldr

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    Homework Helper

    You can't have a one dimensional "hill". You can have a one dimensional line and some constant force or constant acceleration.

    If there is no friction, then there is no angular acceleration, so only the linear acceleration matters.

    If the hill has 2 or 3 dimensions, isn't frictionless, then the rolling resistance will be greater than m*g*cos(θ)*(b/r), because rolling resistance is related to the total force between surfaces, not just the normal component.
     
  7. Aug 9, 2009 #6

    russ_watters

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    Staff: Mentor

    Radius and moment of inertia are needed if the ball is to be spun, but if there is no friction, there is no tangential force applied to the ball and thus no way to spin it.
     
  8. Aug 9, 2009 #7
    To keep the responses to my questions in one place I will post in the other thread.
     
    Last edited: Aug 9, 2009
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