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Ball velocity once rolling

  1. Aug 23, 2009 #1
    1. The problem statement, all variables and given/known data
    A billiards ball (mass m) is hit by a billiard stick on the center of the ball (from the side) with a collision impulse mv0 so that the cues line of travel goes straight through the center of mass of the billiards ball, which in the collision-approximation gives a speed v0 immediately after the collision. The angular velocity is equal to zero directly after the collision (the ball glides, before it starts to roll). The frictioncoefficient is u, and the ball starts at rest.
    Find the relationship between the velocity of the ball when it has just started rolling and the speed v0 directly after the collision.


    2. Relevant equations

    F = ma = m*dv/dt
    small f = friction.
    time ball is hit = t0.
    time ball stops gliding = t1.
    velocity of ball when it is about to start rolling = v1, and velocity when it starts rolling is v2.
    R is the radius of the ball.
    The moment of inertia of the ball I = (2/5)mR².
    the angular acceleration is called alpha = dw/dt, where w = omega meaning the angular velocity.

    3. The attempt at a solution

    Ok so I believe I have solved this question, but i'm sure you can find something I did wrong =).

    the integral of the powers affecting the ball along the x-axis is the integral of friction (f) and the collision impulse mv0.
    Integral from t0->t1:mv0dt = mv0(t1)-mv0(t0)
    Normal N from the ground is mg. friction f = -uN integrated like above over time is -uN(t1-t0). I now assume that the time it glides is very small so that t1-t0 = 0.

    Therefore mv0(t1) = mv0(t0) which means that v0(t1) = v0(t0), which in turn means that v1 = v0.

    keep in mind that mdv/dt = ma = -f for later. (1)

    Now I used Eulers II law stating that the sum of the moment Mz = fR = I*alpha which means that f = I*alpha/R

    (1) mdv = -I*dw/R
    I now integrate from v0 (which remember is equal to v1!) to v2 and I also integrate for the angular velocity from 0 (since it was gliding) to v2/R, because v = w*R. I've also removed m from the equation now since it was on both sides of the = sign.

    (integrate v0 -> v2: dv) = -(2/5)R (integral 0 -> v2/R: dw) => v2 - v0 = -(2/5)v2.
    Solving for v2 I got the answer: v2 = 5v0/7 which means that when you hit the ball, the speed it will have when it starts to roll is 5/7's the speed you hit it with!

    Am I right in these calculations?
     
  2. jcsd
  3. Aug 23, 2009 #2

    Doc Al

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    Staff: Mentor

    Looks good to me. :smile:
     
  4. Aug 23, 2009 #3

    kuruman

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    Here is a much faster way to answer this question. (I already started it before seeing tiny-tim's posting ** Edit: Doc Al not tiny-tim, sorry **.) You start with a ball sliding (but not rolling) with a center of mass velocity v0. You end up with a rolling ball with center of mass velocity v. The angular momentum of the ball relative to the point of contact is conserved. So

    Lbefore = m v0 R
    Lafter = I ω

    where I is about the point of contact.

    [tex]I = \frac{2}{5}m R^{2} + m R^{2}=\frac{7}{5}m R^{2}[/tex]

    Since the ball rolls without slipping in the end ω = v / R. Then

    [tex]m v_{0}R = \frac{7}{5}m R^{2}\frac{v}{R}[/tex]

    [tex]v = \frac{5}{7}v_{0}[/tex]

    By the way, what is Euler's II law? Is that what in other parts of the world is known as Newton's Second Law?
     
    Last edited: Aug 24, 2009
  5. Aug 24, 2009 #4
    From wikipedia:
    "Euler's second law states that the sum of the external moments about a point is equal to the rate of change of angular momentum about that point."

    Common usage is Moment = I*alpha
     
  6. Aug 25, 2009 #5
    I believe that Newtons II law is Eulers I law btw.
     
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