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Billiard Ball backspin (draw English) - Rigid body dynamics

  1. Feb 8, 2006 #1
    I've been presented with the following problems, and would like some help/affirmation:

    SITUATION:
    A billiard ball with mass m and radius r is in rest on a horizontal table. The ball is hit with a billiard cue the height r/3 above the table, and has the velocity v0 immediately after the hit.
    The force from the cue F remains horizontal during the hit and the contact between cue and ball is short-termed. The kinetic coefficient of friction between the table and the ball is mu_k

    PROBLEM a) Determine the ball's angular velocity omega0 immediately after the hit.
    -----
    Since the below-c.m. hit will give the ball backspin (draw English), the ball wont have natural roll. Therefore, I can't use the simple equation v0 = omega0*r, since that relationship only exist in natural roll.

    Instead, I have to use Newton's 2nd law, sum_F = m*a and the rotational equivalent sum_tau = I * alpha, where tau is torque (around the c.m.), I is moment of inertia of the ball and alpha is its angular acceleration.

    (1) sum_F = m*a <=> sum_F = m*alpha*r
    <=> F - mu_k*m*g = m*alpha*r

    (2) sum_tau = I*alpha <=> r*F - 2/3*r*mu_k*m*g = (2/3)*m*r^2*alpha

    The two expressions each have only one unknown variable, alpha. In (1) alpha is the translational contribution to the angular acceleraton, and in (2) its the rotational contribution to the angular acceleration. Sum_alpha can now be determined. The hit has the duration dt, so omega0 = sum_alpha * dt. By substituting the expression for alpha into that equation I ought to have determined omega0. Right?

    PROBLEM b) What's the velocity of the ball when it beings to do natural roll?
    ----
    The roll is natural when v = omega * r. Since the v0 and r is known, I should then be able to determine expressions for v and omega from the kinematic equations for the two types of motion. I guess the rolling friction can be ignored, since the needed coefficient isn't given. Therefore, there is no horizontal force. Right?

    PROBLEM c) Determine the work of the friction force during the ball's movement until it starts rolling on the table.
    ----
    The ball starts at the angle 0 and reaches an angle theta. The torque is constant, since the kinetic friction is constant. Then the work done by the friction force must be W_f = tau*(theta-0) <=> W_f = tau*theta. Right?

    :shy:
     
  2. jcsd
  3. Feb 8, 2006 #2

    Doc Al

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    They want you to find the angular velocity immediately after the impact. Think of it this way. The cue exerts an impulse on the ball ([itex]F \Delta t[/itex]) that imparts the given linear momentum. Now consider the angular impulse that that force exerted to find the angular momentum after impact.

    Yes. Once you've properly found the initial angular and linear speed of the ball, apply Newton's 2nd law for translation and rotation. (What's the only horizontal force acting on the ball?) What you call "natural roll" is also called "rolling without slipping"; you are solving for the speed at which that condition is met.

    I'd find the work done by friction by finding the change in total kinetic energy.
     
  4. Feb 8, 2006 #3
    Problem a)

    The angular momentum after the impact is [itex]L=I \omega[/itex], but what is the angular impulse? The net torque? Otherwise, I haven't found examples or equations with the angular impulse in the textbook.
     
  5. Feb 8, 2006 #4

    Doc Al

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    Just like linear impulse is [itex]F \Delta t[/itex], angular impulse is [itex]\tau \Delta t[/itex]. Given "F", what torque does it exert?
     
  6. Feb 8, 2006 #5
    The que force F exerts the torque [itex] 2/3 r F [/itex], and the kinetic friction force f exerts a resistive torque [itex] -r f [/itex], so that the net torque is [itex] r (2/3 F-f) [/itex], or [itex] r (2/3 F - \mu m g) [/itex].
     
    Last edited: Feb 8, 2006
  7. Feb 8, 2006 #6

    Doc Al

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    Right. But I would assume that the duration of impact is very short; so short that the ball has yet to move. Ignore the kinetic friction for part a.
     
  8. Feb 8, 2006 #7
    Yeah, of course :) That must be the justification for that information.

    Then; the (linear) impulse is [itex] F dt [/itex], the angular impulse is [itex] \tau dt = 2/3 r F dt[/itex] and the angular momentum is [itex] L = I \omega [/itex].
     
    Last edited: Feb 8, 2006
  9. Feb 8, 2006 #8

    Doc Al

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    Right! (Where "r" in this case is really "2r/3". :smile: )
     
  10. Feb 8, 2006 #9
    Yes, I just corrected it. But - isn't the angular momentum the same as the angular impulse?

    Merely using [itex] \tau = dL/dt [/itex] gives [itex] \tau dt = L [/itex] - or doesn't it? This gives [itex] \omega_0 = (2/3 F dt)/(2/5 m r) [/itex].
     
    Last edited: Feb 8, 2006
  11. Feb 8, 2006 #10
    OOOOO. I found it! If you want, a while back I found this paper that has tons and tons of this stuff all on pool. It might be in there.
     
  12. Feb 8, 2006 #11

    Doc Al

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    Yes. [itex]\tau \Delta t = I \Delta \omega = I \omega_0[/itex].

    Simplify that further. (Write [itex]F dt[/itex] in terms of [itex]v_0[/itex].)
     
  13. Feb 8, 2006 #12
    Well, yeah.. I'd like to see it. I've found some PDF's myself too.
     
  14. Feb 8, 2006 #13
    Its called Amateur Phyiscs for the Amateur pool player. (its not amateur)~107 pgs
     
  15. Feb 8, 2006 #14
    Is it neccessary? Since the quantity F is mentioned in the symbolic description, it's considered known.

    I guess it should be simplified via [itex] v_0 = a dt [/itex], and [itex] F = m a = m v_0/dt [/itex]. And also the rotational contribution, [itex] v_0 = r \omega [/itex] and [itex] \tau dt = I \omega_o [/itex].
     
  16. Feb 8, 2006 #15
    http://www.physics.ohio-state.edu/~penningt/262/ps/apapp.pdf#search='amateur%20physics%20for%20the%20amateur%20pool%20player
     
  17. Feb 8, 2006 #16
    Heh, I found that. It's a top Google hit. I didn't find anything directly applicable in it though.
     
  18. Feb 8, 2006 #17

    Doc Al

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    While F is mentioned, I wouldn't call it "known". What you know is the impulse, not the force.

    That won't help, since you don't know [itex]F[/itex] or [itex]\Delta t[/itex]. But you do know that the impulse [itex]F \Delta t = m v_0[/itex]; use that to eliminate those unknowns ([itex]F[/itex] & [itex]\Delta t[/itex]) from your expression for [itex]\omega_0[/itex].
     
  19. Feb 8, 2006 #18
    Then: [itex] \omega_0 = (2/3 v_0)/(2/5 r) [/itex] - ?
     
    Last edited: Feb 8, 2006
  20. Feb 8, 2006 #19

    Doc Al

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    Yes: [itex]\omega_0 = (5/3) (v_0/r)[/itex]
     
  21. Feb 8, 2006 #20
    Thanks for the help.
     
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