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Homework Help: How to relate impulse with speed of the ball

  1. Sep 11, 2010 #1
    1. The problem statement, all variables and given/known data

    A billiard ball initially at rest is given a sharp impules by a cue. The cue is held horizontally at a distance h above the centre line as in figure. The ball leaves the cue with a speed v and eventually acquires a speed 9/7 v. Show that h=4R/5, where R is the radius of the ball

    2. Relevant equations

    3. The attempt at a solution

    The friction force acts in the direction of motion as angular velocity is greater than linear speed (backward slipping occurs).

    considering rotation about central axis,
    fR=I(alpha) where (alpha) is the angular acceleration.
    I can get 2 equations relating linear speed at time t and angular speed at time t.

    But I don't know how to relate impulse with speed of the ball.
    I also wanted to ask that if we consider rotation about the point of contact, there is no torque acting on the sphere. How do we proceed this way?

    Attached Files:

  2. jcsd
  3. Sep 11, 2010 #2
    Re: Mechanics

    Well, f might not be constant (what if the kinetic friction coefficient is not constant?), so you shouldn't integrate that equation to deduce w(t) (angular velocity). Instead, why don't you try to combine fR=I(alpha) and f=ma to derive a relation between v, v(t=0) and w, w(t=0)? :wink: This is for the 2nd stage when the ball already leaves the cue and rolls.

    Likewise, you can deduce the relation between v(t=0) and w(t=0) right after the hit by the cue. Because the force by the hit is quite large, you can ignore other forces. This is for the 1st stage when the ball is in interaction with the cue.

    Then, when does the ball come to stable state, i.e. v doesn't change? :wink:

    It's quite complicated. The ball in this case doesn't rotate about the axis through the point of contact, i.e. there is no angular acceleration about that axis. That makes calculation more complex and tedious, as you cannot apply the equation: torque = I(alpha) (there is another subtle thing, but it's even more complicated, so just make it simple :smile:).
  4. Sep 12, 2010 #3
    Re: Mechanics

    Using fR=I(alpha) and f=ma,
    I got these two equations
    w(t)=w(t=0) - {5(mu)/2R} t
    v(t)=v(t=0) + (mu)gt

    How do I relate v(t=0) and w(t=0)?
  5. Sep 12, 2010 #4
    Re: Mechanics

    One more thing I wanted to ask you,
    Suppose a ball is at rest on a rough horizontal surface and a bullet is fired towards it such that it just grazes the top of the ball (uniform sphere).Assume that the cylinder rolls without slipping and at the instant the bullet leaves, there is no relative motion between the bullet and the cylinder.
    Is the kinetic energy conserved?
    my explanation is- since there is no relative motion between the contact points just after the bullet grazes it, there is no dissipation of energy due to friction (i.e. no work done by friction). Moreover the sphere rolls without slipping (so velocity of contact point is 0). So kinetic energy is conserved. Do you agree with this?
  6. Sep 12, 2010 #5
    Re: Mechanics

    As I told you earlier, f is not constant in general, so avoid integrating directly the equations f=ma and fR=I(apha) (though your method can yield the same result after all, but it can be judged as scientifically wrong because of assuming something not given. Beware of harsh teachers :biggrin:). Instead, you can simply eliminate f by combining those 2 equations. What you get is an equation relating dv and dw. This is how you relate v(at any time t), w(at any time t), w(0), v(0). t=0 is the moment when the ball starts to leave the cue. From t=0 onwards, the ball is affected only by the frictional force, not the force from the cue.

    Then if you go back to when the ball gets hit, you will see that during the process the ball is hit by the cue, the impulse is due to one force: the force F from the cue (the frictional force << F, so it can be ignored). This force changes v from 0 to v(0). Besides, F creates a moment about the center, and this moment changes w from 0 to w(0). Using the same idea as above, you can relate v(0) and w(0).

    I assume that you meant "sphere" or "ball", not "cylinder" :wink:

    Firstly, well, you're mixing the 2 processes. There are 2 discrete processes: First, the ball gets hit by the bullet, and then, the ball moves on its own. The former occurs in a very short time (that's collision by the way), and during this process, the ball is affected by both the ground and the bullet. Under certain circumstances, the ball can be treated as affected by only the bullet (like in the cue problem). After this process, then comes the latter process: the ball leaves the bullet, moves on its own and is affected by only the ground.

    Secondly, the relative velocity = 0 ONLY occurs at the time the bullet leaves the ball; at other moments, no. For example, at first, right before the bullet hits the ball, the ball is at rest and the bullet is moving, so there is relative motion between the bullet and the top point of the ball.

    Though the collision lasts in a very short time, and it seems that the ball don't move at all during the collision, but actually it does. The collision typically lasts in about 0.01 seconds (I'm not sure, maybe 0.1 seconds or so, but it's very small anyway), and that period is way too small for our brain to perceive any difference or movement.
    Last edited: Sep 12, 2010
  7. Sep 12, 2010 #6
    Re: Mechanics

    I did not integrating any thing. Just substituted f=(mu)mg in the two equations, found out 'a' and (alpha), substituted them in equations of motion (v=u+at ans w=w. + (alpha)t

    Moreover, why won't f be constant? coefficient of kinetic friction is constant for a particular surface (unless it is wetted or some other change is done to it)

    haha yes :smile:

    I'm sorry, I meant the relative velocity between contact point (though it is not really a point :wink:) of ground and the sphere.
  8. Sep 12, 2010 #7
    Re: Mechanics

    Well v=u+at and similar formulas ONLY apply to cases where ACCELERATION IS CONSTANT.
    Just try out my hint and see if you can get something :wink:

    Does the question mention anything about the surface? Are you sure that that the surface is uniform? :wink:

    Okay, then there are several points to make about the bullet case:
    1 - In the first process (collision), energy is not conserved. Agree?
    2 - In the second process, energy is conserved provided that the ball has to roll without sliding ALL THE TIME. But does the initial condition (i.e. the condition set up right after the collision) ALLOW that to happen? You may find the answer to that if you solve the cue case first :smile:
  9. Sep 12, 2010 #8
    Re: Mechanics

    It does not mention about non-uniform surface either :wink:
    So why complicate? make is simple :wink:

    I got the answer :smile:
  10. Sep 12, 2010 #9
    Re: Mechanics

    If you can get full mark, and more importantly, full understanding by solving the simple case and leaving out the general case, then that's fine :smile:
    Strictly speaking, as I said earlier, that will be judged as scientifically and logically wrong, unless you provide a subtle point :smile: But even if you can point out that point, that may affect your understanding, especially when you face with a bigger and more complicated problem.
  11. Jan 19, 2011 #10
    Re: Mechanics

    Hi hikaru1221

    I have to revive this thread again, got a doubt while revising :biggrin:.

    What is wrong with this approach to the problem-

    The moment of the linear impulse given by the cue about the COM = mvh
    which is equal to 2/5*mR2ω, where ω is the initial angular velocity acquired by the billiard ball.

    Now, we can conserve the angular momentum of the COM about the point of contact just after the collision and during rotation without slipping motion. We have

    mvR + 7/5*mR2ω = 7/5*mR2(9v/7R)

    Solving the above equation, I got ω=4v/7R

    Substituting this in the first equation,
    h = 8/35*R

    which is different from the correct answer.
    I found that if we take the moment of inertia about the point of contact and not about the centre of mass in the first equation, we get the correct answer.

    i.e. mvh = 7/5*R2ω

    What's wrong?
  12. Jan 20, 2011 #11
    Re: Mechanics

    The bold part is wrong. Let me remind you of the theory:

    Angular momentum (A.M.) of the system about a fixed point O in reference frame R =
    A.M. of COM about O in reference frame R + A.M. of the system about COM in reference frame of COM

    The same thing goes for kinetic energy (KE):
    KE of the system in reference frame R =
    KE of COM in reference frame R + KE of the system in reference frame of COM
  13. Jan 20, 2011 #12
    Re: Mechanics

    Never read that any where :uhh:

    So the equation should be modified by adding "A.M. of the system about COM in reference frame of COM" on the L.H.S.

    i.e. mvR + 7/5*mR2ω + 2/5*mR2ω = 7/5*mR2(9v/7R)

    but even this doesnot give the correct answer :cry:
  14. Jan 20, 2011 #13
    Re: Mechanics

    A.M. of COM about O in reference frame R = mvR (COM only!!!)
    A.M. of the system about COM in reference frame of COM = 2/5*mwR^2.

    P.S.: If you want me to help you review the rigid body dynamics part, I can make up some similar problems to discuss.
    Last edited: Jan 20, 2011
  15. Jan 20, 2011 #14
    Re: Mechanics

    Ohh you were talking about that rule! I know that :redface:
    Its a silly mistake!
    Thank you.
  16. Jan 20, 2011 #15
    Re: Mechanics

    Thank you, but I have 100s of problems which I did not solve (no time!). I already finished the rigid body dynamics, so no need now. This small doubt crept in between.
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