1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Exam question: conservation law

  1. Jul 27, 2010 #1
    http://img198.imageshack.us/img198/1946/blahz.jpg [Broken]

    Uploaded with ImageShack.us

    1. Given data
    Ball: I (moment of inertia), m (mass), R (Radius), V0 (Velocity, as shown ), [tex]\omega_{0}[/tex] (angular velocity, as shown).

    The ball collides instantaneously with the wall and immediately after this collision, the ball ceases to rotate and only moves vertically upwards parallel to the wall, as shown in the figure. The question is to find this velocity.

    2. The attempt at a solution

    I think that this problem relates to conservation of angular momentum since the word "instantaneously" is explicitly written. Now, angular momentum is conserved about the instantaneous point between the ball and the wall, and there I can write that the total angular momentum about this point, after the collision is:

    [tex] \vec{L}_{after} = -R \hat{x} \times mV \hat{y} = -mVR \hat{z} [/tex]

    Now, how can I express the linear momentum before the collision?! I mean, about the ball's center of mass, it is obviously [tex] I \omega_{0} [/tex], however, is it the right expression to be used in the conservation equation? OR, should I used the parallel axis theorem and write [tex] (I + mR^{2}) \omega_{0} [/tex] instead?!

    One more question: Is mechanical energy conserved in this problem?!
    In an inelastic collision, energy is never conserved, however, here i'm dealing with one object hitting and infinitely heavy object so is it treated differently or in any case of inelastic collisions, I mustn't use the conservation of energy?!
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jul 28, 2010 #2
    ummm, anyone? =/
  4. Jul 28, 2010 #3
  5. Jul 28, 2010 #4
    This article relates to a collection of particles while all I have in the problem is one particle, so how will that help me? ... I mean, i know directly where the center of mass is (ball's center) and applying the angular momentum equation gives me one of the results that i wrote previously.

    I'm maily asking of how to "attack" this problem ... or how can I use the angular momentum principle for this problem, as it is the only way I may reach a final solution.
  6. Jul 28, 2010 #5
    Body = continuous series of particles.
    You were asking about how to calculate L before the collision, and I have pointed out how by making a reference to that site. All you have to do is to apply the equation in that site.
  7. Jul 28, 2010 #6
    OK, but isn't angular momentum conserved with respect to a specified point?!
    I think that I can't or mustn't equate two angular momentum expressions where one is taken with respect to the center mass and the other with respect to some other instantaneos point.

    One more thing, now i'm noticing that angular momentum with respect to the "collison" instantaneous point (after collision) is even not conserved because I haven't consudered the weight of the body which is by all means an external force =/

    So how can I treat this problem now?
  8. Jul 28, 2010 #7
    Yes. But that it's conserved has nothing to do with calculating it.
    Correct. But this is unrelated to anything we have discussed from the start.
    First, you found out L after the collision about the axis going through the point of contact, parallel to [tex]\hat{z}[/tex], which is [tex]\vec{L}_{after}=-mVR\hat{z}[/tex].
    Then you asked how to find L before collision about that axis. So apply the equation given in the site, with notice that the 2nd term in the final formula is now [tex]-I\omega _o\hat{z}[/tex] as for the ball which is a rigid body.

    The collision typically happens in a very short time. Therefore, the normal force and friction on the ball during collision is very large, so weight can be ignored.
  9. Jul 28, 2010 #8
    Oh, so what you mean is: (A is the instantaneous point and D is the vector locating the center of mass from A)

    Just Before Collision: [tex] \vec{L}_{A} = \vec{L}_{c.o.m} + \vec{D} \times {\vec{p}} = \vec{L}_{c.o.m} = -I \omega_{0} \hat{z} [/tex]

    Just After Collision: [tex] \vec{L}_{A} = \vec{L}_{c.o.m} + \vec{D} \times {\vec{p}} = \vec{D} \times {\vec{p}} = -mVR \hat{z} [/tex]

    and equating both terms yields:

    [tex] V = \frac{I \omega_{0}}{mR} [/tex]

    Last edited: Jul 28, 2010
  10. Jul 30, 2010 #9
    Correct :smile:
  11. Jul 30, 2010 #10
    thanks pal ;)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook