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Collision and angular velocity

  1. Oct 30, 2013 #1
    1. The problem statement, all variables and given/known data
    The given figure shows a mechanical system free of any dissipation. The two spheres (A and B) are each of equal mass m, and a uniform connecting rod AB of length 2r has mass 4m. The collar is massless. Right above the position of sphere A in figure is a tunnel from which balls each of mass m fall vertically at suitable intervals. The falling ballscause the rods and attached spheres to rotate. Sphere B when it reaches the position now occupied by sphere A, suffers a collision from another falling ball and so on. Just before striking, the falling ball has velocity v. All collisions are elastic and the spheres as well as the falling balls can be considered to be point masses. (Attachment 1)

    (a) Find the angular velocity ##\omega_{i+1}## of the assembly in terms of {##\omega_i##, v, and r} after the ith ball has struck it.

    (b) The rotating assembly eventually assumes constant angular speed ##\omega^*##. Obtain ##\omega^*## in terms of v and r by solving the equation obtained in part (a). Argue how a constant ##\omega^*## does not violate energy conservation.

    (c) Solve the expression obtained in part (a) to obtain ##\omega_i## in terms of {i, v, and r}.

    (d) If instead of a pair of spheres, we have two pairs of spheres as shown in figure below. What would be the new constant angular speed ##\omega^*## of the assembly (i.e. the answer corresponding to part (b)). (Attachment 2)

    2. Relevant equations



    3. The attempt at a solution
    Let the rod be rotating with angular velocity ##\omega_i## before the next collision.

    From conservation of angular momentum about centre of rod:
    $$mvr+I\omega_i=I\omega_{i+1}+mv_1r$$

    From conservation of energy:
    $$\frac{1}{2}mv^2+\frac{1}{2}I\omega_i^2=\frac{1}{2}I\omega_{i+1}^2+ \frac{1}{2} mv_1^2$$

    where I is the moment of inertia of rod+A+B and ##v_1## is the velocity of falling ball after collision.

    Solving the two equations, I get:
    $$\omega_{i+1}=\frac{7}{13}\omega_i+\frac{6v}{13r}$$
    which is correct as per the answer key.

    For part b, the question states that the angular velocity almost becomes constant.
    Hence, ##\omega_{i+1}=\omega_{i}##. Using this, ##\omega^*=v/r##.

    But what should I state for the argument? It looks odd to me that angular velocity becomes constant. The angular velocity should keep on increasing as the collisions take place. :confused:

    Any help is appreciated. Thanks!
     

    Attached Files:

    Last edited: Oct 30, 2013
  2. jcsd
  3. Oct 30, 2013 #2
    What is the velocity of the sphere when the constant angular velocity is attained? What happens during its collision with the ball in that case?
     
  4. Oct 30, 2013 #3
    The velocity is ##v##.
    I am unsure about what you ask me here. If I solve the equations, I get the resulting angular velocity to be v/r again.

    Why this constant angular velocity is attained? :confused:
     
  5. Oct 30, 2013 #4
    Don't solve the equations. Just think about what happens in the reference frame of the sphere.
     
  6. Oct 30, 2013 #5

    haruspex

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    So what is the linear velocity of the sphere when the falling ball hits it?
     
  7. Oct 31, 2013 #6
    Can you please elaborate what you actually ask me here? :confused:

    The linear velocity is ##v##.
     
  8. Oct 31, 2013 #7
    What is the velocity of the ball at the time of the collision with the sphere, relatively to the sphere?
     
  9. Oct 31, 2013 #8
    Zero.
     
  10. Oct 31, 2013 #9
    And what do you think about it? You said earlier "the angular velocity should keep on increasing as the collisions take place". Does that sound plausible if the collision happens at zero relative velocity?
     
  11. Oct 31, 2013 #10
    But my question is how this angular velocity even attained? The collisions keep on taking place, how do we know that at some point of time the collisions take place with zero relative velocity?

    Doing the maths is easy but when it comes to explaining the things, I am completely blank. :(
     
  12. Oct 31, 2013 #11
    First of all, is the situation that ## \omega_i > v/r ## for any ## i ## compatible with the model? Can there be a collision in such a situation?

    Second, if ## \omega_i < v/r ##, what happens at the next collision? Does the system it gain or lose momentum?

    What happens if that goes on for a very long time?
     
  13. Oct 31, 2013 #12
    No, the collisions will never take place.
    The system gains momentum.
    As the collisions take place for a very long time, ##\omega_i## will tend to ##v/r##.

    After that there is no momentum transfer to the system, right?
     
  14. Oct 31, 2013 #13
    Can you show that if ##\omega_i < v/r## then ##\omega_{i + 1} < v/r## and ##\omega_{i + 1} > \omega_i##?

    What does that imply?
     
  15. Oct 31, 2013 #14
    $$\omega_i< \frac{v}{r} \Rightarrow \frac{7}{13}\omega_i<\frac{7}{13}\frac{v}{r} $$
    $$\Rightarrow \frac{7}{13}\omega_i+\frac{6}{13}\frac{v}{r}<\frac{v}{r} \Rightarrow \omega_{i+1}<\frac{v}{r}$$

    But how do I show ##\omega_{i+1}>\omega_i##? :confused:
     
  16. Oct 31, 2013 #15
    Consider ## \omega_{i + 1} - \omega_i ##.
     
  17. Oct 31, 2013 #16
    $$\omega_{i+1}-\omega_i=\frac{7}{13}\omega_i+\frac{6}{13}\frac{v}{r}-\omega_i=\frac{6}{13}\left(\frac{v}{r}-\omega_i\right)>0$$
    $$\Rightarrow \omega_{i+1}>\omega_i$$
     
  18. Oct 31, 2013 #17
    Very well.

    Is there anything still unclear?
     
  19. Oct 31, 2013 #18
    No, thank you very much voko. :)

    Part 3 is easy. I have solved it and reached the correct answer.

    Is the system present in part 4 similar to the system originally given? The answer states that ##\omega^*## for this system is also ##v/r##. I don't think I have to make the equations again and solve (which I can easily do). The final answer is same as before so I guess the examiner wants us to provide some reasoning. How should I start with this one?
     
  20. Oct 31, 2013 #19
    The reasoning is again along the lines of #9 and #11.

    Imagine you have a rotating drum in front of you. You use your hands to spin it up. Can you ever make it spin faster than your hands can move? Does that in any way depend on the moment of inertia of the drum?
     
  21. Oct 31, 2013 #20
    No.
    Yes, less moment of inertia implies faster spinning.

    So the system in part 4 has a higher moment of inertia and it would take more time to reach ##\omega^*##, right?

    But I am unsure what would I write in the answer sheet. Should I simply state the system is identical to the given original system except it takes more time to reach ##\omega^*## due to higher moment of inertia?
     
    Last edited: Oct 31, 2013
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