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Challenging Physics Problem

  1. Oct 19, 2007 #1
    Two heavy balls of equal mass M are attached to a long but light metallic rod standing on the floor. The rod with the balls falls to the floor. Find the velocity of each ball at the moment when the rod hits the ground. Neglect mass of the rod and the friction between the balls and the floor.

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    O This ball is already touching the floor. Assume that the system falls to the right.
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    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 19, 2007 #2

    learningphysics

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    Try to use conservation of energy and also conservation of momentum ideas...
     
  4. Oct 20, 2007 #3
    my solution so far is that as the system falls to the right, the ball touching the floor will start to move to the left a distance equal to the distance the top ball falls; therefore the top ball moves down with velocity sqrt(2gh) and the velocity of the bottom ball is sqrt(2gh)/4, but I am not sure about this at all since it must be more complicated than this
     
  5. Oct 20, 2007 #4

    Doc Al

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    What can you say about the motion of the center of mass of this object?
     
  6. Oct 20, 2007 #5

    learningphysics

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    Just as the higher mass is about to hit the ground, what is the horizontal velocity of each mass?

    There's no force in the horizontal direction, so momentum is conserved horizontally.
     
  7. Oct 20, 2007 #6
    well im not exactly sure what you guys mean since the only force acting on this system is gravity which accelerates the higher mass; however, the final velocity of the higher mass is directed downwards whereas the velocity of the lower mass is to the left. i would greatly appreciate it if you guys could please give a more detailed explanation of a possible solution. thanks.
     
  8. Oct 20, 2007 #7

    learningphysics

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    The main idea I used is that momentum is conserved in the horizontal direction... Let's just look at the horizontal direction. initially the momentum of the system in the horizontal direction is 0.

    so it remains 0 throughout the fall... ie the horizontal momentum of the right mass and the horizontal momentum of the left mass (which is its entire momentum since it doesn't move vertically) are equal and opposite.

    so at the very bottom when the rod is horizontal... the right mass has a horizontal velocity of v. The left mass has a horizontal velocity of -v. What is v? Hint think of the rod being rigid.
     
  9. Oct 20, 2007 #8
    ok i understand what you are saying, but i am still a bit confused with the fact that the upper ball will have its velocity vector pointing downward while the lower ball will have its velocity vector pointing to the left and so the velocities will be perpendicular
     
  10. Oct 20, 2007 #9

    learningphysics

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    Are you saying they're perpendicular when the upper ball reaches the bottom? That's true, but they aren't perpendicular throughout the fall... only at the bottom. So at the bottom the lower ball has its velocity vector to the left, the upper ball has its velocity vector downward... what must the lower ball's velocity be? use conservation of momentum in the horizontal direction.
     
  11. Oct 20, 2007 #10

    learningphysics

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    The idea is that the horizontal velocity of the 2 balls is 0 at the bottom... net linear momentum horizontally is 0...

    But we have a rigid rod... we can't have the balls moving in the direction of the rod...

    So all we have is the vertical velocity of the ball that has been falling...
     
  12. Oct 20, 2007 #11
    ok so if i understand you correctly, then the velocity is just sqrt(2gh) since we just have vertical velocity?
     
  13. Oct 20, 2007 #12

    learningphysics

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    yes. the falling ball has a downward velocity of sqrt(2gh). The other ball has 0 velocity.
     
  14. Oct 20, 2007 #13
    ok. just one last question, why doesn't the other ball move? (i mean there is no friction on the floor and the upper ball is applying a force on it)
     
  15. Oct 20, 2007 #14

    learningphysics

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    that ball moves... it gains velocity to the left initially... reaches some maximum velocity... then slows down back to zero...
     
  16. Oct 20, 2007 #15
    isn't acceleration that rises initially, reaches some max, and then slows down back to zero, but velocity is always increases up to the moment it hits the ground
     
  17. Oct 20, 2007 #16

    learningphysics

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    can you explain your reasoning for the acceleration ? not saying you're wrong there... I just didn't think about that...

    The net kinetic energy is increasing... since potential energy is being converted to kinetic energy... but the velocity of the ball on the ground increases then decreases back to zero... the velocity of the other ball keeps increasing all the way down.
     
  18. Oct 20, 2007 #17
    well the upper ball has mass M and thus gravity pulls down with Mg; however this force changes with the angle as the upper balls falls down (for example, initially Mg is directed straight down, but as the ball falls to the right, Mg starts to play a role in the x component of the tension force of the rod until it hits the floor where Mg has no effect on the other ball)
     
  19. Oct 21, 2007 #18

    learningphysics

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    The acceleration of the upper ball towards the right starts at zero... becomes positive... comes back to zero... and becomes negative... before it hits the bottom.

    reason being that the velocity towards the right, goes up to some max... and comes back down to zero...

    so the force of the rod I think goes from compressive to tensile...
     
  20. Oct 21, 2007 #19
    how exactly does the acceleration ever become negative because Vmax is the moment the ball hits the ground and the acceleration on the upper ball is always just gravity, the acceleration of the lower ball changes due to the angle.
     
  21. Oct 21, 2007 #20

    learningphysics

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    gravity isn't the only force on the upper ball... the rod exerts a force too... the Mg downward force is always acting... but the force the rod exerts is changing... initially it is straight upward... at the bottom, there's no force exerted by the rod...

    acceleration of the lower ball to the left = acceleration of the upper ball to the right...

    do you agree that the lower ball accelerates to the left... then decelerates (since it has to get back to 0m/s) ?
     
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