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I was thinking about a rod colliding with an external wall, if you

  1. Dec 12, 2011 #1
    I was thinking about a rod colliding with an external wall, if you will, and began thinking what would happen if the collision was elastic (not sure if that is correct in this context, but simply the energy of the rod is the same before and after the collision.) I assume if a particle was normal to a wall that the change in its momentum after the collision would be the same in magnitude, but opposite in direction - energy would be conserved. With a rod however (see my diagram), the collision brings about an external torque (from wall) and thus a change in angular momentum. This would mean that the center of mass would then have to rebound at a lower velocity to conserve energy (there is now rotational energy), correct? But maybe that is getting ahead, would the center of mass rebound in all cases that energy is conserved, or could there be cases where the center of mass still continues in the same direction with less magnitude (while angular momentum is changing.) There may be many variables here.

    Nonetheless, I am trying to formulate/relate the rod's momentum/angular-momentum before and after the collision. I may simply be missing to many variables, but if anyone could shed some light on what would happen in a collision like this, I would appreciate it.
     

    Attached Files:

  2. jcsd
  3. Dec 14, 2011 #2
    Re: Impulse

    Once the first edge contacts the wall it will begin to rotate, but the second edge will then also strike the wall and apply another torque to remove the angular momentum of the rod. Find something long and give it a shot (just make sure it isn't rotating before it strikes the wall)
     
  4. Dec 14, 2011 #3
    Re: Impulse

    Oh, I see. :) I think this greatly simplifies it. So I imagine, then, that if there was an initial angular velocity/angular momentum about the center of mass, that this would increase (or decrease depending on rotation direction) the second torque causing it to also rebound with an angular velocity the same in magnitude as before the collision? I also imagine that the after angular velocity will be in the same direction?

    I was also wondering, if a force is applied applied horizontally in a plane (let that be your screen) in the photo at the top of the rod (causing a torque), would the center of mass accelerate the same as if the force was applied at the center of mass, or would it be less? If that is true, then, would this mean that if a force was applied to a rod for x time that it would contain more kinetic energy the further the force is from center of mass?

    I intuitively thought that the center of mass would accelerate slower given that the force is also causing a torque, but I am pretty sure my physics instructor said they are separate and the center of mass accelerates as if the force was applied there.
     
  5. Dec 14, 2011 #4
    Re: Impulse

    The center of mass will always act as if the force was applied directly to it :)
     
  6. Dec 14, 2011 #5
    Re: Impulse

    Ok, I see. I guess I am just thinking that in reality, if I were to hit the center of mass of a rod, I know it would do a lot of translating. But if I were to just graze the edge, it would do a lot of rotating and much less translating.

    I then began thinking and came up with the idea that when applying a force on the edge of a rod, one must push with greater acceleration to achieve a force F equal to a force given by pushing on the center of mass. In other words, as soon as the edge of the rod has a force applied, it begins to rotate which means the direction relative to the applied force changes. Considering that, this should mean that even though I would feel as if I am applying the same force over the same distance when hitting the rod in the center and edge, in reality, I am not - and thus the center of mass does not accelerate the same when I hit the edge as it does when I hit the center. Is this correct?

    If this is true, I guess the difference between this reality and the formulas is that the forces being described in the equations are abstract and the stuff I mentioned in reality are irrelevant. In other words, 1N is 1N no matter how it was achieved.
     
  7. Dec 14, 2011 #6
    Re: Impulse

    If you 'jab' at the end of the rod (so the impact is considered to be over a time almost zero) then you will get a lot of rotation but you will also get exactly as much translation as if you jabbed at the center of mass, I'd give it a go with a ruler if I were you.

    Edit: Found a video demonstrating this: http://ocw.mit.edu/courses/physics/...echanics-fall-1999/video-lectures/lecture-21/

    Just click "Video Index" and go to "Dynamics of a Spinning Rod"
     
  8. Dec 14, 2011 #7
    Re: Impulse

    Yes, that is what I then assumed. I was implying that in my reality (as was described) that the force applied is usually smaller and the time line longer. That being said, hitting the edge would result in over all less work done (as rod would usually be out of path of force before motion of jab is done. And being that rotation by torque is quite sensitive, you get what seems to be little translation and a "well" amount of rotation.

    Nonetheless, I now see that the translational acceleration is the same no matter the contact point. Though something questions me here. How is it that you get what seems to be greater kinetic energy when jabbing the edge of the rod (translational motion + angular motion)? It seems by the work - kinetic energy relation that the work done in both cases is the same yet in one case (jabbing the edge) you get greater kinetic energy. Am I applying this theorm incorrectly? (I am currently interpreting the "jab" as a constant large force over a short time interval and thus short distance giving work = Fd). I must be mixing concepts here. :) thanks.
     
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