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Sum of External Forces, Center of Mass Translation

  1. Oct 17, 2011 #1
    I am a bit confused about a statement I read in my textbook saying the sum of external forces on a body is equal to the body's mass times the acceleration of it's center of mass. It seems to me if a long rod is suspended in the air by a string at its center of mass, and I tap the rod near one of the ends with a certain force, the rod will spin and the center of mass of the rod will accelerate less than if I had tapped it at the center of mass. Yet the external forces on the rod in both situations would be equal. What am I missing?
     
  2. jcsd
  3. Oct 17, 2011 #2

    xts

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    Really? Are you sure? Have you made calculations? Or may you present some reasoning supporting your thesis? Or is it just (possibly misleading) intuition?
    Note that common meaning of 'tap' is not directly related to force.
     
  4. Oct 17, 2011 #3
    There are two types of object which is not in equilibrium state.
    1) Force acting at C.M.
    - In this case, the object will have an acceleration at the direction of force.

    2) Force acting somewhere else
    - In this case, the other quantity you have to consider is torque(or moment) [itex]\tau[/itex]. Torque is the main factor that makes the object 'rotate' with an angular accerelation [itex]\alpha[/itex].

    So, the acceleration of C.M. is depend on two things; force and the distance from C.M.

    EDIT:
    If you suspend a rod at its C.M., THEN C.M. COULDN'T HAPPEN TO BE ACCELERATED.
    Except you suspend it near the C.M., may be you mean this.
     
    Last edited: Oct 17, 2011
  5. Oct 17, 2011 #4
    xts: I haven't made calculations, it's just intuition. It seems obvious. Is that statement incorrect? Imagine trying to pull a heavy object with a rope. You would want the center of mass to be in line with the rope, because otherwise the force you apply would spin the object rather than translate the center of mass. Is that wrong? Can you enlighten me?

    *EDIT* By tap I mean I put a finger on the rod and move my finger forwards, making the rod move. As a side note, can you explain how tap wouldn't be related to force?

    Black Integra: Thats what I supposed, but the text says differently.
     
  6. Oct 17, 2011 #5

    xts

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    It is incorrect...
    If you pull end of the rod, you may move it with smaller force, or using the same force - you may give higher acceleration to this end. If you apply the same force in centre of mass you will give lower acceleration, but to whole body. Those effects cancel out - if you calculate them - the total momentum (mass * centre_of_mass_velocity) will be the same in both cases.

    Take an easy excercise and calculate this yourself!

    *EDIT*
    It is intuitively very difficult to compare the same force (in physical meaning) applied to a resistive body and to one accelerating fast. It is not the same 'tap' to push a one tonne car and throw a one pound ball, even if forces are the same in both cases.
     
    Last edited: Oct 17, 2011
  7. Oct 17, 2011 #6
    I don't understand. I suspend a rod by a string at its center of mass, then I apply a force perpendicular to the string at the rod's center of mass. The CM would accelerate in the motion of a pendulum. What do you mean by "couldn't happen to be accelerated"?
     
  8. Oct 17, 2011 #7
    What do the text say?
     
  9. Oct 17, 2011 #8
    I mean "couldn't happen to be accelerated" in case you give a force at one end.
     
  10. Oct 17, 2011 #9
    Yes, that's what I'm saying. The acceleration of the center of mass of the body will be lower. So I still don't understand the equation in my textbook, which says Fext = Mtotal *Acm. Like you said, if I apply the same force to the center of mass I will get a lower acceleration. However, I still have the same external force, and the same "Mtotal" value.
     
  11. Oct 17, 2011 #10
  12. Oct 17, 2011 #11

    xts

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    Acceleration of the centre of mass will be lower than the acceleration of the end you tap.
    But the same force applied to the end of the rod causes bigger acceleration of this end, than if the force is applied at the centre.
    Those effects cancel each other - the same force always causes the same acceleration of CM.

    Calculate it yourself! You may calculate it not for the rod, but for two balls connected by weightless rigid rod - it is a bit easier, and you may easily extend result for any symmetrical system of masses.
     
  13. Oct 17, 2011 #12
    Ahhhhhhhhhhhhhhhhhh thank you so much. I understand now =)
     
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