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External forces on rigid body

  1. Jun 14, 2014 #1
    I'm having trouble with this line from Goldstein-"the center of mass moves as if the total external force were acting on the entire mass of the system concentrated at the center of mass."

    If a bar is floating in space (at rest in the frame) and a point mass strikes the bar perpendicular to the length of the bar and in the exact middle then you can you use the conservation of momentum/kinetic energy to calculate the movement of the bar, its velocity after the impact.

    If the point mass strikes the bar perpendicular to the bar at the end of the bar, then the bar will move differently than in the first case. The center mass of the bar will move differently in each of the cases so you can't just know how the center mass of the bar will move by just applying the force to the center of mass.

    I think that in the second case the bar will rotate as well as translate so the c.m. can't move in the same way as in the first case becasue then it will have more kinetic energy i.e. the kinetic energy of translation plus the kinetic energy of rotation.

    A little fuzzy on this.
     
  2. jcsd
  3. Jun 14, 2014 #2

    mfb

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    If the force and the duration of the force is the same in both cases (in a regular collision, this is not exactly true), then the center of mass will get the same velocity change. In the second case you get an additional rotation, sure, and more kinetic energy, but that does not influence the motion of the center of mass.
     
  4. Jun 14, 2014 #3
    It is just a metaphorical way of stating this mathematical theorem:

    "In a system of (newtonian) particles, if internal forces are "newtonian", then:

    [tex]\vec{F}_{ext}(t) = \frac{d\vec{P}(t)}{dt} = M \frac{d\vec{V}_{CM}(t)}{dt}[/tex]

    (wrt an inertial frame)"

    For simplicity, let us say that in this case 1, a force [tex]\vec{F}[/tex] is exerted in the exact middle point, perpendicular to the length of the bar, during a given time interval [tex][t_1, t_2][/tex].

    Again, for simplicity let us say in this case 2, a force [tex]\vec{F}[/tex] (of the same magnitude as in case 1 ) is exerted at the end of the bar, perpendicular to the length of the bar, during a time interval [tex][t_1, t_2][/tex] (same time interval as in case 1).



    Yes. In the first case the bar will not rotate (only translation movement). In the second case the bar will rotate (translation and rotation movement).


    No. In both cases ( case 1 and case 2 ) the center of mass of the bar will describe exactly the same trajectory.

    In both cases the center of mass of the bar will describe exactly the same trajectory. That is precisely what is proved in the mathematical theorem I stated above.

    As I said, if the exterior force is exactly the same vector in both cases ( case 1 and case 2, only difference is the point of application of the exterior force ), being applied exactly during the same time interval, then the center of mass of the bar will describe exactly the same trajectory in both cases, BUT (and this is important) the work of this "same" force will be different in case 1 and case 2. (The work is greater in case 2 because the end-particle of the bar where the exterior force is applied in case 2, describes a longer trajectory during that time interval, than the trajectory of the center of mass ( where the exterior force is applied ) in case 1 ). That is why the increment of total kinetic energy is greater in case 2 than in case 1.


    EDIT: sorry, I did not see that mfb had already answered your question correctly.
     
  5. Jun 14, 2014 #4
    Thanks MFN and Matt. That's what doesn't seem correct. That the center of mass will have the same velocity change. Then the bars in both cases will have the same kinetic energy by virtue of the velocity of the cm but in the second case you will have additional kinetic energy by virtue of the rotation. I took this to mean that the velocity of the cm could not be the same otherwise one case has more energy. It occurs to me then that the point mass must have a different velocity after collision in the two cases. The velocity of the point mass in the second case must be less so that the ke of the point mass in the first case after collision equals the ke of the point mass in the second case after collision plus the ke of the bar due to rotation.
     
  6. Jun 14, 2014 #5

    AlephZero

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    If the bar is at rest and there is a perfectly elastic collision with a point mass traveling with velocity v, the force (or more accurately, the impulse, i.e. the change of momentum of the point mass and the bar) is different depending on where the mass hits the bar.

    The book is talking about the situation where the force is the same in both cases.

    If you apply the same force for the same time interval at the center of mass and at some other point along the bar, the velocity of the COM will be the same in both cases. If you apply the force off center, the bar will also rotate about its COM, and the total kinetic energy of the bar will be greater. That doesn't contradict the ideas of work and energy, because in the off-center case the point where you apply the force moves through a greater distance, so the same force does more work.

    "Common sense" is confusing here, because humans don't have the "common sense" ability to judge accurately whether two forces are the same or not. If you actually measured the forces, the results would agree with the textbook.
     
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