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Angular momentume

  1. Dec 21, 2013 #1
    Three equal mass, m are conected to vertices of an equilateral triangle, with edges of length L. Edges are massless.The triangle rotates with constant angular velocity ω about an axis through its center of mass, perpendicular to the triangle plane. At t=0 mass 1 is released, at a very short period from the triangle. To answer use m l ω.
    pi1.JPG
    Calculate the angular velocity of the two masses that stay connected.
    To answer this I need to find the angular momentum at point A (center of mass of the two masses) but why before they released the angular momentume is

    LA=LCM+rxmvcm isn't angular momentum suppose to be calculated about the same point in this ces A? and according to this equation angular momentum is also calculated about the center of mass, is that wrong?

    Thank you in advance!
     
    Last edited: Dec 21, 2013
  2. jcsd
  3. Dec 21, 2013 #2
    Can you explain the setup a bit more, please? Where exactly is the axis of rotation? Does it pass through one of the masses or did you perhaps mean to say that it is perpendicular to the triangle rather than parallel?
     
  4. Dec 21, 2013 #3
    yes indeed "perpendicular to the triangle "! At the beginning the axis of rotation passes through the center of mass, (of the triangle, rcm=(0,L√3/3))
     
    Last edited: Dec 21, 2013
  5. Dec 21, 2013 #4
    So after the mass is released, you now basically have two objects... one is a mass on its own, the other is like a dumb bell of two masses joined together.

    The first thing that I suggest you do is find the CoM velocities of these two objects. Look at the single mass first... its velocity after the separation will be the same as its velocity just before the separation. Having done that, then you can use conservation of linear momentum to find the CoM velocity of the two-mass object. So get these two velocities in terms of the parameters you are provided with: omega and L. (It is somewhat unfortunate that the triangle side length is called L because there is potential to confuse it with angular momentum.)
     
  6. Dec 21, 2013 #5
    My goal is to equate the angular momentum before and after....so I'm asking how to make the equation before the separation...
     
  7. Dec 21, 2013 #6
    To write an expression for the angular momentum of the dumbbell part after separation, you will need to know the velocity of its centre of mass, note the presence of vcm in the equation you wrote in your opening post. I can think of two ways to do this... one is what I described with conservation of linear momentum, the other is actually more simple... its centre of mass lies on the middle of the edge of the triangle, so find the velocity of this point just prior to separation. After separation, this will remain the vcm for this dumbbell piece.

    After you have vcm for the dumbbell, you can equate the angular momentum of it before the separation to its angular momentum after separation... before separation, it is just 2mR2ω (R is the distance from the centre of the triangle to one of its corners) and for after separation, you get it from your equation: LA=LCM+rmvcm. In this equation, the only remaining unknown is Lcm, so you can solve for that.
     
  8. Dec 21, 2013 #7
    But the answer states before it is mL2ω why?
     
  9. Dec 22, 2013 #8
    The answer states that what is mL2ω? The total angular momentum of all three particles before separation? That is correct.

    It is asking you to calculate the angular velocity of the dumbbell, though. So the equation that you want to write down is of the form...

    Angular momentum of dumbbell before separation + Angular momentum of single particle before separation = Orbital momentum of dumbbell after separation + Spin momentum of dumbbell after separation + Angular momentum of single particle after separation.

    As it happens, the angular momentum of the single particle before and after separation is the same, so that will just cancel with itself.

    In the equation you wrote at the start, r×mvcm is the orbital momentum of the dumbbell and LCM is its spin. I think this is inconvenient notation, because there are so many L's going around. I'd call the spin momentum S and come up with some other letter for the side length of the tiangle.
     
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