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Problem with center of mass exercise

  1. Jun 8, 2012 #1
    Hello! I'm here again, now with a new exercise! :biggrin:
    1. The problem statement, all variables and given/known data
    A block of mass m is at rest over this wedge (it's a triangle) of mass M that is at rest over a table. There is no friction. The point P of the block has a height h from the table. Then the block starts to move. What's the velocity of the wedge (triangle) when the point P hits the table? There is a picture from my textbook so you can see the situation.

    2. Relevant equations



    3. The attempt at a solution
    Well, since this exercise is from the Center of Mass chapter I tried to solve this by using center of mass. I know the coordinates of the center of mass they are ##x_{CM}## is ##2a/3## and that ##y_{CM}## is ##2b/3## (imagine a rectangle triangle). But the problem is that I have no ##b## or ##a## so I'm stuck. The only thing I was able to calculate was the velocity of the block when it hits the table ##\sqrt{2gh}=v## but it doesn't help at all. So, I'm stuck so I would like a hint to go on.
    Thanks!
     

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  3. Jun 8, 2012 #2

    vela

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    That's the center of mass of the wedge, but you need to consider the center of mass of the system. What can you say about the velocity of the system's center of mass?
     
  4. Jun 8, 2012 #3

    Doc Al

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    That would be the case if the wedge were fixed, but that's not the case here. Both wedge and block have kinetic energy. Hint: What's conserved?
     
  5. Jun 8, 2012 #4
    OK. The center of mass of the system has the coordinates ##(\frac{mx_1+Mx_2}{m+M},\frac{my_1+My_2}{m+M})##. Everything with 1 is the block and 2 the wedge. Now the velocity of the center of mass is ##v_{cm}=\frac{mv_1+Mv_2}{m+M}##. This isn't very helpful because I don't know ##x_1## etc...

    Oh, in this case I think that the total mechanical energy is conserved because there is no friction.
     
  6. Jun 8, 2012 #5

    gneill

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    Concentrate on the horizontal center of mass; Once the block leaves the wedge, both masses will be traveling horizontally.

    Ask yourself what are the conserved properties or quantities in the problem.
     
  7. Jun 8, 2012 #6
    The only force that makes the wedge move is ##mgcos\theta sen\theta## so the velocity of the wedge is related to this force. *just a random thought*

    Check out my draw below. What I'm trying to say is: if we think about the block as a point (point P), we might use those things I wrote in the draw to calculate the x coordinates of the center of mass of the system. IF I'm correct then the x coordinate of the center of mass would be ##x_{CM}=\frac{m.hcos\theta +M\frac{a}{2}}{m+M}##. But this is not good 'cause I don't know the length ##a## of the wedge.
     

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    Last edited: Jun 8, 2012
  8. Jun 8, 2012 #7

    vela

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    I should clarify I was only referring to the horizontal motion here.
     
  9. Jun 8, 2012 #8

    gneill

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    You shouldn't need to deal with picky details like particular forces, lengths, or locations if you work with the conservation laws :wink:
     
  10. Jun 8, 2012 #9
    But you guys are saying "what's the center of mass of the system" and "what's the velocity of the center of mass". This is what I calculated above right?

    Well, let me try again using the conversation laws. First of all I must know what is conserved. Well, there is no friction so the kinetic energy of each part of the system will be conserved. And because there is no friction the internal mechanical energy will be conserved as well. Is this thought correct?
     
  11. Jun 8, 2012 #10

    gneill

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    Yes, that's a good start. So how much KE will the system obtain (and from where)?
    Which other conservation law applies (ALWAYS applies)?
     
  12. Jun 8, 2012 #11

    HallsofIvy

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    You should look at both "conservation of energy", which you mention, and "conservation of momentum", which you don't.
     
  13. Jun 8, 2012 #12
    The kinetic energy that will be transferred to the system will come from the potential energy of the block. Trying to use ##K_i+U_i=K_f+U_F## where ##U_i=0## and ##K_f## is the kinetic energy of the whole system when the block hit the table and it's the difference in the kinetic energy of the whole and the kinetic energy of the wedge. How wrong is that?

    I was about to say conservation of momentum. But my textbook made me think that conservation of momentum only applies for collisions.
     
  14. Jun 8, 2012 #13

    gneill

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    Not so much wrong as incomplete. You should be able to write an expression for the amount of KE the system as a whole will obtain using the given parameters.
    Conservation of momentum always applies, collisions or not.
     
  15. Jun 8, 2012 #14
    Can I write ##K_i+U_i=K_f+U_f## for the system as a whole? If yes, in this case we would have ##0+mgh=\frac{m(v_{1})^2}{2}-\frac{M(v_{2})^2}{2}##.

    This is what I got trying to use the conservation of momentum ##mv_{1i}+Mv_{2i}=mv_{1f}+Mv_{2f}##. I think its incorrect but to me ##0+0=mv_{1f}+M_{2f}##. I don't know how to proceed.
     
  16. Jun 8, 2012 #15

    gneill

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    I'm not sure how you arrived at the second expression, but both terms on the right hand side are kinetic energies (one for the block, one for the wedge). The total kinetic energy should be the sum of the individual kinetic energies -- kinetic energy is a scalar quantity and you sum the magnitudes. The term on the left hand side is the potential energy which is the source of the kinetic energy.
    Momentum is a vector quantity, so direction matters. Your equation is a statement of the fact that the momenta of the block and wedge must be equal and opposite. In fact, conservation of momentum provides the reason why the trajectory of the center of mass of a closed system is unaffected by internal forces or motions of the system components; if it starts out at rest in some coordinate system, then it remains at rest (fixed).

    Taking your two expressions, one for KE and one for momentum, you have two equations with two unknowns: the final velocities.
     
  17. Jun 8, 2012 #16
    I used both equations to try to calculate the final velocity 2 but my answer is way too different of my textbook. My textbook's answer ##v=\sqrt{2m^2gh(\frac{cos\theta^2}{(M+m)(M+m sin\theta^2)})}## and I got something like ##v=2\sqrt{\frac{mgh}{M(2mM-2)}}##. Lol, I'm terrible at solving exercises....
     
  18. Jun 8, 2012 #17

    vela

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    The motion of the center of mass is closely related to conservation of momentum. The velocity of the center of mass is given by
    $$v_\text{cm} = \frac{mv + MV}{m+M} = \frac{\text{total momentum}}{\text{total mass}}$$ If the momentum of the system changes, the velocity of the center of mass changes and vice versa. In this problem, because there are no external forces acting on the system in the horizontal direction, horizontal momentum is conserved, or equivalently, the horizontal position of the center of mass of the system remains at rest.
     
  19. Jun 8, 2012 #18
    If the horizontal position of the center of mass of the system remains at rest it means that ##v_{CM}=0## right?
     
  20. Jun 8, 2012 #19

    vela

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    Right, which means the momenta are equal and opposite so they cancel.

    EDIT: I keep forgetting to qualify that I'm referring only to the horizontal components. The velocity of the center of mass in the vertical direction obviously changes because the block slides down while the wedge is constrained to move horizontally. The change in the vertical component of the momentum is caused by gravity, which is an external force acting in the vertical direction.
     
    Last edited: Jun 8, 2012
  21. Jun 8, 2012 #20
    As I wrote before ##-mv_{1f}=M_{2f}##. But I don't understand how I use it to solve the exercise.
     
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