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Center of Mass Velocity of rotating object

  1. Dec 2, 2013 #1
    1. The problem statement, all variables and given/known data
    As part of a carnival game, a 0.523-kg ball is thrown at a stack of 19.5-cm tall, 0.227-kg objects and hits with a perfectly horizontal velocity of 9.9 m/s. Suppose the ball strikes the very top of the topmost object as shown to the right. Immediately after the collision, the ball has a horizontal velocity of 3.15 m/s in the same direction, the topmost object now has an angular velocity of 1.64 rad/s about its center of mass and all the objects below are undisturbed. If the object's center of mass is located 13.0 cm below the point where the ball hits, What is the center of mass velocity of the tall object immediately after it is struck?


    2. Relevant equations
    L =pr
    I=MR2
    p=mv
    L=Iω


    3. The attempt at a solution
    p=mv
    p = Iω +mballvball
    v = (Iω +mballvball)/(mball*r + I)

    v = (I*1.64 +0.523*3.15)/(0.523*0.13 + I)
    I have calculated this way. However, the teacher said that the method I used to compute the center of mass velocity is wrong. The I value I calculate was correct tho.I would like to know what I did wrong in the process of solving for the center of mass velocity
     
  2. jcsd
  3. Dec 2, 2013 #2

    haruspex

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    I assume the first equation represents the linear momentum before impact, so m there stands for mball and v for the initial speed of the ball?
    The second one is strange. Iω is an angular momentum. You can't mix that with linear momentum. Is I the moment of inertia of the object about its centre? I don't think it would be merely rotating about its centre after impact.
    I don't see how you get this equation fro the other two.
    In fact, there seems to be some missing information in the OP. How wide are the objects? Or are they to be taken as very thin? If they have width, the velocity you are after will not be purely horizontal.
     
  4. Dec 2, 2013 #3
    yes, that is right. and I is the moment of inertia of the object about its centre. I could calculate I by using the conservation of linear momentum. However, I do not know what they mean by centre of mass velocity.
     
  5. Dec 2, 2013 #4

    haruspex

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    How? As I explained, this equation is dimensionally wrong:
    p and mballvball are linear momenta, whereas Iω is an angular momentum.
    It's the linear velocity of the mass centre. Suppose the ball comes from the left. If the top object is essentially a rectangle (in the vertical plane containing the trajectory of the ball) then when it is struck it will rotate about its lower right corner. If the distance from that corner to its mass centre is d then its instantaneous centre of mass velocity will be dω.
    The OP mentions a diagram. Can you post it? If not, what does the object look like in the diagram?
     
  6. Dec 4, 2013 #5
    Screen_Shot_2013_12_04_at_3_59_17_PM.png

    Here is the image.

    I decided to accomplish it this way, but it still didn't work.

    v = (mball*vball + mtall*0.13*ω)/(mball + mtall)

    I believe I have made it all to the same dimensions
     
  7. Dec 4, 2013 #6

    haruspex

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    Please confirm these definitions:
    • vball is the speed of the ball before impact
    • ω is the angular velocity of the bottle after impact
    If so, what is the basis for your equation? What conservation law(s) are you using? How are you arriving at the various before and after values?
    Seems to me that there are lots of things wrong with your equation. You're adding before and after values together; you're assuming the bottle CoM and the ball have the same velocity after; you're assuming the CoM velocity after is horizontal...

    But I am not at all sure how to solve the question. There is so much missing information. Are we to treat the bottle as zero width? If not, are we to take it that the bottle rotates about its lower right corner, or can it slide?
    If we can take it as zero width the question becomes almost trivial and we don't need much of the information provided. We can take the bottle as free floating in space before impact and just use conservation of linear momentum.
    Are there more parts to the question? If so, maybe that's where some of the other information comes in.
     
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