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What is the maximum height of the square's center-of-mass?

  1. Aug 12, 2014 #1
    1. The problem statement, all variables and given/known data

    A square shaped block of mass m travels to the right with velocity v on a frctionless surface. The block has side-length 2d. The block hits a very small, immovable obstacle on the floor and starts to tip.

    The block has moment of inertia Icm=2/3md^2 about an axis through its center of mass and perpendicular to its face.

    Assume the block is travelling slow enough such that it does not tip over. To what maximum height does the block's center-of-mass rise?

    Please see included picture below. I also uploaded it to imgur (http://imgur.com/qptv9Iw)

    2. Relevant equations

    Conservation of Angular Momentum
    L_i = L_f
    L_i = Iw + mrvsin(theta)


    3. The attempt at a solution

    I took the axis to be the obstacle on the ground. This gave me an initial angular momentum of L_i = mvd. For the final angular momentum, I have I_square*omega. I was thinking that omega would equal v/h. Since this approach was not correct, I attempted to use conservation of energy, even though I don't think conservation of energy applies in this case because I think the obstacle absorbs some of the energy of the square. I solved for omega using the first approach then plugged it into my conservation of energy equation (setting the gain in gravitational potential energy equal to the square's rotational energy). Still have yet to reach a solution.

    This question may be more (or less) complicated than I think but I just can't get it. I'll appreciate any help or tips!


    http://imgur.com/qptv9Iw
     

    Attached Files:

    Last edited: Aug 12, 2014
  2. jcsd
  3. Aug 13, 2014 #2

    ehild

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    You said conservation of angular momentum holds at the collision with the obstacle: mvd=Iω0. Why not use it? What is the moment of inertia with respect to the obstacle? What is the initial angular speed then?

    ehild
     
  4. Aug 13, 2014 #3

    Orodruin

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    Did you account for the fact that the given moment of inertia is for rotations around the center of the square and not around the point the square actually rotates?
     
  5. Aug 13, 2014 #4
    Oh yes! Parallel axis theorem.
    So Li = mvd.
    And Lf = (2/3*m*d2 + m*h20

    v=ωr
    Would ω0 just be v/d?
     
  6. Aug 13, 2014 #5

    Orodruin

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    I am uncertain what you mean by Lf. Note that angular momentum is not conserved throughout the rotation due to gravity and normal force providing an unbalanced torque. However, once rotating, energy (potential + kinetic) will be conserved as the force from the ground on the block is not performing any work.
     
  7. Aug 13, 2014 #6

    Orodruin

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    Ok, so thinking a bit about your last post I have some further comments:
    • Note that the moment of inertia will be fixed and not depend on h. What is the distance between the center of mass and the rotational axis?
    • The initial angular momentum before impact is mvd as you say. What is the angular momentum right after impact?
    • Given the above, what is the rotational energy right after impact?
    • What is the rotational energy at the highest point and what is the potential energy at the highest point? What conclusion do you draw from this?
     
  8. Aug 13, 2014 #7
    Thank you for your tips! I'm trying to apply them, as follows.

    Okay so the square is moving with angular momentum mvd. I'm confused as to the time frame. Because right when it hits the obstacle, it stops. So I would say that right when it hits the obstacle, its angular momentum is still mvd, but since it stops, that is not correct. Right after impact, I think its angular momentum is Iω, but as you pointed out, this is not conserved through the rotation.

    When it begins to rotate, the square's initial translational kinetic energy 0.5mv2 is transferred into rotational energy and gravitational kinetic energy, 0.5Iω2 + mgh?
    0.5mv2 = 0.5Iω2 + mgh, where I=(2/3)md2
     
  9. Aug 13, 2014 #8

    Nathanael

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    It stops moving horizontally, but then it starts rotating right? So why do you say that is not correct?

    You can't say that because there is a collision. Energy is surely (or at least possibly) lost in the collisioin, right?

    You can only use conservation of energy after the collision (to relate Gravitational-PE and Rotational-KE)
     
  10. Aug 13, 2014 #9
    Before the collision, angular momentum is mvd. When the collision happens, angular momentum is Iω. And after the collision, rotational energy gets transferred into gravitational potential energy 0.5Iω2 = mg(h-d)?

    So angular momentum is conserved before and during collision --> mvd=Iω
    And energy is conserved after collision --> 0.5Iω2 = mg(h-d)
     
  11. Aug 13, 2014 #10

    Nathanael

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    Right. By virtue of the location you chose the rotation axis to be, there is no torque on the box, and so mvd=Iω

    As Orodruin pointed out, you'll have to be careful to use the correct rotational inertia.
     
  12. Aug 13, 2014 #11

    ehild

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    Initially the square slips on the horizontal surface, and its translational velocity is v. With respect to the obstacle, the angular momentum is mvd.

    When it hits the obstacle, the square stops its translational motion, but starts to rotate about the axis through the obstacle. At that instant, the square is still upright, its centre is still d distance from the ground. What is the moment of inertia about the axis of rotation? What does the Parallel Axis Theorem say? If you know I, the moment of inertia about the axis of rotation, you can determine the initial angular speed ω0 from conservation of angular momentum mvd=Iω0.

    When the rotation starts, the square does not translate: it rotates about the fixed axis through the obstacle. So the initial kinetic energy is 0.5Iω02.
    The angular momentum about the axis of rotation is not (2/3)md2. You find it applying Parallel Axis Theorem.

    You need to find the maximum height of the centre of the square. What is ω then?


    ehild
     
  13. Aug 13, 2014 #12
    Oh. I didn't realize that. Okay, I'm going to attempt again. Lots of things to keep in mind in these problems!
     
  14. Aug 13, 2014 #13
    Okay, I think I've got it.

    The rotational inertia at the moment of collision is the inertia about the center of mass plus m*(sqrt(2)d)2 --> (2/3)md2 + m*(sqrt(2)d)2 = (8/3)md2

    So conservation of angular momentum for before/during collision gives
    mvd=(8/3)md2ω --> ω = (3v/8d)

    Conservation of energy after collision gives
    0.5Iω2 = mg(h-d)

    In this case, the moment of inertia is (2/3)md2. We then have
    d + (1/3)d2ω2/g= h

    Plugging in ω = (3v/8d), we have
    d + (1/3)d2(3v/8d)2/g= h

    h = d + 3v2/64g
     
    Last edited: Aug 13, 2014
  15. Aug 13, 2014 #14
    Actually, I think I am wrong about the moment of inertia. I think it is also (8/3)md2 after the collision. So conservation of energy after collision gives
    0.5Iω2 = mg(h-d)

    In this case, the moment of inertia is (8/3)md2. We then have
    d + (4/3)d2ω2/g= h

    Plugging in ω = (3v/8d), we have
    d + (4/3)d2(3v/8d)2/g= h

    h = d + 3v2/16g
     
  16. Aug 13, 2014 #15

    Nathanael

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    Looks good.

    In the future it might be easier to not plug in the numbers till the end.

    For example, you could have combined your first two equations [[itex]mvd=Iω[/itex] and [itex]0.5Iω^2=mg(h-d)[/itex]] in the following way:

    [itex]0.5\frac{(mvd)^2}{I}=mg(h-d)[/itex]

    And then solved for h (and then plug in I)


    It just makes it a little easier to see where you went wrong :smile: Doesn't really matter, though.
     
  17. Aug 13, 2014 #16

    ehild

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    Well done ! :smile:

    ehild
     
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