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Conservation of Angular Momentum help needed

  1. Nov 23, 2004 #1

    Can someone please help me with the following problem:

    A solid cube of wood of side 2a and mass M is resting on a horizontal surface. The cube is constrained to rotate about an axis AB. A bullet of mass m and speed v is shot at the face opposite ABCD at a height of 4a/3. The bullet becoes embedded in the cube. Find the minimum value of v required to tip the cube so that it falls on face ABCD. Assume m<<M.

    This is what I got so far:

    I understand that it is a perfectly inelastic collision and that


    L(initial)=I*Omega where I is the combined Inertia for both object. The I for a cube rotating around one of its edge is I=(8Ma^2)/3 and the bullet is a point mass with an r of sgrt((4a/3)^2+(2a)^2).

    omega=V/R where V is the final speed and R is the radius of rotation, which is ....????

    Thank you in advance!
  2. jcsd
  3. Nov 23, 2004 #2

    Doc Al

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    This won't help. Since the cube is constrained, translational momentum is not conserved. But angular momentum is conserved. (Hint: How much energy is needed to topple the "cube + bullet"?)
  4. Nov 23, 2004 #3


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    Careful about this one!
    You cannot use this with respect to the C.M. of the bullet+cube system, since that system is affected by an external force comparable in magnitude of the internal forces.
    That external force acts from the axis to which the cube is constrained, ensuring that the cube's contact velocity (with the axis) remains zero throughout the collision.

    Do you agree so far?

    it's only fair that Doc Al beats me occasionally..:wink:
  5. Nov 23, 2004 #4
    So then


    Bullet K(initial)=1/2mv^2
    Bullet + Block= K(final)=1/2I*omega^2


    where I is the combination of both bullet and block, I(block)=(8M(2a)^2)/3 and I(bullet)=mr^2 where r is sgrt[(4a/3)^2+(2a)^2]. Also, omega is=V/R where V is the final speed of a point on the block.

    Can you please double check...

    But how about R and how do I figure out how much Energy is needed in order to topple?
  6. Nov 23, 2004 #5


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    Why do you think kinetic energy is conserved in the collision?
    Doc Al wrote that the system's angular momentum is conserved (with respect to a point lying on the fixed axis)
  7. Nov 23, 2004 #6

    Doc Al

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    During the collision, angular momentum is conserved; after the collision, mechanical energy is conserved.
  8. Nov 23, 2004 #7
    How about this:

    Angular momentum right before the collision for the bullet is

    mvr where r is the distance from the rotation point.

    After the collision for the bullet


    The block's angular momentum after the collision is


    In both cases omega is the angular velocity of the block, which breaks down to omega=V/R

    So is it mvr=mr^2*omega+I*omega?



    Is R SQRT[8a^2]? How about V?
  9. Nov 23, 2004 #8

    Doc Al

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    Careful: [itex]\vec{L} = \vec{r}\times m\vec{v}[/itex]

    OK. What's the rotational inertia of the cube about its edge?

    Since m << M, you can probably ignore the moment of inertia of the bullet.

    You should be able to find [itex]\omega[/itex] in terms of the bullet speed v. What minimum value of [itex]\omega[/itex] is required to get the cube to tip? (That's where energy conservation will come in.)
  10. Nov 23, 2004 #9
    so [tex]L=rx\omega[/tex] so it is the cross product....

    that means that I need to take the perpendicular component to r which is the [tex]\sin[\theta][/tex] of the bullet's angle of impact to r?

    Rotational Inertia of the block around its edge is:

  11. Nov 23, 2004 #10

    Doc Al

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    Sure. Or you can find the perpendicular distance between the velocity vector and the the axis---that's given to you.

    Doesn't look right to me. I'd say the rotational inertia would be [itex]I = 2/3 Ms^2[/itex], where s is the edge length, in your case s = 2a, but you'd better check it.
  12. Nov 23, 2004 #11
    I thought that [tex]L=\frac{2}{3}Ms^2[/tex]is if the center of rotation is in the center of the cube. I will use the parallel axis formula to move it's rotational axis to see what I get....
  13. Nov 23, 2004 #12

    Doc Al

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    With the axis in the center of the cube, the rotational inertia would be [itex]I = 1/6 Ms^2[/itex]
  14. Nov 24, 2004 #13
    Still in need of understanding

    I am still having trouble finding [tex]\omega[/tex] in terms of the bullet's speed v. I do not understand how conservation of energy comes into play after the collision.

    BTW. You were right about the rotational inertia of the cube around it's edge.

    [tex]I=\frac{2}{3}Ms^2[/tex] since s=2a [tex]I=\frac{2}{3}M(2a)^2

    Thank you, and Happy Thanksgiving!
  15. Nov 24, 2004 #14

    Doc Al

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    This is just conservation of angular momentum: initial angular momentum (due to the bullet, in terms of v) equals angular momentum of rotating cube (in terms of [itex]\omega[/itex]) immediately after the collision.
    Immediately after the collision, the cube+bullet has some kinetic energy. As the cube tips up, KE is transformed to gravitational PE. So, how much energy is needed to make it tip over? (If it doesn't have enough KE, it will just fall back down again. It's just like if you wanted to roll a ball over the top of a hill: how much energy do you have to give it at the bottom to make it all the way to the top and over?)

    I know. :wink:
  16. Nov 24, 2004 #15
    Initial Angular due to the bullet

    [tex]L_{i} =mvr\sin{\theta}[/tex]

    Where [tex]r\sin{\theta}=d[/tex]

    Which is given as [tex]d=\frac{4a}{3}[/tex]

    so [tex]L_{i} =mv\frac{4a}{3}[/tex]

    Final Angular velocity of cube+bullet (bullet can be ignored, as you pointed out)

    [tex]L_{f} =I\omega[/tex]

    Where [tex]I=\frac{8}{3}Ma^2[/tex]

    So [tex]L_{i} =L_{f}[/tex]

    Yields [tex]mv\frac{4a}{3}=\frac{8}{3}Ma^2\omega[/tex]

    Solve it for [tex]\omega[/tex] and plug it into below...



    It has to rise higher than when the cube is it's corner which is
    [tex]2a=\sqrt{h^{2}+h^{2}}[/tex] so [tex]h=\sqrt{2}a[/tex]

    That is how far I got... Is PE=Mgh? Or is it KE=PE so, KE-PE=0...
    Last edited: Nov 24, 2004
  17. Nov 24, 2004 #16

    Doc Al

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    You're almost there. Consider that [itex]\Delta {KE} + \Delta{PE} = 0[/itex]. The change in PE of the cube is [itex]Mg\Delta{h}[/itex], where [itex]\Delta{h}[/itex] is the change in height of the cube's center of mass.
  18. Nov 24, 2004 #17
    My last question for the week...

    The cube has to rotate over it's edge. In order to do that I put the cube on it's corner figured out the height of a side corner to the groud..

    So is h going to be [tex]2a=\sqrt{h^{2}+h^{2}}[/tex]

    yields that [tex]h=\sqrt{2}a[/tex]

    Now I plug all that into the KE+PE=0 equation, solve for v and then make v larger than what I get on the other side. This way if v is larger it should have enough energy to rotate over a height h, which is the max height of a side corner if the cube is stands on one of it's corners.
  19. Nov 24, 2004 #18

    Doc Al

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    I assume you mean: [itex]h =\sqrt{a^{2}+a^{2}}[/itex]

    That's the final height of the center of mass. What is the change in height?

    Right! Be sure to use the change in height, not just the height.
  20. Nov 24, 2004 #19
    The change in height is


    Damn, that is going to be messy...
  21. Nov 24, 2004 #20
    I am getting


    Which I think is wrong, since I got a negative under the SQRT.

    Are we sure that it is [tex]\Delta KE+\Delta PE=0[/tex]

    and not [tex]\Delta KE=\Delta PE[/tex] initial KE transfers to final PE

    which is [tex]\Delta KE-\Delta PE=0[/tex] ???
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