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May we equate linear KE with rotational KE?

  1. Oct 28, 2006 #1
    The question is about a projectile hitting a rod; I need to find how much KE is lost.

    Using [tex]K_f = \frac{1}{2} I_f \omega_f^2[/tex] I came up with 162.253; using [tex]K_i = \frac{1}{2} m v^2[/tex] (m being the mass of the projectile), I came up with 16.9785; Is the answer [tex]K_f - K_i[/tex] in this form?
     
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  3. Oct 28, 2006 #2

    Office_Shredder

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    Yes, but it looks to me like you gained a buttload of free energy. Does the projectile stop moving after it makes contact?
     
  4. Oct 28, 2006 #3

    Hootenanny

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    Yes, total kinetic energy is conserved in an elastic collision.
     
  5. Oct 28, 2006 #4
    The projectile sticks to the end of the stationary rod. [Edited] It seems like KE was gained.
     
    Last edited: Oct 28, 2006
  6. Oct 28, 2006 #5

    radou

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    In that case, it's a plastic collision, and all kinetic energy turns into rotational kinetic energy, since the projectile and the rod rotate together around some point. If we assume that there is no deformation, then it's an ideal plastic collision, and no energy is lost.

    Edit: after seeing what Doc Al wrote and consulting my dynamics textbook, I concluded not to write nonsence. Kinetic energy is not conserved. :smile:
     
    Last edited: Oct 28, 2006
  7. Oct 28, 2006 #6

    Doc Al

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    State the full problem. You'll need to use conservation of momentum to find the final speed (translational and rotational) of the "stick + projectile" system. (Kinetic energy is not conserved.)
     
  8. Oct 28, 2006 #7
    But then [tex]K_i = K_f[/tex] yes? Yet they are different values.
     
  9. Oct 28, 2006 #8

    Doc Al

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    Please state the problem fully. (If a projectile hits a rod and sticks to it, that's an inelastic collision--kinetic energy is not conserved.)
     
  10. Oct 28, 2006 #9
    A projectile of mass m moves to the right with speed v. The projectile strikes and sticks to the end of a stationary rod of mass M and length d that is pivoted about a frictionless axle through the center.

    How much KE is lost in the collision relative to the initial KE?

    I have [tex]I_f = \frac{M d^2}{12} + m(\frac{1}{2} d)^2 = \frac{d^2}{12}(M + 3m)[/tex]

    and I found the [tex]\omega_f[/tex] of the system correctly:

    [tex]\omega_f = \frac{6 m v}{(M + 3m)d}[/tex]
     
    Last edited: Oct 28, 2006
  11. Oct 28, 2006 #10

    Doc Al

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    The initial KE is pure translational KE of the projectile; the final KE is pure rotational KE of the "stick + projectile" system. (Angular momentum is conserved.)
     
  12. Oct 28, 2006 #11
    so [tex]K_i[/tex] is translational KE, [tex]K_f[/tex] is rotational KE; how do I put these into similar units so I can find out how much KE was lost?

    ok, sorry for jumping ahead, we have [tex]K_i = \frac{1}{2} m v^2[/tex] but [tex]\omega = \frac{v_f}{r}[/tex], thus [tex]K_i - K_f = \frac{1}{2} m v^2 - \frac{d^2}{24}(M + 3m)(\frac{v_f}{r})^2[/tex]. ??
     
    Last edited: Oct 28, 2006
  13. Oct 28, 2006 #12
    Even if change [tex]\omega[\tex] into [tex]v[\tex], the KE goes up as where the question implies that KE is lost.
     
  14. Oct 28, 2006 #13

    Doc Al

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    [tex]K_i = \frac{1}{2} m v^2[/tex]

    [tex]K_f = \frac{1}{2} I_f \omega^2[/tex]
     
  15. Oct 28, 2006 #14
    For [tex]K_f[/tex] I end up with a greater KE than [tex]K_i[/tex] yet the question states "How much kinetic energy is lost in the collision relative to the initial kinetic energy"
     
  16. Oct 28, 2006 #15

    Doc Al

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    I don't see how you are getting a final energy greater than the initial. You have the correct values for angular speed and rotational inertia, so just plug them into the expression for rotational KE. Unless you're making an algebraic error, your final KE will be less than the initial. (If you're still stuck, show your work and we'll take a look.)
     
  17. Oct 28, 2006 #16
    Thank you for your help.

    A projectile with m = 1.47 kg moves tothe right with speed v=23.1 m/s. The projectile strikes and sticks to the end of a stationary rod of mass M = 6.25 kg and length d = 1.82 m that is pivoted about a frictionless axle through it's center.

    [tex]K_i = \frac{1}{2} m v^2 = \frac{1}{2} 1.47 (23.1)^2 = 392.203[/tex]
    [tex]K_f = \frac{1}{2} I \omega^2 = \frac{1}{2} \frac{d^2}{12}(M + 3m) (\frac{6 m v}{(M + 3m) d})^2 = 162.253[/tex]

    [edit:]oh my goodness! i was not squaring the velocity of the projectile! that makes all the difference. thank you everyone for your patience.
     
  18. Oct 28, 2006 #17

    Doc Al

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    Don't plug in numbers until the last second. First simplify that expression for final KE.
     
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