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Cube & Pivoting Rod Collision

  1. Apr 19, 2008 #1
    1. The problem statement, all variables and given/known data
    A cube of mass m slides without friction at a speed vo. It undergoes a perfectly elastic collision with the bottom tip of a rod length d and mass 2m. The rod is pivoted about a frictionless axle through its center, and initially it hangs straight down and is at rest. What is the cube's velocity- both speed and direction after the collision?

    2. Relevant equations
    Moment of inertia for rod pivoted about center I=(1/12)mr^2

    3. The attempt at a solution

    I used conservation of energy

    Kcubeinitial = Krotational of rod + Kcubefinal
    I replaced angular velocity with v1/.5d

    .5mvo^2 = .5mvf^2 + .5(1/12)2md^2 * (v1/.5)^2

    Then I used conservation of momentum:
    mvo= mvf + 2mv1

    But substituting these two equations into each other leads to something I can't solve. Should I try angular momentum?
  2. jcsd
  3. Apr 19, 2008 #2


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    Hi bcjochim07,

    The problem with using conservation of linear momentum is that the rod does not have a single linear speed--the tip is going fastest, near the pivot is going slow, halfway along the rod it's going half the speed of the tip, etc.

    But the entire rod has the same angular speed, so conservation of angular momentum is useful here.
  4. Apr 20, 2008 #3
    Ok, so for angular momentum

    mvo*(d/2) = mvf(d/2) + I*omega

    v= omega * r omega= v/(d/2)

    But I haven't been get the expression for the velocity of the cube yet, which is just a multiple of vo
  5. Apr 20, 2008 #4


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    You'll have to use both conservation of kinetic energy and conservation of angular momentum.

    Your conservation of kinetic energy formula was

    .5mvo^2 = .5mvf^2 + .5(1/12)2md^2 * (v1/.5)^2

    but it looks like it's missing a factor of d; omega here is v1/(0.5 d), so the d's will cancel.

    Putting this together with your conservation of angular momentum gives two equations in two unknowns (because the answer should be in terms of d and m).
  6. Apr 20, 2008 #5
    Angular momentum:
    m(d/2)vo = m(d/2)vf + (1/12)(2m)d^2*(v1/.5d)
    (d/2)vo= (d/2)vf + (1/3)dv1
    (1/2)vo = (1/2)vf + (1/3)v1

    Kinetic energy
    .5mvo^2 = .5mvf^2 + .5(1/12)(2m)(d^2)(v1/.5d)^2
    .5vo^2=.5vf^2 + (1/12) (v1/.5)^2
    .5vo^2=.5vf^2 + (1/3) v1^2

    but how do I substitute these into each other to come up with answer, (1/5)vo ?
  7. Apr 20, 2008 #6


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    Hi bcjochim07,

    You have two equations in two unknowns vf and v1. You can solve one of the equations (the momentum would probably be easiest) for v1 for example and then plug it into the other equation.

    (You'll probably have a quadratic equation to solve unless there is a fortuitous cancellation.)
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