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Conservation of Angular Momentum Problem

  1. Oct 18, 2003 #1
    A block of mass 3.0 kg slides from rest down a frictionless surface from a height 0.66 m. The block collides with a uniform vertical rod of mass 4.4 kg and length 2.8 m and sticks to it. Find the angle theta that the rod and block pivot about O before momentarily coming to rest.

    I've uploaded the associated picture here ~> http://www.villagephotos.com/viewpubimage.asp?id_=5655439&selected=550922

    lemme just say, I have NO CLUE how to even approach his problem, but following an example in my horrible, HORRIBLE, text book, I did this:

    Kf = Kinetic Energy after the inelastic collision
    d = length of rod

    Kf = (L^2)/2I

    L = mvd


    Kf = ((mvd)^2)/2(md^2+(1/3)md^2) = 143943.8155

    ..lol..I decided to give up after that since I have No clue at all what to do with that absurdly large number.. At first I thought about dividing it by the weight of the system.. but I dont think I'd get very far in the right direction by doing that..any suggestions??
  2. jcsd
  3. Oct 18, 2003 #2

    Chi Meson

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    THis is a three part problem:

    First, you determine the speed of the block at the bottom of the ramp. Use the conservation of energy (U at the top = K at the bottom). Find " v "

    Then it is a conservation of angular momentum problem, where you have an inelastic collision. The initial " L " is that of the moving block at essentially a radial displacement of " d ". This is the amount of angular momentum that must be consterved. After the collision, you have a 2 piece object. The "moment of inertia" (I) of the block is mass x d^2 . Moment of inertia for the rod rotating about its end is (md^2)/2 .

    THe third part is figuring out how high the center of mass of the system will rise. THis can be done using the conservation of energy again.
  4. Oct 18, 2003 #3
    Um, I thought I did that
  5. Oct 18, 2003 #4
    At least, thats what I think I did now that I've tried tackling similar problems and have a better understanding of the concepts..

    Ok lemme try this again

    Using conservation of Energy:

    v(block)= sqrt(2gh) = 3.598 m/s

    So: L = mvd = 30.2232

    I'm thinkin you meant I for rod is (1/3)md^2 (standard formula of "thin rod about perpendicular line through one end" given by book) .. So:

    I' = md^2 + (1/3)Md^2 where m is the mass of the block and M is the mass of the rod

    I' = 23.52 + 11.4986 = 35.0187

    Rotational Kenetic Energy = (1/2)Iw^2 where w is angular velocity

    so Rotational Kenetic Energy = (Iw^2)/2I = L^2/2I

    So K = (30.2232^2)/2(35.0187) = 15993.7725

    lol hmm.. diff answer, I must've messed up somewhere in the original calculation, but anyways, on to the next step..

    (M+m)gh = K
    h = K/((M+m)g) = 220.31 m ?? that seems still too high, but aiight, moving along

    to find the angle from there I assume I'd have to use the equation d = dcos(theta) + h

    so arccos((d-h)/d) = theta = error

    Last edited: Oct 18, 2003
  6. Oct 19, 2003 #5
    I think you're overcomplicating.
    In the initial and final states, there is no motion.
    So why worry about velocities?
    All you need is the conservation of potential energy.

    Imagine the rod had zero mass. Then, clearly, the block would come to rest at the same height where it has started.

    Now since the rod has actually finite mass, some energy will go into the lifting of the rod's center-of-mass. Since this is located at the center of the rod, it will go up exactly half as much as the block does.
    This info is enough to find the answer from a simple equation.
  7. Oct 19, 2003 #6
    Oops, wait...
    What I said might be wrong. Since when things stick, there is some energy absorbed. So I better step thru your detailed analysis to see if I can find some errors...
    OK. Better: L = mvd = 30.2232 m2kg/s
    OK. Better: I' = 35.0187m2kg
    The I in the middle term's denominator shouldn't be there. However, the final term is OK. Should read:
    Rotational Kenetic Energy = (Iw^2)/2 = L^2/2I
    Here's a bad mistake.
    You used your calculator in a wrong way.
    You probably typed (30.2232^2) / 2 * (35.0187) =
    But you should have typed (30.2232^2) / 2 / (35.0187) =
    Here's another mistake.
    As I said, the center-of-mass of the rod only goes up half as much as the block. Should read:
    (M/2 + m)gh = K
    OK again.
    OK once more.
    Use the new values of K and h, and I guess you will be fine.
    Last edited: Oct 19, 2003
  8. Oct 19, 2003 #7

    Chi Meson

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    You could also find the position of the center of mass of the block-rod system. (See the c-o-m part of the text, usually with linear momentum). Find how high this point must rise (rotational K becomes U), call it "h". Then find the angle in the triangle where the hypotenuse is "d" and the adjacent side is "d-h."

    edit: no no! Don't do that! The hypotenuse is the original c-o-m distance from the pivot (call it "s"), and the adjacent side is "s-h".
    Last edited: Oct 19, 2003
  9. Oct 19, 2003 #8
    Thanks for the help Y'all!! That problem was killin me
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