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Conservation angular momentum, difficult problem

  1. Jul 12, 2009 #1
    1. The problem statement, all variables and given/known data

    In Fig. 11-59, a small 0.340 kg block slides down a frictionless surface through height h = 0.336 m and then sticks to a uniform vertical rod of mass M = 0.424 kg and length d = 1.69 m. The rod pivots about point O through angle θ before momentarily stopping. Find θ

    2. Relevant equations
    image at http://www.lowellphysics.org/beta/T...m/Problems/c11x11_14.xform_files/nw0627-n.gif


    3. The attempt at a solution
    I feel like if manipulated a million different things but just end up with something unsolvable. I tried using conservation of angular momentum to find the initial angular momentum of the system. and then i tried finding the net torque and relating this to the inertia of the system. At the end i have to many unknown quantites to solve for theta. Trying to get rid of angular acceleration, I end up with
    1.376^2=theta(13.3)sin(theta)....and i don't think there is a way to solve that...so there must be another way to do this problem...
    i think if i knew the angular acceleration of the system, i could solve the problem, but i don't. haha...any help would be appreciated.
     
  2. jcsd
  3. Jul 12, 2009 #2

    alphysicist

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    Hi Puchinita5,

    There are three parts to the motion in this problem: the block slides down the ramp, the block and rod have a collision, and the block-bar combination swing upward. For each of these three parts, there is some quantity that is conserved (and it is not the same quantity for all three parts).

    So first you have to decide: what is conserved during the slide? what is conserved during the collision? what is conserved during the swing?

    Then if you write out the conservation equation for each part of the problem, you can get the answer. What do you get?
     
  4. Jul 12, 2009 #3
    i used mgh=.5mv^s to find the block's linear velocity just before the collision. So (9.8)(.336)=.5v^s...solving for V i got 2.57 m/s
    -Then I used conservation of angular momentum to figure out the angular velocity just after the collision. So for L=angular momentum...Lrod2+Lblock2=Lblock1.....(and for symbols I is inertia and w is angular velocity)....I(rod)w +I(block)w=m(block)rv
    ....for the inertia of the rod i got .404 and the inertia of the block i have .971...

    so then i have (.404)w+(.340)(1.69^2)w=(.340)(1.69)(2.57)
    solving for w i get 1.07 rad/s

    after this i get really confused on what to do. Angular momentum is obviously not conserved since there is a Torque from gravity acting on it.

    i would assume at this point there would be conservation of energy...and i know rotational kinetic energy is .5Iw^2...but i have no idea what the formula would be for rotational potential energy, though if i were to guess it would be IgTheta???

    no idea what to do from this point, i tried using torque's relationship to inertia to solve for angular acceleration and then use this in the motion equations to solve for theta...but i cannot solve for torque without theta.
     
  5. Jul 12, 2009 #4

    alphysicist

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    No, that's not correct (you can tell it's not right because it does not have the right units). The potential energy here is just gravitational potential energy. It's the same potential energy as in non-rotational problems (mgy or mgh or whatever variables you use), but remember that the position in the potential energy formula is for the center of masses.
     
  6. Jul 12, 2009 #5
    hmm, okay...so would i set mgh=.5Iw^2 or would i set mgh=.5Iw^s+.5mv^2?
     
  7. Jul 12, 2009 #6
    well, those m's would not be the same...but you know what i mean? i guess the kinetic energy should be disregarded though because it was "before" the collision? so just mgh=.5Iw^2???
     
  8. Jul 12, 2009 #7

    alphysicist

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    I would just use two potential energy terms on the left side, one for the block and one for the rod. (You don't have to, though.)

    The rotational kinetic energy term (.5 I w^2) is accounting for all of the kinetic energy here that there is right after the collision, and so it is all you need for the kinetic energy.
     
  9. Jul 12, 2009 #8
    hmm i don't know if maybe i made a mistake somewhere, but when i do this i dont' get the right answer. I had m(block)gh+m(rod)gh=.5Iw^2
    and the mass of the block is .340
    mass of rod=.424
    height for block (1.69-1.69cos(theta))
    height for rod (1.69-1.22cos(theta))
    Inertia of rod and block=1.37
    w= 1.07

    do all those look correct? because when i use these numbers i end up with cos theta equaling a number greater than 1, which can't happen...

    not sure where i went wrong??
     
  10. Jul 12, 2009 #9

    alphysicist

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    I have not checked all of your numbers, but this does not look right to me. For the rod, you want the height of it's center of mass, so the length in your height equation will be L/2 instead of L.
     
  11. Jul 12, 2009 #10
    ahh u are right...i accidentally took the center of mass of the block rod system instead of just the rod...so it should be (1.69-.845cos(theta))
    however, even when i correct this i still get the wrong answer

    anything else look fishy?? lol...this has turned into quite an irritating problem for me!
     
  12. Jul 12, 2009 #11

    alphysicist

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    No, I don't think you should have two different numbers here; the 1.69 should be 0.845. Remember, when you have the energy equation in this form:

    (final potential enegy)=(initial kinetic energy)

    then you are saying that there is zero potential energy at the initial time (immediately after the collision). That means both block and rod have to be at h=0, so the h=0 position is different for each object (which is okay).


    As an alternative, if you want to have just one h=0 point (at the bottom of the swing), then remember that rod will have initial potential energy. (Which is okay too, but you'll just need an extra term on the right hand side.)
     
  13. Jul 12, 2009 #12
    fantastic!!! gosh i don't think i ever would have caught that mistake...thank you so much for your help!
     
  14. Jul 12, 2009 #13

    alphysicist

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    Glad to help!
     
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