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Angular momentum/ energy

  1. May 1, 2014 #1
    1. The problem statement, all variables and given/known data

    A thin rod of mass M and length L is suspended vertically from a frictionless pivot at its upper end. A mass m of putty traveling horizontally with a speed v strikes the rod at its CM and sticks there. How high does the bottom of the rod swing?



    2. Relevant equations



    3. The attempt at a solution

    First I must look at the collision:

    [itex] v = ωR = ω\frac{L}{2} [/itex]

    [itex] ω = \frac{2v}{L} [/itex]

    [itex] I_m = mR^2 = m(\frac{L}{2})^2 = \frac{mL^2}{4} [/itex]

    [itex] I_M = \frac{1}{12}ML^2 [/itex]

    using angular momentum:

    [itex] I_m (\frac{2v}{L}) = (I_m + I_M)ω [/itex]

    [itex] \frac{mL^2}{4} (\frac{2v}{L}) = (\frac{mL^2}{4} + \frac{1}{12}ML^2)ω [/itex]

    [itex] \frac{mLv}{2} = (\frac{mL^2}{4} + \frac{1}{12}ML^2)ω [/itex]


    Have I set this up right so far? I am trying to find the angular velocity at the end of the collision, convert it to linear velocity, and plugging that into an energy equation
     
  2. jcsd
  3. May 1, 2014 #2

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    You need to use a different formula for the moment of inertia of the rod. Here, the rod is pivoting at its end, not at its center.

    Other than that, it looks good so far. :smile:
     
  4. May 1, 2014 #3
    So it would be this instead?

    [itex] \frac{mLv}{2} = (\frac{mL^2}{4} + \frac{1}{3}ML^2)ω [/itex]
     
  5. May 1, 2014 #4

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    Yeah, that looks right. :approve:
     
  6. May 1, 2014 #5
    ok so:


    [itex] \frac{v}{2} = (\frac{L}{4} + \frac{L}{3})ω [/itex]


    [itex] \frac{v}{2} = \frac{7L}{12} ω [/itex]

    [itex] v = \frac{14L}{12} ω [/itex]

    [itex] v = \frac{7L}{6} ω [/itex]

    [itex] ω = \frac{6v}{7L} [/itex]

    to get linear velocity use v1=Rω

    [itex] v_1 = (\frac{L}{2}) \frac{6v}{7L} [/itex]

    [itex] v_1 = \frac{3v}{7} [/itex]


    plugging this into an energy equation:


    [itex].5(m+M)(\frac{3v}{7})^2 = (m+M)gh[/itex]

    is this the correct set up for the energy equation to find h?
     
  7. May 1, 2014 #6
    wait i canceled out masses where I couldnt have. one sec
     
  8. May 1, 2014 #7
    [itex] \frac{vm}{2} = (\frac{mL}{4} + \frac{ML}{3})ω [/itex]

    [itex] \frac{vm}{2L} = (\frac{m}{4} + \frac{M}{3})ω [/itex]

    [itex] \frac{vm}{2L(\frac{m}{4} + \frac{M}{3})} = ω [/itex]



    to get linear velocity use v1=Rω

    [itex] v_1 = (\frac{L}{2}) \frac{vm}{2L(\frac{m}{4} + \frac{M}{3})} [/itex]

    [itex] v_1 = \frac{vm}{4(\frac{m}{4} + \frac{M}{3})} [/itex]


    plugging this into an energy equation:


    [itex].5(m+M)(\frac{vm}{4(\frac{m}{4} + \frac{M}{3})})^2 = (m+M)gh[/itex]

    is this the correct set up for the energy equation to find h?
     
  9. May 1, 2014 #8

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    I'm good with everything up to this point. :approve:

    But I don't think that's quite right. I wouldn't convert to linear velocity just yet. It's not really valid since different parts of the rod are moving at different speeds, and it's not valid to model the kinetic energy of the rod simply by the motion of its center of mass; it has rotational energy too.

    Instead, determine the initial kinetic energy in the form of [itex] \frac{1}{2}I \omega^2 [/itex] There's no reason to move away from angular coordinates just yet.

    You then have the choice of how you want to represent the final height and the final potential energy. Above, you used [itex] (m + M)gh [/itex], which is fine. But realize here that h is the change in height of the system's center of mass. The problem asks you to find the final height of the bottom of the rod, not the center of mass. So you'll have to double h before submitting your final answer.
     
  10. May 1, 2014 #9
    ok so the energy equation would be :

    [itex] (\frac{mL^2}{4}+\frac{ML^2}{3})(\frac{vm}{2L(\frac{m}{4}+\frac{M}{3})})^2 = (m+M)2gh [/itex]


    ??
     
  11. May 1, 2014 #10

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    Yes. At least that's the same equation that I got. :approve:

    Of course, some of the terms cancel, which makes it a little easier.

    So solve for h, double it, and I think you're done. :smile:
     
  12. May 1, 2014 #11
    ahh so making it 2h wouldnt make it right? i should have left it as h, solved for h and THEN doubled it?
     
  13. May 1, 2014 #12

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    Well, I thought your 2h on the right side came from the 1/2 in the [itex] \frac{1}{2}I \omega^2 [/itex] on the left side (it seemed to me that you multiplied both sides of the equation by 2).

    But yes, solving for h and then doubling it at the end should work. If instead you want to define h as the final height of the end of the rod (not the center of mass), from the get-go, the potential energy would be (m + M)g(h/2), would it not?
     
  14. May 1, 2014 #13
    ah right i forgot about that 1/2 by accident. so the 2 would cancel out with the h/2 leaving

    [itex] (\frac{mL^2}{4}+\frac{ML^2}{3})(\frac{vm}{2L(\frac{m}{4}+\frac{M}{3})})^2 = (m+M)gh [/itex]

    and solving for h here would give me the height?
     
  15. May 1, 2014 #14

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    Yes, I believe that is correct. No need to double h if you use the above equation.
     
  16. May 1, 2014 #15
    thank you!
     
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