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Getting frequency from angular momentum?

  1. Jun 21, 2016 #1
    1. The problem statement, all variables and given/known data

    I messed up the original thread in terms of information, so I made a new better one. Ok here goes:

    I have a problem where a right circular cylinder with diameter .115meter and a mass of .03kg is rotating at 300 rev per min. Now an Ant which weighs .01kg collides with the rotating cylinder and lands on the rim of the cylinder. The Ant is treated as a point mass.

    So I am looking for the final frequency in rev/min of the cylinder/Ant system? Which is part a of the problem?

    This is the farthest I got, and I wanted to know if my work is good so far and how to use the angular momentum to get the final frequency?


    2. Relevant equations
    I(inertia)=mr^2 this is for point mass
    w=v/r I used v=300 rev/min(not sure about this)
    angular momentum is L=I*w

    3. The attempt at a solution
    I=(.01kg)*(.0575m^2), so I=33.06*10^-6 inertia for the point mass after it landed on the cylinder
    w=300/(.0575m), w=5.21k rad/sec(this part I am not sure about)
    L= (33.06*10^-6 )*(5.21k rad/sec)
     

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    Last edited: Jun 21, 2016
  2. jcsd
  3. Jun 21, 2016 #2

    haruspex

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    Your conversion from revs/min to rad/sec is wrong. It has nothing to do with the radius.
    How many revs/sec is it? How many radians in one revolution?
     
  4. Jun 22, 2016 #3

    CWatters

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    Read Simon's post on that other thread again. Don't start with equations, think about principles. See if you can reply to this without mentioning a single equation :-) If you get that right, then you can think about which equations might be relevant.

    So which principle should you apply here?
     
  5. Jun 22, 2016 #4
    @haruspex 300 rev/min =31.42 rad/sec(300/60=5 5*2pi= 31.42rad/sec), also since you stated it has nothing to do with the radius this means w=31.42 rad/sec, which saves me some work. The thing I don't get is how do I get the wfinal which is after the ant landed so that I can then convert into the final frequency's rev/min ? I mean even if I get L(angular momentum) how would that help me get Ffinal or wfinal, especially since the w I am using is using the initial frequency of 300 rev/min?

    @ CWatters the principle I am using is related to rotational quantities or to be more specific the angular momentum. Since the kinetic energy changes its not conservation of kinetic energy. But I don't see the relationship with getting the angular momentum and wfinal or the final frequency?
     
    Last edited: Jun 22, 2016
  6. Jun 22, 2016 #5

    gneill

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    The problem is a rotational version of a linear collision problem. The same concepts apply as for linear collisions, only using rotational quantities rather than linear ones (even the equations are the same).

    What type of collision is portrayed? What is conserved in that type of collision?
     
  7. Jun 22, 2016 #6
    I was doing some thinking an made the conclusion that the conservation of angular momentum is conserved therefore I1*w1=I2*w2. Where I is the inertia So w2=(r1/r2)^2*w1 Where w2 is the final angular velocity an I can get the final frequency from that. My problem is since the radius does not change its basically w1 *1 which gives the final angular velocity = to the initial. Is this a correct assumption?
     
  8. Jun 22, 2016 #7

    haruspex

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    I1 is for just the cylinder, but I2 includes the ant.
     
  9. Jun 22, 2016 #8
    So for getting I2 the mass will be the cylinder mass + the ants mass correct? And my intial assumption about w2=w1 cause the radiuses does not change is that correct? Since w2=(r1/r2)^2*w1?
     
  10. Jun 22, 2016 #9

    haruspex

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    No, the moment of inertia, I2, will be the cylinder's moment of inertia about its axis plus the ant's moment of inertia about that axis.
    No, the angular velocity will change because the total moment of inertia changes while angular momentum stays constant.
     
  11. Jun 22, 2016 #10

    CWatters

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    Momentum/Angular Momentum is always conserved.
     
  12. Jun 22, 2016 #11
    Alright beer with me on the math becasue I want to check my answers to both part a and b

    Part A)
    I1*w1=I2*w2

    I1=.5*m*r2=.5*(.03*(.0575meter2)) = 49.5*10-6
    w1= 300/60=5 5*2pi= 31.42rad/sec

    The ants inertia is I=(.01*(.0575^2))= 33.06*10-6

    I2=(49.5*10-6 + 33.06*10-6) = 82.56-6

    w2=(49.5*10-6*31.42)/82.56*10-6

    w2=18.8 rad/sec, Ffinal=179.8 rev/min

    Am I good so far?

    Part B)
    Change in kinetic energy is the initial - final,

    Kinitial=.5*I*w2=.5*(49.5*10-6)*(31.422)=24.43*10-3

    Used I2 as I for k final
    Kfinal=.5*I*w2=.5*(82.56*10-6)*(18.8^2)=14.59*10-3

    The change in k is trivial so I wont compute it, but is everything good?
     
    Last edited: Jun 22, 2016
  13. Jun 22, 2016 #12

    haruspex

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    No. You got this right in your original post.
     
  14. Jun 22, 2016 #13
    thank you, I fixed it now, the only thing that concerns me is for the Kfinal. I can just add both the ant and cylinder mass together, or must I find each of their kinetic energy separately then add them to get kfinal?
     
  15. Jun 22, 2016 #14

    haruspex

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    What formula leads to that calculation? And why do you want find the peripheral velocity? How do you normally calculate rotational KE?
     
  16. Jun 22, 2016 #15
    Ah I used the wrong stuff, KE rotational is .5*I*w2, I corrected it. And for KE final i used I2 calculated value. Thanks guys.
     
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