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Find final angular velocity.

  1. Mar 5, 2012 #1
    1. The problem statement, all variables and given/known data
    A solid disk of radius 5m and mass 8 kg rotates clockwise at 1.5 rad/s. Above this disk is a hoop of radius 2.5 m and mass 8 kg , rotating counterclockwise at 3 rad/s. The hoop drops down onto the disk, and friction causes them to rotate together. Find the final velocity. Find the kinetic energy lost in this collision.


    2. Relevant equations

    ωf=ωi+αt and θf=θi+ωit+(1/2)αt2 I=1/2 Mr^2 I=Mr^2

    3. The attempt at a solution

    I think that by starting to find the acceleration of each of the disk i will be able to find the velocity and times of each one of them.
     
  2. jcsd
  3. Mar 5, 2012 #2

    gneill

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    Since you don't have a coefficient of friction to work with, you'll have a difficult time trying to find the forces and accelerations. Is there another approach?
     
  4. Mar 5, 2012 #3
    I was thinking in angular momentum Li=Lf
     
  5. Mar 5, 2012 #4

    gneill

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    Yup. Much better!
     
  6. Mar 5, 2012 #5
    (mr^2w^2)initialhoop+(1/2mr^2w^2)finaldisk=(mr^2w^2)finalhoop+(1/2mr^2w^2)initaldisk but than i find that i have two different angular velocities so i cannot factor.
     
  7. Mar 5, 2012 #6

    gneill

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    You have numerical values for all the parameters, so why not begin by calculating the important quantities separately?

    I1 = ? (Moment of inertia for the disk)
    I2 = ? (Moment of inertia for the hoop)
    L1 = ? (Angular momentum of the disk)
    L2 = ? (Angular momentum of the hoop)

    and you'll need

    KE1 = ? (Initial rotational KE of the disk)
    KE2 = ? (Initial rotational KE of the hoop)

    Keep in mind that angular momentum has a "direction", it can be positive or negative depending upon the sense of rotation.
     
  8. Mar 5, 2012 #7
    ok give me a minute.
     
  9. Mar 5, 2012 #8
    I1 = 1/2mr^2=0.5(8kg)(5m)^2=50kgm^2 (Moment of inertia for the disk)
    I2 = Mr^2=(8kg)(2.5)^2=100kgm^2 (Moment of inertia for the hoop)
    L1 = Iw=(100kgm^2)(1.5rad/s)=-150Nm/s (Angular momentum of the disk)
    L2 = Iw=(50kgm^2)(3rad/s)=150Nm/s (Angular momentum of the hoop)

    KE1 = 1/2Iw^2=1/2(1/2mr^2)(v/r)^2=1/4mv^2=1/4(8kg)(1.5 rads)^2=4.5 J (Initial rotational KE of the disk)
    KE2 =1/2Iw^2= 1/2(mr^2)(v/r)^2=1/2mv^2=1/2(8kg)(3rad/s)^2=36 J (Initial rotational KE of the hoop)

    Is that right?
     
  10. Mar 5, 2012 #9

    gneill

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    I think you swapped the values for the moments of inertia when you copied them for I1 and I2, but I see that you used the correct ones for the angular momenta, and they look good.

    The KE calculations look very strange (and the results are incorrect). What are those v terms? Angular momentum is simply (1/2) I ω2. And you have values for I already calculated and ω is given to you for each object.
     
  11. Mar 5, 2012 #10
    well the v terms are w=v/r and doesnt the KE rot cancel out the radius when you turn w into (v/r)^2?? was that wrong ??
     
  12. Mar 5, 2012 #11

    gneill

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    Well, there's no v involved anywhere! You're dealing with angular velocities only. You really only need to deal with v's or a's (linear stuff) when there's some interface between an angular motion part of a system and a linear motion part.
     
  13. Mar 5, 2012 #12
    so you are saying that instead of doing 1/2(mr^2)(v/r)^2 i should just leave 1/2(mr^2)w^2 instead and solve for KE rot?? also leaving the r alone too??
     
  14. Mar 5, 2012 #13

    gneill

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    Sure, if you wish. But you've already calculated the moments of inertia so I don't know why you're repeating the effort when calculating the KE. Why not just plug in your already calculated values? ##KE = \frac{1}{2} I \omega^2##. Not a v in sight :smile:
     
  15. Mar 5, 2012 #14
    XD True.
     
  16. Mar 5, 2012 #15
    KE1 = 1/2(100kg/m)(1.5)^2=112.5 J (Initial rotational KE of the disk)
    KE2 =1/2(50kg/m)(3rad/s)^2=225 J (Initial rotational KE of the hoop)

    is that now correct??
     
  17. Mar 5, 2012 #16

    gneill

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    The results are correct.

    This doesn't affect the numerical results here, but watch your units: the moment of inertia has units kg*m2.
     
  18. Mar 5, 2012 #17
    so from here how do I find the total angular velocity??
     
  19. Mar 5, 2012 #18

    gneill

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    How would you do it for a linear system if you knew the total angular momentum? This is the angular version of a linear inelastic collision.
     
  20. Mar 5, 2012 #19
  21. Mar 5, 2012 #20

    gneill

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    Zero is a perfectly respectable magnitude :smile:

    If you have two equal mass objects traveling with the same speeds in a head-on collision and they stick together, what's their final velocity?

    By the way, the equation shown in your link is for elastic collisions, not inelastic. In a perfectly inelastic collision the two objects fuse -- m1+m2.
     
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