1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Rotational Energy

  1. Nov 9, 2005 #1
    I am having problems with a rotational energy problem.

    It reads...
    a think stick of length 1.6m is dense at one edn than at teh other: it's mass desity is p=.4kg/m-.070kg/m^2x, where x is the distance from the heavier end of teh stick. The stick rotates about an axis perpindicular to the heavier end with a period of 1.1s. Determine teh Rotational kinetic energy of teh stick.

    I used K=1/2Iw^2. I found w by using the period. However, when I try to use the integral I=(integral)pR^2 dV, I cannot find R becuase it is not stated. How do I find R or is there another way around the problem?
     
  2. jcsd
  3. Nov 9, 2005 #2

    Fermat

    User Avatar
    Homework Helper

    R is the length of the stick.

    And your integral is wrong. It should be I=(integral)pr² dr, not I=(integral)pR^2 dV,
     
    Last edited: Nov 9, 2005
  4. Nov 9, 2005 #3

    Päällikkö

    User Avatar
    Homework Helper

    I disagree: it should be dV (I can't think why it shouldn't, enlighten me :smile:).

    (The distance from the rotational axis) R2 = x2 + y2. As the stick's thin: y,z [itex]\to[/itex] 0.
     
    Last edited: Nov 9, 2005
  5. Nov 9, 2005 #4
    R is the radius of the stick, I am sure of this. And the integral has to be with respect to volume, it's a 3d object
     
  6. Nov 9, 2005 #5

    Fermat

    User Avatar
    Homework Helper

    We have a stick rotating about one end. A small element δm at a distance r giving a small inertia δI = r²δm. Puting δm = ρ*δr and substituting gave me my final result. Hmm, how is it derived with dV, I'm not familiar with that ?
     

    Attached Files:

    Last edited: Nov 10, 2005
  7. Nov 9, 2005 #6

    Päällikkö

    User Avatar
    Homework Helper

    I feel this should be:
    δm = ρ*δV

    Then again as the stick's thin, δm = ρ*δV basically reduces to δm = y*z*ρ*δr, where y and z are the thicknesses into different directions (I don't think you can jus cancel them out).

    The attachement will take a while before it is approved (thus I cannot view it), maybe host the file on ImageShack? (EDIT: You already did :smile:)

    If you use your method to derive the I for a uniform thin rod, you will come up with (1/3)ρR3, which is wrong. Either of the methods I've mentioned should give the correct (1/3)MR2

    EDIT: Lots and lots of edits :smile:, small typos of all kinds.
     
    Last edited: Nov 9, 2005
  8. Nov 9, 2005 #7

    Fermat

    User Avatar
    Homework Helper

    ρ is the linear density, in kg/m
     
  9. Nov 9, 2005 #8
    The integral is still from 0 to 1.6 though right?
     
  10. Nov 9, 2005 #9

    Fermat

    User Avatar
    Homework Helper

    Yeah, I think we would both agree on that :smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Rotational Energy
  1. Rotational energy (Replies: 1)

  2. Rotational Energy (Replies: 3)

  3. Rotational energy (Replies: 1)

  4. Rotational Energy (Replies: 1)

  5. Rotational energy (Replies: 4)

Loading...