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Kinetic energy of a hollow cylinder

  1. Jan 24, 2016 #1
    Hello
    1. The problem statement, all variables and given/known data

    In a uniform solid cylinder of radius r, mass m and height h we emptied a cylindrical cavity of radius r/4 tangent and parallel to the symmetrical axis z of the original cylinder.
    The cylinder is rolling without slipping on a horizontal plane. Find the kinetic energy in terms of θ and θ' where θ is the angle between vertical and the line joining the center of the original solid cylinder and the center of the cavity.

    2. Relevant equations
    K= 1/2 Icmw²+1/2mvcm²
    K=1/2 I w²

    3. The attempt at a solution
    I found the moment of inertia I=17/32 m r²
    and Icm=17/32 m r² - md²
    To find d² I found the z component of cdm = h/2

    I don't know which formula is easier to use the first one using Icm and vcm or the second one ?
    It is rolling without slipping so w and vcm are related...
    But I don't know how yo find w, is it w=θ' * r ?
    Thanks
     
    Last edited: Jan 24, 2016
  2. jcsd
  3. Jan 24, 2016 #2

    BvU

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    Hello Lilphy, :welcome:

    Could you show the steps taken to find ##I = {17\over 32} mr^2##.
    I wonder what ##I_{\rm cm} is once you found I. :rolleyes:

    And I see variables in your equations and solution attempt that I don't see in the problem statement... d, h,

    My side view of the cylinder (which you have drawn for yourself, I should hope) is as folllows:

    upload_2016-1-24_23-42-45.png

    ##\omega## is indeed ##\dot \theta r ## as you expect.
    [edit] oh boy! see post #3 o:)

    Do you think the mass of the rolling object is ##m## as in the problem statement, or is it ##{15\over 16}m## ?
     
    Last edited: Jan 25, 2016
  4. Jan 24, 2016 #3

    haruspex

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    I think you mean ##v_{cm}=r\omega=r\dot \theta##
     
  5. Jan 24, 2016 #4
    thanks for the answer !

    H is the height, d is just the variable used in the definition of the theorem of parallel axis.

    To find 17/32 m r2, I just calculated the triple integral
    I= 2 π ρ h ∫ r3 dr for r from a/4 to a.
    And replacing for ρ=M/(πhr2 15/16). But my M is false so ρ should be=m/(πhr2). So I would be now = 255 r2/512.

    But i am not sure this is the good answer because the cylinder is not emptied in the center, so i don't know if this result holds.

    To find the distance d to determine Icm, I don't know if it is correct but I calculated the z coordinate of the center of mass and found h/2 so d=h/2. coordinatez= ρ/M 2π ∫∫ zr dr dz r from a/4 to a and z from 0 to h. Then replaced M by ρ*h*π*r2 15/16.
    :oldconfused::oldconfused:
     
  6. Jan 24, 2016 #5

    haruspex

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    You don't need to any integration for this problem, just apply standard results.
    I would not bother finding the common mass centre. Just treat the motion as rotation about the rolling point, so find the moments of inertia about that point.
     
  7. Jan 24, 2016 #6
    I am not sure I understand what you mean by rolling point ?
    So I just have to calculate E=1/2 I w2 with w=θ'
    And the moment of inertia would be the moment of inertia of a solid cylinder but with the 15/16 of the mass ?
     
  8. Jan 24, 2016 #7

    haruspex

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    I mean about the point of contact with the ground.
    No, I didn't say that.
    Treat the hollow as a negative mass cylinder, same density, combined with a normal complete cylinder.
    Each has a moment of inertia about the point of contact with the ground, and you can find these using standard formulae.
     
  9. Jan 24, 2016 #8
    Okℤ
    For the solid cylinder K=1/2(1/2Mr2θ'2)
    I think that for the hollow cylinder there will be a dependance in θ, but in the sketch of bvu, the center of the solid cylinder and of the cavity are aligned with the vertical, so i don't "see" what theta represents ?
    Thank you !
    Edit: Now i know what theta is i forgot that the cylinder is rolling
     
  10. Jan 24, 2016 #9

    haruspex

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    No, you need the moment of inertia about the point of contact with the ground.
     
  11. Jan 24, 2016 #10
    Right, so I have to use parallel axis theorem.
    I=1/2Mr2+Mr2 = 3/2 M r2 ?
     
  12. Jan 24, 2016 #11
    and for the hollow it would be
    I=1/2m(r/4)2+mr/4sinθ with m=-ρlπr2/16 ?
     
  13. Jan 24, 2016 #12

    haruspex

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    Where M is the mass of the completed cylinder, yes.
    A few problems with that term. How far is the centre of the cavity from the point of contact?
    (It's not clear from the problem statement whether theta is measured from the upper radius or the lower radius. I.e. at theta=0, is the cavity above or below the middle of the cylinder?)
     
  14. Jan 24, 2016 #13
    The centre of the cavity using cos law is at a distance of sqrt(5r2/4 -r2/2 cosθ)
    So I=m/2 (r/4)^2+ msqrt(5r2/4 -r2/2 cosθ)

    Then the total inertia moment at the point of contact is the sum of both.
    And K=1/2 Itotθ'2
     
  15. Jan 24, 2016 #14

    haruspex

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    That's not what I get. Please show your working.
     
  16. Jan 25, 2016 #15
    Oups I made an error in calculating the distance.
    It is sqrt(15r2/16-r2/2 cosθ)
     
  17. Jan 25, 2016 #16

    haruspex

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    That's nearer to what I get, but still two differences.
    Please clarify whether you are measuring theta from the radius that goes up from the centre or from the radius that goes down from the centr.
     
  18. Jan 25, 2016 #17
    http://imageshack.com/a/img911/1145/CawDxl.jpg [Broken]
    This is an approximate sketch of the situation and how i calculate the distance. C is the contact point.
    BD= r/4 sinθ
    AD=r/4cosθ
    So BC=sqrt( (AC-AD)^2+BD^2 ) = sqrt( (r-r/4cosθ)^2 + (r/4 sinθ)^2 ) = sqrt( r2-2r2/4 cosθ + (r/4)2) = sqrt( 17r2/16 -r2/2 cosθ)
     
    Last edited by a moderator: May 7, 2017
  19. Jan 25, 2016 #18

    haruspex

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    Good, you got 17/16 this time, which is what I have. The other difference was because I was measuring theta from the upper vertical radius, whereas you show it from the lower.
    On to the next step.
     
    Last edited by a moderator: May 7, 2017
  20. Jan 25, 2016 #19
    The total moment of inertia at the contact point is Itot= 3/2Mr2+mr2/32+17mr2/16-mr2/2 cosθ = 3/2Mr2+35mr2/32-mr2/2cosθ
    m=-1/16M
    Itot=3/2Mr2-35Mr2/512+Mr2cosθ/32
    Is my Itot correct ?
     
  21. Jan 25, 2016 #20

    TSny

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    I think that's right. Of course, you can combine the first two terms.
     
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