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Moment of inertia and a gyroscope

  1. Jan 15, 2008 #1
    1. The problem statement, all variables and given/known data
    A gyroscope consists of a uniform disc of mass M and diameter d attached at its centre to a light spindle of length l perpendicular to the plane of the disc.

    i) Show that the moment of inertia, I, of the disc around the axis of the spindle is Md²/8.

    2. Relevant equations

    I=MR² Moment of inertia

    3. The attempt at a solution

    I cant understand why it becomes I=Md²/8 as R=d/2 and R²=d²/4 not d²/8

    or am i missing something?
     
  2. jcsd
  3. Jan 15, 2008 #2

    malawi_glenn

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    I = MR^2/2 for a circular plate.
     
  4. Jan 15, 2008 #3
    ok i'll ask my lecturer about that tomorrow as in her solutions to this she has written

    I=MR²

    thanks for the help
     
  5. Jan 15, 2008 #4

    malawi_glenn

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  6. Jan 15, 2008 #5
    The MOI of a disc rotating around its centre is certainly I = MR^2/2

    the quatity MR^2 is used when applying the parallel axis theory.
     
  7. Jan 15, 2008 #6

    malawi_glenn

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    It is very hard to see how this gyroscope is built and how it rotates :)
     
  8. Jan 16, 2008 #7
    yea, I=MR^2 is for used on a particle or a person in some examples.
    I=(MR^2)/2 is for a disc
     
  9. Jan 17, 2008 #8

    Shooting Star

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    Must be a careless mistake. Let the OP clarify with the teacher.

    What sort of persons in which examples?
     
  10. Jan 17, 2008 #9
    I was given two very similar questions (which I still cant tell apart)

    Question 1

    A uniform horizontal disc of mass M and radius R rotates about its vertical axis with angular frequency w. Find and expression for the moment of inertia.

    Solution

    Mass per unit area (lambda) = M/piR^2

    Mass of small ring of thickness dr (dm) = lambda*2pi*r*dr

    I = integral (0->R) r^2 dm
    I = integral (0->R) r^2*lambda*2pi*r dr
    I = 2pi*lambda integral (0->R) r^3 dr
    I = 2pi*lambda*(R^4)/4
    I = 2pi*(M/pi*R^2)*(R^4)/4
    I = (MR^2)/2

    Question 2

    A gyroscope consists of a wheel of mass M and radius R attatched to a light central rod of length l that is perpendicular to the plan of the wheel. If the wheel is a uniform ring with light spokes determine an expression for the moment of inertia of the wheel about the axis of the spindle.

    Solution

    I = MR^2
     
  11. Jan 17, 2008 #10

    Shooting Star

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    Both correct.

    What do you mean you can't tell them apart? The first is a disk, where the mass is distributed from the centre upto the circumference. The second is a ring, where all the mass is at a distance of R from the centre.
     
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