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Moment of inertia of rod about an axis inclined

  1. May 19, 2015 #1
    1. The problem statement, all variables and given/known data
    What is the Moment of inertia of rod about an axis inclined?
    It is given as ## \frac{ML^2sin^2θ}{12} ## but how?

    2. Relevant equations
    I know MOI for axis passing through center along y axis as
    ## \frac{ML^2}{12} ##

    3. The attempt at a solution
    Shouldn't it be ## \frac{ML^2}{12} = Icosθ ##
    by taking components?
    Snapshot4.jpg
     
  2. jcsd
  3. May 19, 2015 #2

    BvU

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    Hello Raghav,

    What is the axis of rotation in your picture :smile: ?
     
  4. May 19, 2015 #3
    Sorry for the diagram,
    The horizontal one is the rod.
    The axis of rotation is the one which I have denoted by I.
     
  5. May 19, 2015 #4

    BvU

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    The customary choice for ##\theta## is the angle between rod and axis of rotation. Perpendicular ##\Rightarrow I = {ML^2/12}## and ##\theta = 0 \ \ \Rightarrow \ \ I = 0\ \ ##.

    You can check by working out ##I = \int r^2 dm ##
     
  6. May 19, 2015 #5
    Yeah, the angle should be between rod and axis of rotation.
    Because if we know the formula ML2sin2θ/12 . Then when θ= 90° the MOI is ML2/12.

    Now in the first place how to derive that formula ## I = ML^2sin^2θ/12 ## ?
    I know derivation of axis passing through center and perpendicular to rod.
    The derivation goes like this.
    Let ## dm = \frac{Mdx}{L} ##

    $$ I = \int_{\frac{-L}{2}}^{\frac{L}{2}} x^2dm $$
    Therefore on simplifying and integrating we get I = ML2/12
    But how to go when inclined at angle θ?
     
  7. May 19, 2015 #6

    BvU

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    You work out

    Rod_inertia.jpg ##I = \int r^2 dm ##
     
  8. May 19, 2015 #7
  9. May 19, 2015 #8

    BvU

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    Yes and no. You should really do some work and check that yourself, instead of asking what to do !

    One of the checks would be that ##M = \int dm## should work out OK :rolleyes: !
     
  10. May 19, 2015 #9
    Okay, getting dm = 2Mdx/L, as the length from centre to one end is half.
    So $$ I = \int_0^{\frac{L}{2}} \frac{2x^2tan^2θMdx}{L} $$
    But by this the answer coming is wrong.

    And I have checked$$ M = \int_0^{\frac{L}{2}} \frac{2Mdx}{L} $$
     
  11. May 19, 2015 #10

    BvU

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    Last check fails miserably! x does not run from 0 to L/2
     
  12. May 19, 2015 #11
    It can run from 0 to L/2.
    By all this we would get half the MOI and then we can multiply it by 2.
     
  13. May 19, 2015 #12

    BvU

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    No.
     
  14. May 19, 2015 #13
    So are you trying to say from -L/2 to L/2. ?
    Now you would say check it,:cool: but I am still arriving at wrong answer.
     
  15. May 19, 2015 #14

    BvU

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    Take a ruler and measure half the length of the rod on your screen.
    Then measure from 0 to the upper bound of the x integral.
     
  16. May 19, 2015 #15
    So, you are trying to say one is base and other the hypotenuse?
    Are you telling me to do trigonometry in a twisted manner?
    Your every reply is looking a riddle, Don't know why it is not helping me.

    Would take time to go to market and buy a ruler as at these times it is closed. Have not used a ruler from a long time.
     
  17. May 19, 2015 #16

    BvU

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    It can also be made clear with two pencil marks on a piece of paper that x does not run from 0 to L/2. If ##\theta=0## it doesn't run at all !

    Yes! ##\theta## plays a role and you need it to get the integrals for M and also for I right !
     
  18. May 19, 2015 #17
    Judging from your post 6th diagram cosθ = dm/dx
    But I know it is wrong dimensionally.
    I am not getting the approach for dm in terms of θ.
    Can you tell finally?
     
  19. May 19, 2015 #18
    Okay, got it myself. Your diagram was confusing me of taking the triangle forming 90°

    dm = Mdx/L
    sinθ = r/x
    r = xsinθ

    $$ I = \int_{\frac{-L}{2}}^{\frac{L}{2}} \frac{x^2sin^2θMdx}{L} $$ Why were you telling not to take -L/2 to L/2 earlier ?
    Solving we get the desired result,
    I = ## \frac{ML^2sin^2θ}{12} ##
     
  20. May 19, 2015 #19

    BvU

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    By now I have enough faith in you that I start to think it must be me who is so completely unclear.
    How can you possibly say this ? Anyone can see that the piece called dm is longer than the piece called dx !!! So, if anything at all, then ##dx = dm \cos\theta## and not vice versa !
    Of course the dimension is wrong, but that you fix by changing it to mass: multiply by ##\rho## where ##\rho## has the dimension of mass/length. In short, ##\rho## is M/L and then ##\displaystyle dm = \rho \; dl = {M\over L} dl = {M\over L} {dx\over \cos\theta}##.
    And if ##l## runs from 0 to L/2 in the integration, then ##x## runs from 0 to ... ?

    because it is not correct ##x## does not run from 0 to L/2 or from -L/2 to L/2.

    Shame on you ! You had the correct " ##r = x \tan\theta## " in post #7 !!

    Your question was
    and the answer in post 8 was and still is

    Yes and no.
     
  21. May 19, 2015 #20
    That was my big mistake.
    Yes you are correct and clear. I also have faith in you and I admit cosθ = dm/dx was my typo.
    Actually I typed it fast as it was looking dimensionally incorrect that how we are dividing length and mass.
    But you told me the reason in above post by converting dm to length part.

    Okay x should run from 0 to Lcosθ.

    So $$ I = \int_0^{Lcosθ} \frac{x^2tan^2θMdx}{Lcosθ} $$ -----1)

    And I have checked$$ M = \int_0^{Lcosθ} \frac{Mdx}{Lcosθ} $$

    Integrating ---1) I am getting,

    ## I = \frac{ML^2sin^2θ}{3} ##
    But I am supposed to get 12 in denominator.
     
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