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Homework Help: Rod - Moment of inertia

  1. Mar 12, 2016 #1
    1. The problem statement, all variables and given/known data
    2cflusI.png
    2. Relevant equations
    I = (1/3)ML2
    Parallel axis theorem: Inew = Iold + mr2

    3. The attempt at a solution
    I already solved part a) and I got (1/12)ML2. For b), I get that I have to use the parallel axis theorem but I don't know how to go from there. We have not gone over this concept yet in class (not even moment of inertia) so bear with me...
     
  2. jcsd
  3. Mar 12, 2016 #2

    Simon Bridge

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    ... start from the equation for the parallel axis theorem. You can look it up: the meanings of the symbols I mean.
    I think you were supposed to use it for part (a) so how did you do that?
     
    Last edited: Mar 12, 2016
  4. Mar 12, 2016 #3
    how did you do the part (a)-- if you have doneit correctly then (b) should not be a problem?

    the above is not relevant -did you use it?
     
  5. Mar 12, 2016 #4

    Simon Bridge

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  6. Mar 13, 2016 #5
    All I did for part a) was that since the rod is bent at its center, that means both segments of the rod have masses of M/2 and lengths of L/2. So for one rod, I = (1/24)ML2. For 2 rods, then it's just 1/2(ML2)
     
  7. Mar 13, 2016 #6

    SammyS

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    Is that a typo ?

    I thought that you got 1/12 ML2 , (which is correct.)
     
  8. Mar 14, 2016 #7
    Yes, sorry, that was a typo. That is what I got. But how am I able to use the parallel axis theorem for b)? We did not go over it in class completely yet...
     
  9. Mar 14, 2016 #8

    SammyS

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    What is the location of the center of mass ?
     
  10. Mar 16, 2016 #9
    The location of the center of mass for each segment of the rod is the midpoint of each segment, correct?
     
  11. Mar 16, 2016 #10
    I don't understand the axis that it is being rotated. Is it the midpoint between the 2 ENDS of each segment of the rod? So does it boil down to a trig problem?
     
  12. Mar 16, 2016 #11

    SammyS

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    So then where is the center of mass of the whole bent rod?
     
  13. Mar 16, 2016 #12
    The midpoint of the rod, at the bent center?
     
  14. Mar 16, 2016 #13

    SammyS

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    No, not at the bend.

    Make a sketch.
     
  15. Mar 16, 2016 #14
    Is it the midpoint of the ends of the two segments if they were connected?
     
  16. Mar 16, 2016 #15

    SammyS

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    No. That's the pivot point for part (b).

    It should be mid-way between CM of each segment, Right?
     
  17. Mar 16, 2016 #16
    So like 1/4 of the way of each segment?
     
  18. Mar 16, 2016 #17

    SammyS

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    No.

    Start with the following:

    That's a correct statement.

    The C.M. for the bent rod is mid-way between the center of mass for each of the two segments
     
    Last edited: Mar 16, 2016
  19. Mar 16, 2016 #18
    I have the same problem as OP has, and while I know where the C.M. of the bent rod is, how can you use the center of mass of the two segments of the rod to algebraically obtain the center of mass of the whole rod?

    What I did was determine the distance of the center of mass from the rotational axis and use that as the L for the moment of inertia equation.

    My attempt at finding ## I_{CM} ##:

    ## I_{CM} = m*(\sqrt{((L / 2) * cos30)^{2} + ((L / 2) * sin30)^{2}})^{2} ##

    This is for part (a) of the problem.
     
  20. Mar 16, 2016 #19

    SammyS

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    If you don't understand that, then it's doubtful that do know the correct location of the center of mass of the whole rod .


    Simplifying what you have for ##\ I_\text{CM}\ ## gives:
    ##\displaystyle \ m\cdot\left(\sqrt{((L / 2)^2 \cdot \cos^2(30^\circ) + ((L / 2)^2 \cdot \sin^2(30^\circ)}\right)^{2} \ ##

    ##\displaystyle \ = m\cdot\left(\sqrt{(L / 2)^2 \cdot( \cos^2(30^\circ) + \sin^2(30^\circ))}\right)^{2} \ ##

    ##\displaystyle \ = m\cdot\left(\sqrt{(L / 2)^2 }\right)^{2} =mL^2/4\ ##​

    It's not at all clear, how that is the moment of inertia about any point.
     
  21. Mar 16, 2016 #20
    Isn't the center of mass of each segment of the rod at their midpoint? Which means that the distance between the center of mass for each segment of the rod and the axis of rotation is ## L / 4 ##?
     
  22. Mar 16, 2016 #21
    Sorry about that. Yeah, I had no idea what I was doing with that. I guess I still don't know how to figure out the center of mass for this bent rod and use it in the appropriate moment of inertia equation.

    In my second attempt, I applied the parallel axis theorem to each segment and then added their resulting ## I_{NEW} ##s and I got the correct answer: ## \frac{1}{12} * m * L^2 ##.

    My work:
    ## I_{NEW} = I_{OLD} + m * d^2 ##
    ## I_{NEW} = ((1/12) * (m/2) * (L/2)^2) + (m/2) * (L/4)^2 ##
    ## I_{NEW} = (1/96) * m * L^2 + (1/32) * m * L^2 ##
    ## I_{NEW} = (1/24) * m * L^2 ##
    ## 2 * I_{NEW} = (1/12) * m * L^2 ##
     
  23. Mar 16, 2016 #22

    SammyS

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    Yes. Of course, that's what OP got, but didn't really explain it.
     
  24. Mar 18, 2016 #23

    Simon Bridge

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    If we label the endpoint ##P## and ##Q##, and the bend ##R## then we can say (from post #1) ##|PR|=|RQ|=L/2## (may need to sketch as I describe.)
    Initially the axis of rotation is point ##R## ... which nobody seems to have had trouble with.

    The axis of rotation is shifted to a new point exactly half way along the segment ##\overline{PQ}## ... call that point ##S##.
    The angle ##\angle PSR## is a right angle. Since we also know the other angles it is tempting to try trig and/or pythagoras ... but is there any need?

    If the com of the segment ##\overline{PR}## and ##\overline{RQ}## are ##A## and ##B## respectively, then we notice that ##\triangle PAS## and ##\triangle SBQ## are iscoseles triangles with ##|PA|=|AS|=|SB|=|BQ|## and we know that ##|PA|=|QB|=L/4##

    Each segment therefore has ##I=\frac{1}{96}ML^2## about com, which is a distance ##L/4## from the center of rotation, and there's two of them... so:

    $$I = 2\left(\frac{ML^2}{96} + M\left(\frac{L}{4}\right)^2\right)$$ ... and simplify.

    Which I think Arkud noticed earlier. If I read the above correctly, that was the method used: well done.
     
    Last edited: Mar 18, 2016
  25. Mar 18, 2016 #24

    Simon Bridge

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    How about the other approach - where the moment of inertia about the center of mass of the bent rod is calculated and the parallel axis theorem is applied to that?
    This is how it would normally done IRL: you have a thingy that may get rotated about an arbitrary axis, so you work out the moment of inertia about the com, with the position of the com, and put that information in a set of tables or in the documentation that you sell with the thingy. Some other engineer can. then. work out the details for their particular application.

    Using the above labelled diagram, the combined COM is (from symmetry) half way along the segment ##\overline{AB}## call it ##C## the distance we need is:
    ##\qquad \qquad d=|AC|=|CB|=|AR|\sin(\theta /2)##
    ## \qquad \qquad \qquad \qquad## ... where ##\theta## is the angle of the bend from the problem statement, and ##|AR|=L/4##
    Notice how the angle given was chosen to have nice sines and cosines?

    There are a number of other ways of finding ##d##.
    You end up with ##I_{C} = 2M\left( \frac{1}{96}L^2 + d^2 \right)##

    The distance from C to S can also be found by pythagoras ... ##|CS|=|RC|=\frac{1}{2}|RS|##.
    You should be able to work through the steps and check that it gives the same answer as above.
     
    Last edited: Mar 18, 2016
  26. Mar 18, 2016 #25

    SammyS

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    @Simon Bridge ,

    You are using M as the mass of each segment, while in the OP, M is the mass of the entire rod.
     
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