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Homework Help: Moment of Inertia of slender rod

  1. Jul 23, 2009 #1
    1. The problem statement, all variables and given/known data

    A thin uniform rod 50.0 cm long with mass 0.320 kg is bent at its center into a V shape, with a 70.0 0 angle at its vertex. Find the moment of inertia of this V-shaped object about an axis perpendicular to the plane of the V at its vertex.

    2. Relevant equations

    Moment of Inertia of slender rod

    3. The attempt at a solution
    In solution manual, he has considered these V-shaped object has two slender rods and he has applied formula I= 1/3 ML^2 for both rods. He has also given that moment of Inertia is independent of the angle between the two sides of the V

    But I tried to solve the problem using Centre of mass concept. First, i found the moment of inertia about its centre of mass and and then applied parallel axis theorem. I think it certainly depends on the angle between the rods because the Moment of inertia depends on the distribution of mass.

    So i am little bit confused.... kindly help...
  2. jcsd
  3. Jul 23, 2009 #2


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    Staff: Mentor

    Seems like it should depend on the angle to me as well. If the rods are straight out, what is the moment of inertia? Does the answer match?
  4. Jul 23, 2009 #3
    Ya, if i apply his method then moment of inertia is same as if it is bent by 70 degrees.... do u feel the moment of inertia should be same or different??
  5. Jul 23, 2009 #4


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    How did you set this up, and what did you get? I believe you should find that the angle will cancel out when you do the calculation.
  6. Jul 23, 2009 #5
    First i calculated the position of the centre of mass for the system(If i consider origin at the vertex then the position of centre of mass is (0, 0.25). Then i used the below formula to calculate the moment of inertia about the centre of mass:

    I = m1r12 + m2r2 2

    where r1 and r2 are the perpendicular distance of the centre of mass of the two rods from the centre of mass.

    Then i used parallel axis theorem to calculate the moment of inertia about the given axis. My soln: 0.020 and as per soln manual : 6.667 * 10^-3
  7. Jul 23, 2009 #6


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    No, I don't think that's right. The center of mass will depend on the angle. If the angle was zero degrees (so that the two sides were on top of one another), then the distance would be 1/4 of the total length (away from the origin, between the rods). If the angle were ninety degrees, the center of mass would be the center of the rod (zero distance away from the origin.

    This applies to particles; but the moment of inertia depends on the distribution of the mass. So you need to use:

    \int dm\ r^2

    where r is the distance that the element dm is away from the axis you are calculating for.

    The reason the angle does not matter is because the rough idea is that it is the radial distribution of the mass away from the rotation center that matters.
  8. Jul 23, 2009 #7
    Thanks :smile: I understood the Moment of inertia part however in the above comment do u mean to say that the postion of centre of mass is wrong...
  9. Jul 23, 2009 #8


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    Yes, it looks wrong to me. The total rod is 50 cm, so each 'arm' of the rod is 25 cm. So in this case the center of mass should be between 0cm and 12.5 cm away from the vertex, depending on the angle.

    (Unless I misunderstood your center of mass response. When you said (0,.25) I interpreted that to mean .25 m away from the vertex, in the y direction.)
  10. Jul 23, 2009 #9


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    Hey guys! :smile:

    It obviously doesn't depend on the angle …

    there are two rods of length L, and the axis goes through the end of both …

    so the MI of the pair is the sum of the individual MIs.

    (I expect you've used the wrong formula for the MI about the combined c.o.m. … I don't understand how you found that without also using the parallel axis theorem :confused:)
  11. Jul 23, 2009 #10
    Plz check my below calculation of centre of mass:

    I have taken origin at the vertex and +ve x- axis in the right direction and +ve y-axis upward direction. Then,

    x = 0 ( here we don't have any problem)
    y=( (0.160 * 0.204) + (0.160 * 0.204) ) / (0.160+0.160) = 0.204
    (here i got 0.204 using y1=(sin 55) * (0.25)
    therefore, c.o.m = (0,0.204) ???
  12. Jul 23, 2009 #11


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    Hi tiny-tim,

    As I mentioned in post #6, once you find the position of the center of mass, you can get it from a direct application of

    \int dm\ r^2

    with r being the distance from dm to the center of mass. I think everything is quite straightforward, except perhaps the need to convert from ds (along the rod) to dx (along the line between the arms of the rod). But since the angle of inclination of ds is constant along each arm, ds and dx are proportional to each other, and it turns out to be not bad at all.
  13. Jul 23, 2009 #12
    Ya, i have completely messed up the problem. Actually i considered the c.o.m of two rods as two particles and found the MI about c.o.m instead of using integral form as mentioned by alphysicist and then i used Parallel axis theorem to find the MI about the given axis and also i went wrong in assuming that MI depends on the angle...:frown:
  14. Jul 23, 2009 #13


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    I'm not sure where the 55 degree angle came from. Also remember that each arm is only 25 cm long, so the center of mass of each arm is only 12.5 cm away from the vertex, and that the angle is relative to the line between the arms, so the cosine function gives the component of the arm along that line.
  15. Jul 23, 2009 #14
    If i use cosine also i would get same answer:
    y1= cos 35 * 0.25 = 0.204
    and i m using length in meters instead of cm
  16. Jul 23, 2009 #15


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    Here is how I am visualizing the problem (sorry for the image quality; I can't quite get a symmetric V-shape!):

    http://img11.imageshack.us/img11/741/bentwire.jpg [Broken]

    Each leg of the V-shape has a center of mass that is 12.5cm from the vertex, so you just need to find the y-component of the right triangle that has a hypotenuse of 12.5cm.
    Last edited by a moderator: May 4, 2017
  17. Jul 23, 2009 #16
    Thank you very much :smile:
  18. Jul 23, 2009 #17


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    Sure, glad to help!
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