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Total Moment of Inertia of Two Rods

  1. Nov 28, 2015 #1
    1. The problem statement, all variables and given/known data
    The rods of length 2 meters and and mass 20 kg are joined at their ends to form a V shape. What is the total moment of inertia measured from the reference point perpendicular to the plane of the paper and at the point where the two rods are joined. (So find total moment of inertia at the edge of the V shape, but with respect to the plane perpendicular to the page). Also the rods are 60 degree apart.

    2. Relevant equations
    Moment of Inertia at the edge of a rod = 1/3 ML^2

    3. The attempt at a solution
    Can I not just add the two moments of inertia to get 2/3 ML^2 as the total?

    EDIT: I think the correct way to do this is to find the CoM of the system (which will be the mid point of the line connecting the midpoints of the two rods). And then use parallel axis theorem to find the total moment of inertia about the CoM. And then apply parallel axis theorem again to find the MI about the the point connecting the two rods. Can anyone confirm this?
     
    Last edited: Nov 28, 2015
  2. jcsd
  3. Nov 28, 2015 #2

    gneill

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    Either method should bring you to the same result. I know which way I'd choose if I was in a hurry :smile:
     
  4. Nov 28, 2015 #3

    SteamKing

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    The only problem with the latter plan is how to calculate the MOI of the rods about the C.o.M. Because the rods are fixed together at one end, I think you have to calculate the MOI using integration or use the formula for calculating the MOI of a rod about a rotated coordinate axis.
     
  5. Nov 28, 2015 #4

    gneill

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    Hence my comment above...
     
  6. Nov 29, 2015 #5
    Schaum's- 3000 Solved problem in Physics. page 212.
    Rule: Moments of inertia about an axis are added algebraically.
     
  7. Nov 29, 2015 #6

    SteamKing

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    I don't think this applies to this particular case.

    The individual rods are not joined at their centroids, but at one end of each rod and at an angle to one another, in the shape of a V when looking parallel to the presumed axis of rotation.

    You can calculate the MOI of each rod about one end, but combining the two is a bit trickier than you are led to believe by this 'Rule'.
     
  8. Nov 29, 2015 #7
    11.33
    Four coplanar, large, irregular masses are held by a rigid frame of negligible mass, as shown in figure 11-6. Taking an axis through P and perpendicular to the page, show that I=I1 +I2+I3+I4 where I1 is the moment of inertia of object 1 alone about the axis and similarly for the others.

    p is intersection of lines joining the masses.
     
    Last edited: Nov 29, 2015
  9. Nov 29, 2015 #8

    SteamKing

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    It's not clear what I1 thru I4 represent. Are they supposed to be the moments of inertia for the individual masses referred to the axis thru P?

    How exactly to you 'add algebraically' the MOI of one mass which is placed at an angle to another mass?
     
    Last edited: Nov 29, 2015
  10. Nov 29, 2015 #9

    gneill

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    We can go to the basics for moment of inertia about an axis and write the integral for the given scenario. In the figure the z-axis projects out of the page:
    Fig1.png
    L is the length of each rod of mass M, the distance from the z-axis of a mass element is r. The domain of r is 0 to L, and a differential mass element is ##dm = (M/L)dr##.
    $$ I = 2\int_0^L r^2 \frac{M}{L}~dr$$
    It should be clear from the symmetry that the moment of inertia will be twice that of a single rod.
     
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