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Inertia of two masses m2 connected to a rod.

  1. Apr 8, 2014 #1
    I have never dealt with moment of inertia before, this is a physics lab i need to do some pre planning for which involves topics we have never covered and are expected to learn. I've been busy working two jobs and am struggling to get time to pick up this before my lab tomorrow so some help would be extremely helpful.

    We have a rod with length 2d, spinning at its centre. With 2 masses on either side of mass m2, the masses are spherical with radius r. What is the moment of inertia?

    The answers is:

    [itex] I=2 m_2 (d^2 + \frac{2}{5} r2) [/itex]

    By my logic the moment of inertia is the sum of each component, we have two masses m2 and their centre of gravity is d+r away from the centre hence each sphere will contribute, based on the inertia of a sphere:

    [itex] I_(spheres) =2 \frac{2}{5} m_2 (r + d)^2 [/itex]

    And then add the component of the rod, which I believe is ignored as we don't have any value for the mass of the rod. Obviously my assumption is wrong as it isn't the same as the answer. Any tips would be choice. Cheers.
     
    Last edited: Apr 8, 2014
  2. jcsd
  3. Apr 8, 2014 #2
    By the way I should emphasise that the first equation is not 4m(etc), it's 2 m2, meaning mass 2. Just thought i'd clarify just in case don't know how to subscript.
     
  4. Apr 8, 2014 #3

    Doc Al

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    Staff: Mentor

    The spheres have radius r, not r + d. You'll take the moment of inertia of each sphere about its center (which you know) and then compute its moment of inertia about the center of the rod using the parallel axis theorem: Parallel Axis Theorem
     
  5. Apr 8, 2014 #4
    So how come the rods contribute moment of inertia about themselves and around the centre too? that seems slightly confusing.
     
  6. Apr 8, 2014 #5
    Ah i see because the axis running through the sphere is parallel to the axis running through the centre of the rod, the parallel axis theorem says that the I through the rod will be equal to the moment of inertia through the axis through the sphere plus the mass multiplied by the distance between the two parallel axis? Is this interpretation correct? So I get.

    [itex] I_t = I_c + m_2 (d+r)^2 [/itex]

    Where [itex] I_c [/itex] is moment of inertia with the axis through the centre of the sphere. Is this correct, i used d+r squared because the distance between the two axis would be d+r?
     
    Last edited: Apr 8, 2014
  7. Apr 8, 2014 #6

    Doc Al

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    Multiplied by the distance squared, I think you mean.

    Is the distance from the center of the sphere to the axis of rotation r+d or d? If the spheres are just tacked on to the ends of the 2d rod, then you are correct.
     
  8. Apr 8, 2014 #7
    Okay so to get the answer they arrive at I must therefore assume that somehow the spheres are not just glued on a distance d from the centre but the centre of the spheres are connected to the rod with a hole or something and that the centres of The spheres are actually d away from the central axis?
     
  9. Apr 9, 2014 #8

    Doc Al

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    Yes, exactly. (Did they provide a diagram?)
     
  10. Apr 9, 2014 #9
    Nope. Yeah I guess at poorly worded to question. But pretty trivial now I know the theory. Thanks so much.
     
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