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Moment of inertia problem

  1. Apr 1, 2017 #1
    Hi everyone,
    I have just started the chapter in my Physics book that deals with calculations of Moments of momentum for various extended bodies. I have come across a problem which involves, as part of the overall problem, finding the moment of momentum of the centre of mass of the object and axis, and I am confused as to how I calculate this term.
    The overall problem is as follows :-
    A uniform bar has 2 small balls glued to it's ends. The bar is 2.00 m long and has mass 4.00 kg, while the balls each have mass 0.500 kg and can be treated as point masses.Find the moment of inertia of this combination about the following axes.
    a) an axis perpendicular to the bar through it's centre
    b) an axis perpendicular to the bar through one of the balls
    c) an axis parallel to the bar through both balls
    d) an axis parallel to the bar and 0.500 m from it

    now I have successfully completed parts a), b) and c) above, but for part d) I decided to use the 'Parallel Axis Theorem to solve it ..
    The equation I used was ..
    I = I(cm) + M d(squared)
    where I(cm) is the moment of inertia of the centre of mass of the bar and axis,
    M is total mass of bar and balls, d is distance from bar to axis

    My problem is that I am not sure how to work out 'I(cm)' in this case.

    Can anyone help give me an explanation of how to work this value out or let me know if there is an alternative method of solving the problem ?
     
  2. jcsd
  3. Apr 1, 2017 #2
    Hi jack:

    I am not certain about the notation, but my guess is that "I(cm)" is the "c" answer, and "cm" means "the center of mass" with its axis parallel to the specified axis.

    Hope this helps.

    Regards,
    Buzz
     
  4. Apr 1, 2017 #3
    I am a bit confused by the statement. If this is a uniform bar, and the masses glued to its ends are to be considered as point masses, ##I_\text{cm}## through an axis that is parallel to the bar and goes through the two masses should be ##0##, right? If that is the case, the parallel-axis (aka Steiner's) theorem should be enough to solve part d).
     
  5. Apr 1, 2017 #4
    Hi JoePhysics,
    Thank you for your input .. I wasn't aware that the Icm through the axis parallel to the bar was zero .. can you explain why this is the case ?
     
  6. Apr 1, 2017 #5
    The bar has no radial dimensions, and the two masses are to be taken as point masses, so in the equation ##I_\text{CoM,parallel} = \int r^2 dm##, the ##r## is vanishing, as I understand the situation.
     
  7. Apr 1, 2017 #6

    Nidum

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    Science Advisor
    Gold Member

    The Icentre of mass values for balls and rod are indeed all zero .

    Parallel axis theorem may not be not clearest way to think about this problem .

    Really you just have three masses all at same distance R from the axis of rotation and total MoI about this axis is just sum of the M R2 terms for the three masses .
     
    Last edited: Apr 1, 2017
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