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Disk with hole

  1. May 23, 2005 #1
    A uniform circular disk has radius 35 cm and mass 350 g and its center is at the origin. Then a circular hole of radius 11.667 cm is cut out of it. The center of the hole is a distance 17.5005 cm from the center of the disk. Find the moment of inertia of the modified disk about the origin.

    First i find the moment of inertia of the whole disk and then i find the inertia of of the hole .
    Then i subtract circular hole disk form the uniform circular disk inertia
    but i am not getting right answer
     
    Last edited: May 23, 2005
  2. jcsd
  3. May 23, 2005 #2

    Gokul43201

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    Did you find the MI of the hole with respect to the center of the disk ?
     
  4. May 23, 2005 #3
    no i didn't how do we do that
     
  5. May 23, 2005 #4

    Gokul43201

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    Look up the "parallel axis theorem"
     
  6. May 23, 2005 #5
    but how do we find the mass of the cutout part.
    Is this correct
    350/(35)^2 * area of the cutout part
     
  7. May 25, 2005 #6

    Gokul43201

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    What is the mass of the whole disk ? What is the area of the whole disk ? What is the area of the cut-out part ?
     
  8. May 25, 2005 #7
    Ok divide this problem into two parts .

    First find the moment of inertia of complete disc (without any part cut) about the origin.

    Now find MI of the cut out part about the origin using parallel axis theorem.

    For the second part , you will have to find out mass of cutout part,this can be easily done as the disc is uniform , use unitary method.

    Now add the MI of both parts , remembering that the mass of the cutout part that you calculate has to be taken negative.
     
  9. May 26, 2005 #8
    Call me an old fashioned person, but I just don't like "negative mass"--its like one of those tricks invented by non-theorists who wanted to finish problems (sorry if I sound offensive but its kinda true isn't it!!). Instead it is easier to regard the new arrangement as a SUPERPOSITION of the cut mass and the mass without the cut portion. Their sum must equal the original MI. Remember to use the same axis for reference.

    So this really is an application of one of the most fundamental ideas in physics: Superposition. I like to keep it that way. Negative mass, "pseudo forces" and some other things should be kept away and used only when needed (lets use pure unadelterated physics please!).
     
  10. May 26, 2005 #9
    Actually the negative mass method is the same as the superposition principle.
    In superposition method , we first calculate MI due to uncut par and then we subtract the MI due to cut part . And by taking negative mass, we need not subtract the two MI's.These are two different ways of saying the sane thing and do not adulterate physics!
     
  11. May 26, 2005 #10
    Dr. Brain if you've been studying in some coaching school, then you will be expected to know some of these tricks but when you get into mainstream physics (if at all thats your pursuit) you will discover the fallacies in some of these tricks. It is very easy to point similarities between two methods one of which seems clever and useful but is deceptive, but it isn't as easy to figure out the intricacies of physics by using tricks instead of resorting to hardcore time-tested principles.

    I do not think it appropriate therefore, to argue with you about pure physics and the adulterated version you are trying to justify, as this would be a waste of time and would stray us away far into the shallowness of argument :-D. Anyway, I am only against the mechanization of such "tricks" replacing PHYSICS.

    Cheers
     
    Last edited: May 26, 2005
  12. May 26, 2005 #11
    Not extending the argue further, I agree with you fully...But there's no harm using the tricks as per me..
     
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