Moment of Inertia of two small bars

  • Thread starter jaredmt
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  • #1
jaredmt
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Homework Statement


A uniform bar has two small balls glued to its ends. The bar is 2m long and has mass 4kg, while the balls each have mass .5kg and can be treated as point masses. Find the moment of inertia of this combination about each of the following axes: (a) an axis perpendicular to the bar through its center; (b) an axis perpendicular to the bar though one of the balls; (c) an axis parallel to the bar through both balls; (d) an axis parallel to the bar and .5m from it.


The Attempt at a Solution



i tried the formula: (1/12)(ML^2) but that got the wrong answer and i assume because the balls at the end changed it. I am not sure what to do... any help would be very much appreciated
 

Answers and Replies

  • #2
Lord Crc
341
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Yes, the balls change the total moment of inertia. If you take a look at how the moment of inertia is defined for a rigid body of point masses then you should find the answers to your questions. If you don't find it in your book, I can recommend wikipedia :)
 
  • #3
jaredmt
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ok i believe i found the answers but I am not entirely sure if i did it all correctly so please correct me if my proceedure was wrong:

a) I = (1/12)(4kg)(2m)^2 + 2(.5kg)(1m)^2 = 1.33 + 1 = 2.33 kgm^2
the first part i got the formula as i described above. then for the last part: 2 is because there is a total of 2 balls (1 on each side) and .5kg is the mass of each then 1m is the distance the balls are from the center

b) I = (1/3)(4kg)(2m)^2 + .5kg(2m)^2 = 5.33 + 2 = 7.33 kgm^2
the first part i got from this formula: (1/3)(mass)(distance)^2
the second part(for the ball) i just have as (mass)(distance)^2

also, one more question: is there a way that i could throw the center of mass into a formula to find the moment of inertia? or do i have to break it up into 2 parts like this? because i found that the center of the mass, going from the center-point to the end-point, is .6m from the center. but i plugged it into formulas and it didnt come out right
 
Last edited:
  • #4
CrazyIvan
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also, one more question: is there a way that i could throw the center of mass into a formula to find the moment of inertia? or do i have to break it up into 2 parts like this? because i found that the center of the mass, going from the center-point to the end-point, is .6m from the center. but i plugged it into formulas and it didnt come out right

The Moment of Inertia for N point masses is defined as
[tex]I = \sum^{N}_{i=1} m_{i} r_{i}^{2}[/tex]

The Center of Mass for N point masses is
[tex]\vec{R} = \frac{\sum m_{i} \vec{r}_{i}}{\sum m_{i}}[/tex]

Considering that [tex]\vec{R}[/tex] is a vector and [tex]I[/tex] is a scalar, I can't think of any easy way to define one in terms of the other.

And considering that you are only dealing with 3 masses, it's probably easiest to just chug through the old-fashioned way.
 
  • #5
jaredmt
121
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ok, just makin sure I am not taking any long steps when there is a much shorter step. but i don't think finding the cm and plugging in a formula would make it any shorter anyways come to think of it.

but did i find the answers correctly? or did i just somehow luck out with the right answer using the wrong formulas? the 1st parts of each formula seemed right, i just had to plug the numbers in. but the second part, i not 100% sure. i don't see how else i could do it so I am 90% sure i have it all right
 
  • #6
CrazyIvan
45
0
Both calculations look correct.

Just keep plugging along with that Parallel Axis Theorem :cool:
 

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