Moment of inertia of a set of spinning disks

  • #1

Homework Statement


The shows an arrangement of 15 identical disks that have been glued together in a rod-like shape of length L = 1.0000 m and (total) mass M = 100.0 mg. The disks are uniform, and the disk arrangement can rotate about a perpendicular axis through its central disk at point O. (a) What is the rotational inertia of the arrangement about that axis? (b) If we approximated the arrangement as being a uniform rod of mass M and length L, what percentage error would we make in calculating the rotational inertia?



M = 100.0 mg = 1×10-4 kg
L = 1.0000 m
# of disks = 15

Homework Equations



Inertia of a disk about a central diameter
I = (MR2)/2

Inertia of set of particles
I = ∑miri2

The Attempt at a Solution


Itotal = (MR2)/4 + ∑miri2
Itotal = ML2 / 4⋅302 + (2M/15)(1 + 4 + 9 + 16 + 25 + 36 + 49)(L/15)
Itotal = 8.298 kg - m2

Did I have the right solution?
 
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Answers and Replies

  • #2
BvU
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Depends. They want you to compare it with ##I## for a rod with the same M and L. What did that give you ?
Oh, sorry, that's part b. Still: a nice check...

[edit] and for part a) I would check my dimensions if I were you. And the problem statement: are the disks really that light ?
 
  • #3
Simon Bridge
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Hint: parallel axis theorem.
 
  • #4
BvU
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Hi Simon !
I think the parallel axis theorem is there, but the work is a bit sloppy. M and m , numerator squared, denominator not, etc.
 
  • #5
Hi Simon !
I think the parallel axis theorem is there, but the work is a bit sloppy. M and m , numerator squared, denominator not, etc.
I tried to separate the rotational inertia of the center disk and the remaining fourteen disks. Is that how it is done? I'm still having trouble finding the rotational inertia of the arrangement.
 
  • #6
BvU
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Ah, yes. So SImon was right. Do you know about the parallel axis theorem ? You'll need it here.
 
  • #7
Ah, yes. So SImon was right. Do you know about the parallel axis theorem ? You'll need it here.
Yeah, I do know about it.
So I should just find the moment of a single disk then add it to its md2 for each disk? Is that correct?
Making the solution for a as:
0.5mdiskrdisk2 + 0.5mdiskrdisk2(22 + 42 + 62 + 82 + 102 + 122 + 142)
 
  • #8
BvU
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Go slower. There are terms missing your last line.
the moment of a single disk then add it to its md2 for each disk? Is that correct?
yes, if by ##d## you mean the distance from the axis to the center of mass.

"moment of a single disk" for each disk is what I'm missing ...
 
  • #9
Go slower. There are terms missing your last line.
yes, if by ##d## you mean the distance from the axis to the center of mass.

"moment of a single disk" for each disk is what I'm missing ...
Yeah, what I'm referring to d is the distance of one disk to the axis of rotation O.
What I mean about moment of a single disk is its moment when spun at its center.
So for example, the outermost disk has a distance of 14r from its center, then can I consider its moment to be 1/2 mr^2 + m(14r)^2.
Then next to that disk is a disk 12r from the axis of rotation, then can I use 1/2 mr^2 + m(12r)^2.
and so on... adding those all together in the end.
 
  • #10
BvU
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Agreed. But I don't see the ##I## for each disk around its own center, except for the disk in the middle.
 
  • #11
Agreed. But I don't see the ##I## for each disk around its own center, except for the disk in the middle.
upload_2017-2-17_0-6-39.png

This is what I'm referring to... The first term is for the center disk while the remaining is for the next ones, times two because there are rods from the left and right of the axis.
This is letting m be the mass of each disk and r be the radius of each disk too.
 
  • #12
BvU
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upload_2017-2-17_0-6-39-png.113311.png


is the same as

0.5mdiskrdisk2 + 0.5mdiskrdisk2(22 + 42 + 62 + 82 + 102 + 122 + 142) ???
 
  • #13
upload_2017-2-17_0-6-39-png.113311.png


is the same as

0.5mdiskrdisk2 + 0.5mdiskrdisk2(22 + 42 + 62 + 82 + 102 + 122 + 142) ???
Sorry, I was confused at first... nevermind the former equation: 0.5mdiskrdisk2 + 0.5mdiskrdisk2(22 + 42 + 62 + 82 + 102 + 122 + 142)
Is the other equation, the one in the image, the correct one I should use? Is that how parallel axis theorem applied in this problem?
 
  • #14
BvU
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Seems to me the picture comes from the solution manual ?:) and we can't argue with that, can we :rolleyes: ?

But already before we had
So I should just find the moment of a single disk then add it to its md2 for each disk? Is that correct?
answered with "yes" ! So the thing to do -- now that spoiler is in existence -- is check that your plan and the picture are in agreement. (And hopefully also see why I kept asking for the ##I## of the off-center disks around their own center :smile:)
 
  • #15
Seems to me the picture comes from the solution manual ?:) and we can't argue with that, can we :rolleyes: ?

But already before we had
answered with "yes" ! So the thing to do -- now that spoiler is in existence -- is check that your plan and the picture are in agreement. (And hopefully also see why I kept asking for the ##I## of the off-center disks around their own center :smile:)
Found the inertia using that equation to be 8.352*10^-6 kg-m^2
When it is considered as a thin rod of length 1 m, the inertia becomes 8.333*10^-6 kg-m^2
The percent error becomes: 0.2217% (pretty low)

BTW, I didn't get the image from a manual =( I used a math editor 'cause I don't know how to use LATEX here.
 
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  • #16
BvU
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BTW, I didn't get the image from a manual =( I used a math editor 'cause I don't know how to use LATEX here
So much the better !

You sure these disks are so light ? And then you still find a total ##I## of
8.352*10^6 kg-m^2 for such a small rod ?
 
  • #17
So much the better !

You sure these disks are so light ? And then you still find a total ##I## of
8.352*10^6 kg-m^2 for such a small rod ?
Oh, sorry, the exponent are raised to a negative power :smile:
 
  • #18
BvU
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##\LaTeX## is easy ! Much faster than [ SUB] [ /SUB] etc.:

$## ##$ {1\over 2} mr^2 + 2 \; {1\over 2} mr^2 \left ( 1 + 2^2 \right ) + ... $## ##$ yields$$
{1\over 2} mr^2 + 2 \; {1\over 2} mr^2 \left ( 1 + 2^2 \right ) + ... $$

Learn by right-clicking over a formula and Pick Show math as ..##\TeX## commands
 
  • #19
30
1
a rod of mass M and length L has moment of inertia,I=ML2/12
but it is only for a thin rod
here the rod is a line passing through centers of disks,it's M.I constitute only small part of total M.I of the arrangement
here,the moment of inertia the arrangement is comparable to the M.I of a rod of length L and mass M and width=2r=L/15
 
  • #20
a rod of mass M and length L has moment of inertia,I=ML2/12
but it is only for a thin rod
here the rod is a line passing through centers of disks,it's M.I constitute only small part of total M.I of the arrangement
here,the moment of inertia the arrangement is comparable to the M.I of a rod of length L and mass M and width=2r=L/15
So I should still consider the radius when calculating the moment of the rod? Why so?
 
  • #21
30
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the actual M.I is not like that
it must calculated by calculating M.I of each disks about o
it may be near to the M.I of the rod with width L/15 given above ,(answer to 2nd question) while the rod has no gaps that seen in the arrangement b/w disks
you have to calculate M.I by I=Σmiri and parallel axis theorem.ie by calculating M.I of each disk about o that what you are still doing.
 
  • #22
the actual M.I is not like that
it must calculated by calculating M.I of each disks about o
it may be near to the M.I of the rod with width L/15 given above ,(answer to 2nd question) while the rod has no gaps that seen in the arrangement b/w disks
you have to calculate M.I by I=Σmiri and parallel axis theorem.ie by calculating M.I of each disk about o that what you are still doing.
Isn't that what I've done? Using the parallel axis theorem, I calculate each flat disk's rotational inertia. Thereafter, I've added each of their inertia to find the real moment of inertia of the arrangement. Thus, the inertia of the whole arrangement equals to the rotational inertia of the central disk plus the rotational inertia of the remaining fourteen disks which are place beside it. I didn't use the summation of mi times ri2 because it was not a set of particles, instead it was a set of disks which has a certain rotational inertia of (M*R2)/2
 
  • #23
BvU
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Hi,
I think you have. Not clear to me either what akshay is referring to .
 
  • #24
Hi,
I think you have. Not clear to me either what akshay is referring to .
Yeah, I was pretty confused too.

Just a follow-up question; what if the arrangement is not in a straight line. For example, the central disk has 7 adjacent disks to its left and 7 adjacent disks on top of it, making the arrangement form a 90 degree angle? Will the arrangement's rotational inertia be the same?
 
  • #25
TSny
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Just a follow-up question; what if the arrangement is not in a straight line. For example, the central disk has 7 adjacent disks to its left and 7 adjacent disks on top of it, making the arrangement form a 90 degree angle? Will the arrangement's rotational inertia be the same?
upload_2017-2-17_9-42-46.png


Yes, you should be able to see that the rotational inertia will be the same for 90 degrees (or any other angle).
 

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