Rotational Inertia: Calc. & Guidance Needed

[Riman643]

Homework Statement

The figure shows an arrangement of 15 identical disks that have been glued together in a rod-like shape of length L = 1.6900 m and (total) mass M = 104.0 mg. The 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? Give your answer to four significant figures. (b) If we approximated the arrangement as being a uniform rod of mass M and length L, what percentage error would we make in using the formula in Table 10-2e to calculate the rotational inertia?

Relevant Equations

I = 1/12 ML^2
I = 1/2 mr^2

This question is ridiculously complex for me. I know how to find the Inertia if it was a rod

$I =\frac{1}{12} ML^{2} = \frac{1}{12}(0.000104)(1.6900)^2 =0.00002475 kg * m^2$

To find the Inertia for the individual disks I am having trouble. Not sure how to add them all up together. Guidance would be greatly appreciated.

 

[haruspex]

To find the Inertia for the individual disks I am having trouble

Consider the nth disk from O, to the right, say.
What is its MoI about its own centre?
How far is its centre from O?
Using the parallel axis theorem, what is its MoI about O?

 

[Riman643]

I think I get where you are going. Every disk will be:

$I = \frac{1}{2} mr^{2} = \frac{1}{2} (\frac{0.000104}{15})(\frac{1.6900}{\frac{15}{2}})^{2} = 0.000000011 kg * m^2$

Then using the parallel axis theorem it would be:
$I = \frac{1}{2}mr^{2} + md^{2}$

but which disk do I use for the $md^{2}$ part? And would the distance to let's say the one to the right and left of the center bed twice the radius?

 

[haruspex]

Every disk will be

How many radii make up length L?

would the distance to let's say the one to the right and left of the center bed twice the radius?

Yes. What about the next ones along?

 

[Riman643]

How many radii make up length L?

15? Would that then make the parallel axis theorem in this particular case:

$I_{O} = 15(\frac{1}{2}mr^{2}) + md^{2}$

Yes. What about the next ones along?

The next ones along would be 4 times the length of the radius. So I should use the furthest one away? 14 times the length of the radius?

 

[haruspex]

15?

No, try again.

The next ones along would be 4 times the length of the radius.

Yes.

So I should use the furthest one away?

No, each makes the contribution it makes. Sum the series.

 

[Riman643]

No, try again.

Oops. It's 30 that will make up the entire length

No, each makes the contribution it makes. Sum the series.

So would a correct formula with a series of distances be:

$I_{O} = 15mr^{2} + 2mr^{2}(2 + 4 + 6 + 8 +10 +12 + 14)$

 

[haruspex]

Oops. It's 30 that will make up the entire lengthSo would a correct formula with a series of distances be:

$I_{O} = 15mr^{2} + 2mr^{2}(2 + 4 + 6 + 8 +10 +12 + 14)$

Close...
There's something you left out on the 2, 4, 6 etc.

 

[Riman643]

Close...
There's something you left out on the 2, 4, 6 etc.

Oops. took out the square when I took out $r^{2}$. I got the answer right for the first part. But for some reason I am struggling getting the percentage with the rod wrong. For the rod I get $24.75 mg * m^{2}$ and for the actual Inertia I get $25.02 mg * m^{2}$. Then I take $(1 - \frac{24.75}{25.02}) * 100 = 1.067 %$. Not sure what I am doing wrong with that.