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Old Apr 6th 2017, 03:38 AM   #1
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Moment of inertia of a compound body

Good morning all,
I am having trouble with a problem to calculate the Moment of Inertia of a body made up of two distinct but linked extended bodies. I have used the generic equations for these two bodies and then added the values to obtain what I thought was the solution but my answer does not match the answer given in the book and I am at a loss as where I have gone wrong.

The problem as stated in the book is as follows -

A compound disk of outside diameter 140.0 cm is made up of a uniform solid disk of radius 50.0 cm and area density 3.00 g/cm(squared) surrounded by a concentric ring of of inner radius 50.0 cm, and outer radius 70.0 cm, and area density 2.00 g/cm(squared).
Find the Moment of inertia of this object about an axis perpendicular to the plane of the object and passing through it's centre.

Now I took the object in this problem to be an inner solid cylinder surrounded by an outer hollow cylinder. From this definition of the object and since the there is only one axis of rotation I then proceeded to find the Moment of Inertia for each cylinder in turn using generic equations for such objects. Then finally adding each of these calculated values to give the overall Moment of Inertia for the object as a whole.
Note - before I used the values of mass given in the stated problem I converted the values from g/cm(squared) to kg/m(squared) using the conversion factor of 1 g/cm(squared) = 10 kg/m(squared) and also converting the values for the 3 radii from 'cm' to 'm'.

The equations I used were ..
For the inner (solid) cylinder .. I = 1/2 MR .. (M= 30 kg/m(squared and
R=0.70 m)
For the outer (hollow) cylinder .. I = 1/2 M[(R(1)(squared) + R(2)(squared)]
(M=20 kg/m(squared) and
R(1)=0.70m and R(2)=0.50m)
This gives ..

I = 1/2((30(0.50(squared)) + ((1/2(20)((0.70(squared))+(0.50(squared))

I = 15x0.25 +10(0.49+0.25) = 3.75 +7.40 = 11.15 kg.m(squared)

Now the answer in the book is .. I = 8.52 kg.m(squared)

Can someone let me know where I have gone wrong because I just can't see it. Thanks.
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Old Apr 6th 2017, 05:35 AM   #2
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$I_1=\dfrac{1}{2}(7.5\pi \, kg)(0.5^2 m^2) = \dfrac{15\pi}{16} \, kg \, m^2$

$I_2 = \dfrac{1}{2}(9.8\pi \, kg)(0.7^2 \, m^2) - \dfrac{1}{2}(5\pi \, kg)(0.5^2 \, m^2) = \dfrac{222\pi}{125} \, kg \, m^2$

$I_1+I_2 = \pi \left(\dfrac{15}{16}+\dfrac{222}{125}\right)= \dfrac{5427\pi}{2000} \approx 8.52 \, kg \, m^2$
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Old Apr 6th 2017, 08:13 AM   #3
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Moment of inertia of a compound body

Hi Skeeter,
Thanks for taking the time to help me with the problem .. but I am puzzled, can you tell me how you have calculated the masses of each disk as ..
7.5 pi kg, 9.8 pi kg and 5 pi kg respectively .. I don't understand how you arrived at these values for the masses ?
Regards,
Jackthehat
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Old Apr 6th 2017, 09:56 AM   #4
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For the complete disk ...

mass = (area)(area density)

$m_1 = \pi(50 \, cm)^2(3 \, g/cm^2) = 7500\pi \, g = 7.5\pi \, kg$

for the concentric ring, I used a subtraction technique to calculate its rotational inertia.

$I_{ring}$ = ($I$ of a whole disk if the "hole" were filled) - ($I$ of the smaller disk "hole")

mass of the "whole" disk = (area)(area density)

$M = \pi(70 \, cm)^2(2 \, g/cm^2) = 9800\pi \, g = 9.8\pi \, kg$

mass of the "hole" disk = (area)(area density)

$m = \pi(50 \, cm)^2(2 \, g/cm^2) = 5000\pi \, g = 5\pi \, kg$

$I_{ring} = \dfrac{1}{2}MR^2 - \dfrac{1}{2}mr^2$
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Old Apr 6th 2017, 10:47 AM   #5
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Moment of inertia of a compound body

Hi again Skeeter,
I see it now ... that makes things much clearer to me now. I should have known that mass = area times area density from the units of the area density given. The book has stressed that you should always write down the units when working out the calculations from the very first chapter as it is good practice and makes things clearer. I should have followed that advice, but I'm afraid bad habits die hard. I will be more careful of this in the future.
Well thank you again for taking the time out to help me and indeed for the clear explanation of the method. It has been a good lesson to me and pointed out to at least one bad habit I have gotten into, it should stand me in good stead for the future, I hope.
Regards,
Jackthehat.
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