Physics Help Forum Moment of inertia of a compound body
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 Apr 6th 2017, 05:35 AM #2 Member     Join Date: Aug 2008 Posts: 68 $I_1=\dfrac{1}{2}(7.5\pi \, kg)(0.5^2 m^2) = \dfrac{15\pi}{16} \, kg \, m^2$ $I_2 = \dfrac{1}{2}(9.8\pi \, kg)(0.7^2 \, m^2) - \dfrac{1}{2}(5\pi \, kg)(0.5^2 \, m^2) = \dfrac{222\pi}{125} \, kg \, m^2$ $I_1+I_2 = \pi \left(\dfrac{15}{16}+\dfrac{222}{125}\right)= \dfrac{5427\pi}{2000} \approx 8.52 \, kg \, m^2$
 Apr 6th 2017, 08:13 AM #3 Member   Join Date: Jan 2015 Posts: 75 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
 Apr 6th 2017, 09:56 AM #4 Member     Join Date: Aug 2008 Posts: 68 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) $m = \pi(50 \, cm)^2(2 \, g/cm^2) = 5000\pi \, g = 5\pi \, kg$ $I_{ring} = \dfrac{1}{2}MR^2 - \dfrac{1}{2}mr^2$
 Apr 6th 2017, 10:47 AM #5 Member   Join Date: Jan 2015 Posts: 75 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. skeeter likes this.

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