How to Calculate the Moment of Inertia for a Half Disk System?

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SUMMARY

The moment of inertia for a half disk system can be calculated using the subtraction method, where the moment of inertia of a larger half disk with radius r2 is subtracted from that of a smaller half disk with radius r1. The moment of inertia for a half disk is defined as (1/2)mr², leading to the equations: I(r2) = (1/2)(1/2)ρπr2² * r2² and I(r1) = (1/2)(1/2)ρπr1² * r1². The mass M of the cut-out section is given by M = (1/2)ρπ(r2² - r1²), which is essential for accurate calculations.

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ajaysabarish
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in this video

http://www.physicsgalaxy.com/lectures/1/44/234/Solved-Example-2#12(see only the question)

the method illustrated is integration but i thought of an alternate method,

moment of inertia of half disc with radius r2 is 1/2mr2^2 and that of half disc with radius r1 is 1/2mr1^2.so the moment of intertia of system must be equal to difference of these moment of inertia since half disc with radius r2 is cutoff from radius r1,so shouldn't moment of inertia be subtracted too?
 
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That is because you have have to be careful with the definition of the mass M.
To use the subtraction method, we have to begin by considering a half disk of radius r_{2} with mass (\frac{1}{2}\rho \pi) r_{2}^{2}. This can be decomposed into
(1) A half disk of radius r_{1} with mass (\frac{1}{2}\rho \pi) r_{1}^{2}, and
(2) The "cut-out" system of interest, which has mass M = (\frac{1}{2}\rho \pi) (r_{2}^{2} - r_{1}^{2})

Then the moment of inertia of our system is simply given by the difference of the moment of inertias of the two half disks,
\frac{1}{2}\left(\frac{1}{2}\rho \pi r_{2}^{2}\right) r_{2}^{2} - \frac{1}{2}\left(\frac{1}{2}\rho \pi r_{1}^{2}\right) r_{1}^{2}

Finally, use M = (\frac{1}{2}\rho \pi) (r_{2}^{2} - r_{1}^{2}) to recast the equation.
 
Fightfish said:
That is because you have have to be careful with the definition of the mass M.
To use the subtraction method, we have to begin by considering a half disk of radius r_{2} with mass (\frac{1}{2}\rho \pi) r_{2}^{2}. This can be decomposed into
(1) A half disk of radius r_{1} with mass (\frac{1}{2}\rho \pi) r_{1}^{2}, and
(2) The "cut-out" system of interest, which has mass M = (\frac{1}{2}\rho \pi) (r_{2}^{2} - r_{1}^{2})

Then the moment of inertia of our system is simply given by the difference of the moment of inertias of the two half disks,
\frac{1}{2}\left(\frac{1}{2}\rho \pi r_{2}^{2}\right) r_{2}^{2} - \frac{1}{2}\left(\frac{1}{2}\rho \pi r_{1}^{2}\right) r_{1}^{2}

Finally, use M = (\frac{1}{2}\rho \pi) (r_{2}^{2} - r_{1}^{2}) to recast the equation.
great,thank you,completely cleared my doubt,i forgot about the mass.
 

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