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

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The discussion focuses on calculating the moment of inertia for a half disk system using both integration and a subtraction method. The moment of inertia for a half disk with radius r2 is expressed as 1/2mr2^2, while that for radius r1 is 1/2mr1^2. The key point is that the moment of inertia of the system can be determined by subtracting the moment of inertia of the smaller half disk from that of the larger one, taking care to define the mass correctly. The mass of the "cut-out" section is defined as M = (1/2)ρπ(r2^2 - r1^2). The clarification provided in the discussion resolves doubts regarding the mass consideration in the calculation.
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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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