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Introductory Physics Homework Help
Moment of Inertia of a Disk
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[QUOTE="Jefffff, post: 5891851, member: 584164"] [h2]Homework Statement [/h2] A disk of radius R has an initial mass M. Then a hole of radius (1/4) is drilled, with its edge at the disk center. Find the new rotational inertia about the central axis. [I]Hint[/I]: Find the rotational inertia of the missing piece, and subtract it from that of the whole disk. You’ll find the parallel-axis theorem helpful. Express your answer in terms of the variables M and R. [h2]Homework Equations[/h2] I = I[SUB]cm[/SUB] + Md[SUP]2[/SUP] Moment of Inertia of a Disk: 1/2 MR[SUP]2[/SUP] [h2]The Attempt at a Solution[/h2] Based on the formula for mass of the disk, let M' be mass of the drilled out smaller disk. M = 2πr[SUP]2[/SUP]hρ Since all factors are constant here except r, where the new radius is 1/4 of the original, thus (1/4)[SUP]2[/SUP] = (1/16) M so M' = (1/16)M Plugging in values to find the moment of inertia of the drilled out disk piece: I = I[SUB]cm[/SUB] + Md[SUP]2[/SUP] I = (1/2)(1/16M)(R/4)[SUP]2[/SUP] + (M/16)(R/4)[SUP]2[/SUP] This yields I = (3/512) MR[SUP]2[/SUP] The new moment of inertia is (1/2)MR[SUP]2[/SUP] - (3/512)MR[SUP]2[/SUP] = (253/512) MR[SUP]2[/SUP] The problem is the correct answer for this question according to the quiz was: (1/2)MR[SUP][/SUP] - (1/256)MR[SUP]2[/SUP] which is different from what I got. Not sure where I went wrong. Anything helps! [/QUOTE]
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Moment of Inertia of a Disk
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