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Isha Lohan posted an Question
August 26, 2020 • 20:53 pm
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30 points
30 points
- IIT JAM
- Physics (PH)
Q.12 consider a uniform thin circular disk of radius r and mass m.a concentric square or side is cur ne disk (see figure). what is the moment of inertia of the
Q.12 Consider a uniform thin circular disk of radius R and mass M.A Concentric square or Side is Cur ne disk (see figure). What is the moment of inertia of the re eSde R2 Is cut esuitant dtsK about an axIS passing through the centre of the disk and perpendicular to it? ()I- - (8) I|- () I-|
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Chandra prakash
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Moment of inertia of the circular disc about a line perpendicular to the circular disc through it's center=MR^2/2 .now if we cut out a square lamina of side R/2 then moment of inertia of this square lamina about a line perpendicular to the lamina and passing through the center=W/3((R^2/4)+(R^2/4)) where w=4.(R/4)(R/4).density of the squre lamina now the required moment of inertia = M(R^2)/2)_(R^4/48.2).density. ( by putting the value of w). again M=πR^2.density now putting the value of density we get the results=(MR^/2)(1-(1/48π))
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Hope you get it
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sorry I didn't get it. please explain inertia of square
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Ruby negi
first, find the moment of Inertia of disc(assuming that no square is cut out from disc) then find the moment of inertia of square and subtract the two. u will get ur result...
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Chandra prakash
Moment of inertia of the circular disc about a line perpendicular to the circular disc through it's center=MR^2/2 .now if we cut out a square lamina of side R/2 then moment of inertia of this square lamina about a line perpendicular to the lamina and passing through the center=W/3((R^2/4)+(R^2/4)) where w=4.(R/4)(R/4).density of the squre lamina now the required moment of inertia = M(R^2)/2)_(R^4/48.2).density. ( by putting the value of w). again M=πR^2.density now putting the value of density we get the results=(MR^/2)(1-(1/48π))