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A thin disc of mass M and radius R has mass per unit area sigma( r) = kr^(2), where r is the distnace from its centre. Its moment of inertia about an axis going through its centre of mass and perpendicular to its plane is xMR^(2), where x is _________ |
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Answer» M.I. of ELEMENT ring of radius r and infinitesimally small thickness dr about the axis is `dI=(DM).r^(2) ={(kr^(2))2pi rdr}r^(2)`  `rArr I=2pi kR^(6) //6` ...........(1) Also, MASS of ring, `M = int dm = int_(r=0)^(R) (kr^(2))(2pi r dr)` Or, `M=2pi k R^(4)/4`........(2) From (2) and (2) `I= 4/6 MR^(2) = 2/3 MR^(2)` |
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