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A solid sphere of mass M and radius R having moment of inertia I about its diameter is recast into a solid disc of radius r and thickness t. The moment of inertia of the disc about an axis about the edge and perpendicular to the plane is I. Then the radius r of the disc is given by : |
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Answer» `r=SQRT((2)/(15))R` Moment of inertia of disc about an AXIS PASSING through its EDGE and perpendicular to its plane. `I=I_(CM)+Mr^(2)=(Mr^(2))/(2)+Mr^(2)=(3)/(2)Mr^(2)` `implies (2)/(5)MR^(2)=(3)/(2)Mr^(2)orr=(2)/(sqrt(15))R`. |
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