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A solid sphere of mass M, radius R and having moment of inertia about an axis passing through the centre of mass as J, is recast into a disc of thickness t, whose moment of inertia about an axis passing through its edge and perpendicular to its plane remains J. Then, radius of the disc will be |
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Answer» `(2R)/(sqrt(15))` Moment of inertia of circular disc of radius r and mass M about an axis passing through the centre of mass and perpendicular to its plane `=1/2Mr^(2)` `therefore` Using theorem of PARALLEL AXES, moment of inertia of disc about its edge is `I.=1/2Mr^(2)+Mr^(2)=3/2Mr^(2)` GivenI=I. or, `2/5MR^(2)=3/2Mr^(2)` or `r^(2)=4/15R^(2)` or, `r=(2r)/(sqrt(15))` |
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