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the radius of gyration of a hemi sphere of mass 'm'and radius 'r'
about an axis parallel to diameter at a distance '3r/4'is given by [center of mass of hemisphere lies at a height '3r/8' from base]
?
Dear bhargav ,
Since the center of mass of a solid hemisphere is 3/8 R above the circular base; so reducing it to a one particle system, and the moment of inertia of that one particle system is M(3/8 R)^2=M(9/64)R^2. Now moment of inertia of the particle about an axis parallel to the original one but at a distance of 3/4 R can be obtained using the parallel axis theorem New M.I. = MR^2(45/64) So Radius of gyration= R(45/64)^1/2
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