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CBSE Class 12-science Answered

4 solid spheres  of mass m and radius r are placed on a circular disc of mass 2m and radius 2r.
the system is rotated about an axis passing throughe the center of the ring and perpendicular 
to the plane what is the radius of gyration of the system.
Asked by Pallavi Chaturvedi | 18 Nov, 2014, 12:01: AM
answered-by-expert Expert Answer
begin mathsize 14px style Let space 4 space solid space sphere space of space mass space straight m space and space radius space straight r space are space placed space over space the space circular space disc space of space mass space 2 straight m space and space radius space 2 straight r. Whole space system space is space rotated space about space an space axis space passing space through space the space center space of space the space disc. The space moment space of space inertia space of space disc space about space an space axis space passing space through space the space center space is space given space as comma straight I subscript straight d equals 1 half mr squared space equals 1 half cross times 2 straight m cross times left parenthesis 2 straight r right parenthesis squared straight I subscript straight d equals 4 mr squared space The space moment space of space inertia space of space solid space sphere space about space an space axis space passing space through space the space center space is space given space as comma straight I subscript straight s equals 2 over 5 mr squared space since space 4 space solid space sphere space placed space over space the space each space other space on space dis comma total space mass space of space solid space disc space would space be space 4 straight m space with space radius space straight r Hence comma straight I subscript straight s equals 2 over 5 cross times 4 mr squared Total space moment space of space inertia space of space system straight I equals space straight I subscript straight d plus straight I subscript straight s straight I equals 4 mr squared space plus 2 over 5 cross times 4 mr squared straight I equals 14 over 5 cross times mr squared formula space for space radius space of space gyration space is space given space as comma straight I equals mk squared 14 over 5 cross times mr squared equals mk squared straight k squared equals 14 over 5 straight r squared straight k equals square root of 14 over 5 end root straight r space end style
Answered by Priyanka Kumbhar | 18 Nov, 2014, 11:10: AM

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