Question
two point masses m and 3m are placed at distance r.The moment of inertia of the system about an axis passing through the centre of mass of system and perpendicular to the line joining the point masses is
ans-3\4mr2
Read 1 Solution.
as mass of both of the object r not equal thus cm (centre of mass ) will be more towards 3m than m
Let dist of cm from mass m be x & that from 3m be (r-x)
mx = 3m (r-x)
on solving u will get x= 3r/4 and r-x will = r/4
now cm of the system will be = 3m(r-x)2 + mx2 = ( 3mr2/16)+ (m9r2/16) = 3/4 mr2 ANS ( on putting x = 3r/4 )
I THINK ANS SUBMITTED BY U AGAIN HAS PROBLEM
SARIKA SHARMA 11 year ago
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