Questions
Moment of inertia of a thin ring of mass m and radius r about an axis passing through its centre and perpendicular to its plane is I. Three such identical rings are joined together as shown in the figure. Then moment of inertia of the system about axis AB is
detailed solution
Correct option is D
I=mR2 (R=Radius)M.O.I. of a ring about a diameter =12mR2=I2d = Distance of centre of ring 3 from axis AB =R+2Rsin60o=3+1R∴IAB=mR22+mR2 x 2+mR22+m3+12R2=72mR2+22+3 x mR2=15+432 x ITalk to our academic expert!
Similar Questions
A rectangular loop has mass M and sides a and b. An axis passes tlrrough the centre C of the loop and is parallel to side a (lie in the plane of the loop). Then the radius of gyration of the loop, for the axis is
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