Question

$I$ is moment of inertia of a thin circular plate about its natural axis. The moment of inertia of a circular ring whose mass is half of mass of plate but radius is twice the radius of plate about an axis passing through any tangent of ring in its plane is

• A. $3I$
• B. $4I$
• C. $6I$
• D. $1.5I$

Answer 1

Answer: (C)

$6I$

Moment of inertia of Disk $I=M{R}^2/2$ ,

Now the mass of ring is M/2 and Radius is 2R,

So MOI about the axis in a plane of ring is $I_{cm}=\left(M/2\right)(2R{)}^2/2=M{R}^2$

Now, using parallel axis theorem $I=I_{cm}+\left(M/2\right)d^2$ , here D=2R

So $I=M{R}^2+\left(M/2\right)(2R{)}^2=3M{R}^2=6I$