Question

The moment of inertia of a ring of mass $2kg$ about a tangential axis lying in its own plane is $3kg/m^2$. The radius of the ring is:

• A. $1m$
• B. $2m$
• C. $3m$
• D. $6m$

Answer 1

Answer: (A)

$1m$

Let the moment of inertia of the ring about any of the two perpendicular axes passing through the diametral plane be $I$. Using perpendicular axis theorem we get $2I=M{R}^2$ or $I=\frac{M{R}^2}{2}$. Now using parallel axis theorem we get the moment of inertia about tangential axis will be $\frac{M{R}^2}{2}+M{R}^2=\frac{3}{2}M{R}^2$. Now substituting the values we get
$3=\frac{3}{2}\left(2R^2\right)$
or
$R=1m$