Two circular rings have masses in the ratio 2 : 3 and diameters in the ratio 3 : 2 , then the ratio of their moment of inertia is {both the axis are passing through the centre of the respective rings

Question

Two circular rings have masses in the ratio 2 : 3 and diameters in the ratio 3 : 2 , then the ratio of their moment of inertia is {both the axis are passing through the centre of the respective rings and perpendicular to the plane of the respective ring}

TestHub · Accepted Answer

The moment of inertia of a circular ring about an axis passing through its centre and perpendicular to its plane is given by $I = M R^{2}$ .

Given the ratio of masses is $\frac{M _{1}}{M _{2}} = \frac{2}{3}$ and the ratio of diameters is $\frac{D _{1}}{D _{2}} = \frac{3}{2}$ .

Since the radius is proportional to the diameter, the ratio of radii is $\frac{R _{1}}{R _{2}} = \frac{3}{2}$ .

The ratio of their moments of inertia is:

$\frac{I _{1}}{I _{2}} = \frac{M _{1} R _{1}^{2}}{M _{2} R _{2}^{2}} = (\frac{M _{1}}{M _{2}}) (\frac{R _{1}}{R _{2}})^{2}$

Substituting the given values:

$\frac{I _{1}}{I _{2}} = (\frac{2}{3}) (\frac{3}{2})^{2} = \frac{2}{3} \times \frac{9}{4} = \frac{3}{2}$

Thus, the ratio of their moments of inertia is $3 : 2$ .

Answer: $3 : 2$

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