Question

the diameter of a ring is 4 cm and its mass is 20 gram what will be its moment of inertia about an Axis passing through its centre and perpendicular to its plane​

Answer 1

Moment of inertia of ring will be $8\times 10^{-7}kg-m^2$

Explanation:

We have given diameter d = 4 cm = 0.04 m

So radius $r=\frac{d}{2}=\frac{0.04}{2}=0.02m$

Mass of the ring m = 20 gram = 0.002 kg

We have to find the moment of inertia of ring passing through its center and perpendicular to the axis

Moment of inertia is given by

$I=MR^2=0.002\times 0.02^2=8\times 10^{-7}kg-m^2$

So the moment of inertia of ring will be $8\times 10^{-7}kg-m^2$

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