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.
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Answer 1

Moment of inertia is $I = 8\times 10^{-5}$ $kg - m^2$

Explanation:

Given ,

Diameter of ring d = 4  cm

Radius of ring , r = 2 = 0.02 m

Mass , m = 0.02 kg

We know that,

Moment of inertia,

$I =m\times r^2$

$I = 0.02\times 0.02^2$

$I = 0.02\times 0.0004$

I = 0.000008

$I = 8\times 10^{-5}$ $kg - m^2$