Question

Moment of inertia of and c g of hollow circular section

Answer 1

for a hollow circular ring moment of inertia perpendicular to its plane passing through its centre is
I = MR²
While about its diameter is
I = MR²/2
And about the tangent perpendicular to its plane is
I = 2MR²
And about the tangent in its plane
I = 3MR²/2
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Answer 2

# Given-

m = 1.7×10-27 kg

r = 4×10^-10 m

# Solution-

Here,

Total mass M = 2m

Radius of rotation R = r/2

Moment of inertia of hydrogen molecule is given by-

M.I. = MR^2

M.I. = (2m)(r/2)^2

M.I. = mr^2/2

M.I. = 1.7×10-27 × (4×10^-10)^2 / 2

M.I. = 1.36×10^-46 kgm^2