Question

Moment of inertia of the ring about the tangent to the circle of the ring

Answer 1

Moment of inertia of ring perpendicular to plane through its centre of mass is MR²/2.
The centre of mass and parallel to the plane is MR²/4.

About a tangent is MR²+MR²/4
5MR²/4

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