Question

Moment of inertia of a thin ring of radius R about an axis passing through any diameter is 1/2MR²

1. What is the radius of gyration of the ring about an axis passing through any diameter.

2. A thin metal ring of radius 0.25m and mass 2kg starts from rest and roll down an inclined plane. If the linear velocity on reaching the foot of the plane is 2m/s, calculate its rotational kinetic energy at that instant.

Answer 1

1. \(\frac{MR^2}{2}\) =\( M(\frac{R^2}{2})\)

But I = MK²

^∴ K² = \(\frac{R^2}{2}\)

or K = \(\frac{R}{\sqrt{2}}\)

2. KE = \(\frac{1}{2} tω^2\)

Answer 2

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