Question

what is the moment of inertia and kinetic energy of thin uniform ring of mass 2 kg and diameter 1 metre rotating about the axis passing through its centre of centre and perpendicular to the plane of ring​

Answer 1

As we know that

Moment of inertia of ring=MR^2

Kinetic energy=1/2 ×I×W^2,where,W is the angular velocity....

now,here

radius of ring=0.5

mass of ring=2kg

moment of inertia of ring=2×0.5×0.5=.5

Now, kinetic energy of the ring=1/2×I×W^2

=1/2×0.5×W,^2=>.25W^2

{hope it helps you}