Question

What is the moment of inertia of a
(i) uniform circular ring of mass 2 kg about its 4 m diameter ?
(ii) a thin disc about an axis coinciding with a diameter ?

Answer 1

The moment of inertia of a uniform ring about an axis passing through its centre and perpendicular to it is given by, `I=MR^(2)`
According to the theorem of perpendicular axis, `I_(z),I_(x)+I_(y)`

Now x and y-axes are along the diameter of the disc nad by symmetry
`I_(x)=I_(y)rArrI_(z)=2I_(x)rArr I_(x)=MR^(2)`
The moment of inertia of ring about of its diameter
`I_(x)=(MR^(2))/(2)=((2)(2)^(2))/(2)=4 "kg-m"^(2)`
(ii) Similarly, moment of inertia about the diameter of disc
`I=(1)/(2)((1)/(2)MR^(2))=(1)/(4)MR^(2)`.