Find the rotational kinetic energy of a ring of mass 9 kg and radius 3 m rotating with 240 rpm about an axis passing through its centre and perependicualr to its plane.

Explanation

The mass M=3K g radius, `R=50 cm =50xx10^(-2) m=0.5m`
(i) The moment of inertia (ii) about an axis passing through the center and perpendicular to the plame of the disc is
`I=(1)/(2)MR^(2)`
`I=(1)/(2)xx3xx(0.5)^(2)=0.5xx3xx0.5xx0.5`
`I=0.375 kg m^(2)`

(ii) The moment of inertia (i) about an axis touching the edge and perpendicular to the plane of the disc by parallel axis theorem is,
`I=I_(C)+Md^(2)`
where `I_(C)=(1)/(2) MR^(2) and d=R`
`I=(3)/(2)MR^(2)+MR^(2)=(3)/(2)MR^(2)`
`I=(3)/(2)xx3xx(0.5)^(2)=1.5x3xx0.5xx0.5`
`I=1.125 kg m^(2)`

(iii) The moment of inertia (i) about an axis passing through the center and lying on the
`I_(Z)=I_(X)+I_(Y)`
Where `I_(X)=I_(Y)=I and I_(Z)=1)/(2) MR^(2)`
`I_(Z)=2I, I=(1)/(2) I_(z)`
`I=(1)/(2)xx(1)/(2) MR^(2)=(1)/(4)MR^(2)`
`I=(1)/(4)xx3xx(0.5)^(2)=0.25xx3xx0.5xx0.5`
`I=0.1875 kg m^(2)`