A flywheel is a mechanical device specifically designed to efficiently store rotational energy. For a particular machine it is in the form of a uniform 20 kg disc of diameter 50 cm, able to rotate about its own axis. Calculate its kinetic energy when rotating at 1200 rpm. Use `pi^(2)=10`. Calculate its moment of inertia, in case it is rotated about a tangent in its plane.

Explanation

(I) As the flywheel is in the form of a uniform disc rotating about its own axis, `I_(z)=(1)/(2)Mr^(2)`
`therefore` Rotational kinetic energy
`=(1)/(2) I omega^(2) =(1)/(2) ((1)/(2)MR^(2)) 4 pi^(2) n^(2)`
`therefore` Rotational kinetic energy
`= M pi^(2) (Rn)^(2) =20xx10xx(0.25xx20)^(2)=5000 J`
(II) Consider any two mutually perpendicular diameters x and y of the flywheel. If the flywheel rotates about these diameters, these three axes (own axis and two diameters) will be mutually perpendicular and concurrent. Thus, perpendicular axes theorem is applicable. Let `I_(d)` be the moment of inertia of the flywheel, when rotating about its diameter. `therefore I_(d) = I_(x) =I_(y)`
Thus, according to the theorem of perpendicular axes,
`I_(z) =(1)/(2) MR^(2) =I_(x)+I_(y) =2I_(d)`
`therefore I_(d) =(1)/(4) MR^(2)`

As the diameter passes through the centre of mass of the (uniform) disc, `I_(d) =I_(C)` Consider a tangent in the plane of the disc and parallel to this diameter. It is at the distance h= R from the diameter. Thus, parallel axes theorem is applicable about these two axes.
`therefore I_("T, parallel") =I_(o) =I_(c) +Mh^(2)=I_(d) +MR^(2)`
`=(1)/(4) MR^(2)+ MR^(2)=(5)/(4) MR^(2)`
`therefore I_("T, parallel") =(5)/(4) MR^(2)=(5)/(4) 20xx0.25^(2)`
`=1.5625" kg m"^(2)`