State an expression for the moment of inertia of a solid uniform disc rotating about an axis passing through its centre, perpendicular to its plane. Hence derive an expression for the moment of inertia and radius of gyration.
`(i)` about a tangent in the plane of the disc, and
`(ii)` about a tangent perpendicular to the plane of the disc. In a set, `21` tuning forks are arranged in a series of decresing frequencies. Each tuning fork produces `4` beats per second with the preceding fork. If the first fork is an octave of the last fork , find the frequencies of the first and tenth forks.

Explanation

Moment of inertia for a solid uniform disc rotating about an axis passing through its centre, perpendicular to its
plane, `I_(C )=(MR^(2))/(2)`
`(i)` `M.I.` about diameter `=(MR^(2))/(4)`
`I_(t)=I_(d)+MR^(2)` [Using parallel axis theorem]
`=(MR^(2))/(4)+MR^(2)=(5MR^(2))/(4)`
Radius of gyration , `k=sqrt((I)/(M))=sqrt((5MR^(2))/(4M))`
`k=(sqrt(5)R)/(2)`
`(ii) I_(t)'=I_(c )+MR^(2)` [Using parallel axis theorem]
`I_(t)'=(MR^(2))/(2)+MR^(2)=(3MR^(2))/(2)`
`Mk'^(2)=(3MR^(2))/(2)`
`k'=sqrt((3)/(2))*R`
Numerical :
Let, `n_(1)=`frequency of first tuning fork
`n_(21)=`frequency of last tuning fork
Difference in frequencies `(-d)`
`=-4` beats/second
`P=` number of tuning forks
`n_(P)=n_(1)+(P-1)d`
`n_(21)=n_(1)+(21-1)(-4)`
`n_(21)=n_(1)-80`........`(i)`
`n_(1)=2(n_(21))` (Given)
`:.(n_(1))/(2)=n_(21)`
`:. (n_(1))/(2)=n_(1)-80` [From equation `(i)`]
`n_(1)=160Hz`
`n_(10)=` frequency of `10th` tuning fork
`n_(10)=n_(1)+(10-1)d`
`n_(10)=160+9(-4)`
`n_(10)=124Hz`