Find the period of small torsional oscillational of a system consisting of two discs slipped oon a thin rod with rod with torsional coefficient `k`. The moments of inertia of discs relative to the rod's axis are equal to `I_(1)` and `I_(2)`.

Explanation

Suppose the disc 1 rotates by angle `theta_(1)` and the disc 2 by angle `theta_(2)` in the opposite sense. Then total torsion of the rod `= theta_(1)+theta_(2)`
and torsional `P.E.=(1)/(2) k ( theta_(1)+theta_(2))^(2)`
The K.E. of the system `(` neglecting the moment of inertia of the rod `)` is
`(1)/(2)I_(1) dot(theta_(1))^(2)+(1)/(2)I_(2)dot (theta_(2)^(2)`
So total energy of the rod
`E=(1)/(2)I_(1) dot(theta_(1))^(2)+(1)/(2)I_(2)dot (theta_(2)^(2)+(1)/(2)k(theta_(1)+theta_(2))^(2)`
We can put the total angular momentum of the rod equal to zero since the frequency associated with the rigid rotation of the whole system must be zero `(` and is known `)`.
Thus `I_(1)theta_(1)=I_(1)dot (theta_(2))` or `(dot(theta_(1)))/(1//I_(1))=(dot(theta_(2)))/(1//I_(2))=(dot(theta_(1)))/(1//I_(1))(+dot (theta_(2)))/(+1//I_(2))`
So `dot(theta_(1))=(I_(2))/(I_(1)+I_(2))(dot(theta_(1))+dot(theta_(2)))` and `dot (theta_(2))=(I_(1))/(I_(1)+I_(2))(dot (theta_(1))+dot (theta_(2)))`
and `E=(1)/(2) (I_(1)I_(2))/(I_(1)+I_(2))(dot(theta_(1))+theta_(2))^(2)+(1)/(2)k(theta_(1)+theta_(2))^(2)`
The angular oscillation, frequency corresponding to this is
`omega^(2)=k//(I_(1)I_(2))/(I_(1)+I_(2))=k//I^(')` and `T=2pisqrt((I^('))/(k))`, where` I^(')=(I_(1)I_(2))/(I_(1)+I_(2))`