Torsional Pendulum: Deriving an expression with time Period, Suspension Length, Moment of Inertia and Rigidy Modulus Constant I have been looking for the derivation/place where I can quote the formula $T=\frac{2\pi}{r^2}\sqrt{\frac{2IL}{\eta\pi}}$ from as I remember seeing in class but can't seem to find it online to quote/derive it for my self.
Clarification:
For
$$T=\frac{2\pi}{r^2}\sqrt{\frac{2IL}{\eta\pi}}$$
$I$ is the moment of inertia, $L$ is suspension length, $\eta$ is the rigidity modulus of the material and $T$ is the time period.
Any help would be greatly appreciated.
 A: For relatively small (see Note) angular twists of the wire, the torque $\tau$ which restores the wire to its untwisted position is proportional to the angle $\theta$ through which the end of the wire is rotated : $$\tau=-\kappa \theta$$
$\kappa$ is called the torsion constant of the wire. The minus sign indicates that the direction of the torque is opposite to the direction in which angle $\theta$ is increasing.
The torque causes rotational acceleration $\ddot\theta$ of mass at the end of the wire. Newton's 2nd Law for this acceleration is $$\tau=I\ddot\theta$$ where $I$ is the moment of inertia of the mass. This is the rotational equivalent of $F=ma$.
Combining the above two equations, the equation of motion for the torsional pendulum is $$\ddot\theta+\frac{\kappa}{I}\theta=0$$
This has the form $\ddot x+\omega^2 x=0$ which describes Simple Harmonic Motion. Here $\omega=2\pi f$ is the angular frequency of the periodic motion (radians per second) and $f$ is frequency (cycles per second; one cycle is $2\pi$ radians). Period and frequency are related by $f=\frac{1}{T}$ and $\omega=\sqrt{\frac{\kappa}{I}}$ therefore $$T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{I}{\kappa}}$$ 
This can be compared with the period of a mass on a spring : $$T=2\pi\sqrt{\frac{m}{k}}$$ The moment of inertia $I$ is equivalent to mass $m$ and the torsion constant $\kappa$ is equivalent to the force constant $k$ or stiffness of the spring. 

The final difficult step is to relate torsion constant $\kappa$ to the dimensions of the wire and its shear modulus $\eta$. This calculation is done in Deriving the Shear Modulus S From the Torsion Constant κ. The result is $$\kappa=\frac{\eta \pi r^4}{2L}$$ in which $r$ is the radius of the wire and $L$ its length. Substituting into the equation for the period we get $$T=2\pi\sqrt{\frac{2LI}{\eta \pi r^4}}$$ which is the formula you were given.

Note
The angle $\theta$ through which the end of the wire is rotated is not the same as the angle of twist $\psi$ along the length of the wire. Whereas $\theta$ is measured by the rotation of a radius in the base of the wire, $\psi$ is measured by the twist of a line parallel to the axis. The two are related by $$L\psi=r\theta$$ 
To ensure that the elastic limit of the wire is not exceeded $\psi$ should be small, typically no more than $10^{\circ}$. However since $L\gg r$ the angle $\theta$ can be quite large. The base can be rotated a whole circle without exceeding the elastic limit.  
