A uniform metallic circular disc, mounted on frictionless bearings, is rotating at an angular frequency $\omega$ about an axis passing through its centre and perpendicular to its plane. The coefficient of linear expansion of the metal is $\alpha$. If the temperature of the disc is increased by $∆t$, the angular frequency of rotation of the disc will

The moment of inertia of the disc about the given axis of rotation is

$I=\frac{1}{2}M{R}^2$                  (1)

where M is the mass of the disc and R its radius. If the disc is heated, it expands. Hence R increases. The resulting increase in I is obtained by partially differentiating (1).

$$\begin{gathered} ∆I=\frac{1}{2}\times M\times 2R∆R (\because M=cons\tan t) \\ or ∆I=MR∆R \\ But ∆R =R\alpha ∆t. Therefore, \\ ∆I=M{R}^2\alpha ∆t \left(2\right) \end{gathered}$$

Now, the angular momentum of the disc is given by 
$J=I\omega \left(3\right)$                         
Since no external torque acts, J remains constant. Partially differentiating (3), we have 

$$\begin{gathered} I∆\omega +\omega ∆I=0 or \\ ∆\omega =-\frac{\omega ∆I}{I} \left(4\right) \end{gathered}$$

Using (1) and (2) in (4), We get

$∆\omega =-2\alpha \omega ∆t$

The negative sign indicates that the angular frequency decreases due to increase in temperature.