A solid body of any shape possesses a moment of inertia I_0 at 0^ ^...

The moment of inertia of a solid body is given by $I_0 = \sum m_i r_i^2$.

When the temperature of the body increases by $\theta$, the distance of each mass element from the axis of rotation increases due to thermal expansion. The new distance $r_i'$ is given by:

$r_i' = r_i(1 + \alpha \theta)$

where $\alpha$ is the coefficient of linear expansion.

The new moment of inertia $I$ at temperature $\theta$ is:

$I = \sum m_i (r_i')^2$

$I = \sum m_i [r_i(1 + \alpha \theta)]^2$

$I = (1 + \alpha \theta)^2 \sum m_i r_i^2$

$I = I_0 (1 + \alpha \theta)^2$

Since $\alpha$ is very small, we can use the binomial approximation $(1 + x)^n \approx 1 + nx$ for $x \ll 1$ and neglect higher-order terms of $\alpha$:

$I = I_0(1 + 2\alpha \theta)$