Piecewise Circle Moment

As the mass is uniformly distributed from the origin $(0,0)$ , the arcs can be considered as point masses with masses $M_1$ , $M_2$ and $M_3$ respectively and $M_1 + M_2 + M_3 = M$ . The mass of an arc is proportional to its length. Let the density per length be $\sigma$ , then we have:

$\begin{aligned} M_1 & = \sigma r_1 \theta_1 = 1\left(\frac {2\pi}3 - 0\right) \sigma = \frac {2\pi}3 \sigma \\ M_2 & = \sigma r_2 \theta_2 = 2\left(\frac {4\pi}3 - \frac {2\pi}3\right) \sigma = \frac {4\pi}3 \sigma \\ M_3 & = \sigma r_3 \theta_3 = 3\left(2 - \frac {4\pi}3 \right) \sigma = 2 \pi \sigma \end{aligned}$

$\begin{aligned} \implies M & = M_1 + M_2 + M_3 = \left(\frac 23 + \frac 43 + 2 \right) \pi \sigma = 4 \pi \sigma \\ M_1 & = \frac 16 M \\ M_2 & = \frac 13 M \\ M_3 & = \frac 12 M \end{aligned}$

The moment of inertia of the object is thus given by:

$\begin{aligned} I & = \sum_{k=1}^3 r_k^2 M_k \\ & = r_1^2 M_1+r_2^2 M_2 + r_3^2 M_3 \\ & = \frac {1^2}6 M + \frac {2^2}3 M + \frac {3^2}2 M \\ & = 6 M \end{aligned}$

$\implies \alpha = \boxed{6}$