You must be consistent to what you define as positive. Given the picture above, the sum of moments about the pivot point is (considering the positive direction the clockwise direction):

$$ \circlearrowright \sum M=I_o\alpha\\ mg3a\sin \theta+2mg6a\sin \theta+mg8a\sin \theta=(78ma^2+2m(6a)^2)\alpha\\ 23mga\sin \theta=150ma^2\alpha $$

But $\theta$ is defined positive counter-clockwise, so $\alpha=-\ddot{\theta}$.

Therefore, the equation of motion is:

$$ \ddot{\theta} = -\frac{23mga}{150ma^2}\sin \theta=-\frac{23g}{150a}\sin \theta $$

If you define the positive direction the ccw direction, then the LHS of the equation is negative, but $\alpha=\ddot{\theta}$. The final result, however, does not change.

You must be consistent to what you define as positive. Given the picture above, the sum of moments about the pivot point is (considering the positive direction the clockwise direction):

$$ \circlearrowright \sum M=I_o\alpha\\ 23mga\sin \theta=150ma^2\alpha $$

But $\theta$ is defined positive counter-clockwise, so $\alpha=-\ddot{\theta}$.

Therefore,

$$ 23mga\sin \theta=-150ma^2\ddot{\theta} $$

The resultant is:

$$ \ddot{\theta} = -\frac{23g}{150a}\sin \theta $$

If you define the positive direction the ccw direction, then the LHS of the equation is negative, but $\alpha=\ddot{\theta}$. The final result, however, does not change.