# Angular acceleration and the moment of inertia

The wheel is made from a 10-kg thin ring and four 0,5-kg slender rods. If the torsinal spring attached to the wheel's center has a stiffnes $$k = 2 N*m/rad$$, so that the torque on the center of the wheel is $$M = (2* \theta)*N*m$$, where $$\theta$$ is in radians. The wheel is rotated two full turns, $$\theta = 4*\pi$$ Determine the angular velocity when the $$\theta = 0$$ and determine the angular acceleration immediately after release, just before the movement begins. Determine the angular acceleration at the moment when the wheel is back to its original position.

I have calculated the angular velocity but I have problems calculating the angular acceleration. Where to start?

$$w = 10.9 rad/s$$ and $$\alpha_1 = 9.5 rad/s^2$$ and $$\alpha_2 = 0$$  For the acceleration after releasing, \begin{align} \sum M &= I\alpha \\ 2*4\pi &= 2.667 kg m^2*\alpha_1 \\ \alpha_1 &= 9.42 rad/s^2 \\ \end{align} (Not sure how they rounded up to 9.5.)
At the original position (no torsion in spring) of a simple harmonic oscillator, the velocity is a maximum (magnitude) and the acceleration is a minimum (magnitude). Therefore, $$\alpha_2=0$$.