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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 $\alpha_2 = 0$

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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, $\alpha_2=0$.

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You need to find the moment of inertia of the wheel; not enough info is shown. Then the acceleration is a=T/j at every positional angle.

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