\( \theta=-arccos\left(\dfrac{\mathbf K-L^2}{2\mathbf Q}\right) \)
\( \omega= \dfrac{d\theta}{dt}=\dfrac{Lv}{\mathbf Q\sqrt{1-\left(\dfrac{\mathbf K-L^2}{2\mathbf Q}\right)^2}} \)
\( \alpha= \dfrac{d\omega}{dt}=\dfrac{( v^2 +L\cdot a) \left[1-\left(\dfrac{\mathbf K-L^2}{2\mathbf Q}\right)^2\right] - \dfrac{L^2v^2\left(\dfrac{\mathbf K-L^2}{2\mathbf Q}\right)}{2\mathbf Q^2}}{\mathbf Q\sqrt{1-\left(\dfrac{\mathbf K-L^2}{2\mathbf Q}\right)^2}^3} \)
como \( \cos(-\theta)=\dfrac{\mathbf K-L^2}{2\mathbf Q} \)
\( \alpha=\dfrac{( v^2 +L\cdot a) \left[1-\cos^2(-\theta)\right] - \dfrac{L^2v^2\cos (-\theta)}{2\mathbf Q^2}}{\mathbf Q\sqrt{1-\cos (-\theta)^2}^3} \)
\( \alpha=\dfrac{( v^2 +L\cdot a) \sin^2(-\theta) - \dfrac{L^2v^2\cos (-\theta)}{2\mathbf Q^2}}{\mathbf Q\sin^3 (-\theta)} \)
como \( \cos(-\theta)=\cos(\theta) \) y \( \sin(-\theta)=-\sin(\theta) \)
queda
\( \alpha=-\dfrac{( v^2 +L\cdot a) \sin^2(\theta) - \dfrac{L^2v^2\cos (\theta)}{2\mathbf Q^2}}{\mathbf Q\sin^3 (\theta)} \)
\( \alpha=\dfrac{L^2v^2\cos \theta-2\mathbf Q^2( v^2 +L\cdot a) \sin^2\theta }{2\mathbf Q^3\sin^3 \theta} \)
Saludos
