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# The uniform slender rod of mass m and length L is released from rest in the inverted vertical position. Neglect friction and the mass of the small end roller and find the initial acceleration of A. Evaluate your result for $\theta=30^{\circ}$.

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To solve this, let's observe the sum of moments around point $A$, with the given picture, which is equal to 0 since only weight is acting on $A$ and it doesn't produce any moment. The moment of inertia around the center of mass $G$ can be found in various mathematical tables and manuals to be $I_G = \dfrac{mL^2}{12}$.

Using the general moment equation (Eq. 6/5) we have:

\begin{align*} & + \circlearrowleft \sum M_A = I_G \alpha + \sum mad \\ & 0 = \dfrac{mL^2}{12} \alpha + m \dfrac{L}{2} \alpha \dfrac{L}{2} - ma_A \cos \theta \dfrac{L}{2} \\ & \dfrac{mL^2}{12} \alpha + \dfrac{mL^2}{4} \alpha - \dfrac{mL}{2} \cos \theta a_A \\ & \dfrac{L}{3} \alpha - \dfrac{1}{2} \cos \theta a_A \\ & \dfrac{L}{3} \alpha = \dfrac{1}{2} \cos \theta a_A \\ & \alpha = \dfrac{3a_A}{2L} \cos \theta \end{align*}

Next, for the sum of forces in $x$ we have:

\begin{align*} & + \nearrow \sum F_x = ma_x \\ & W \sin \theta = ma_A - m \dfrac{L}{2} \alpha \cos \theta \\ & mg \sin \theta = ma_A - m \dfrac{L}{2} \alpha \cos \theta \\ & g \sin \theta = a_A - \dfrac{L}{2} \alpha \cos \theta \\ & \dfrac{L}{2} \alpha \cos \theta = a_A - g \sin \theta \\ & \alpha = \dfrac{2(a_A - g \sin \theta)}{L \cos \theta} \end{align*}

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