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Question

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 θ=30\theta=30^{\circ}.

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Answered 2 years ago
Answered 2 years ago
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To solve this, let's observe the sum of moments around point AA, with the given picture, which is equal to 0 since only weight is acting on AA and it doesn't produce any moment. The moment of inertia around the center of mass GG can be found in various mathematical tables and manuals to be IG=mL212I_G = \dfrac{mL^2}{12}.

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

+MA=IGα+mad0=mL212α+mL2αL2maAcosθL2mL212α+mL24αmL2cosθaAL3α12cosθaAL3α=12cosθaAα=3aA2Lcosθ\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 xx we have:

+Fx=maxWsinθ=maAmL2αcosθmgsinθ=maAmL2αcosθgsinθ=aAL2αcosθL2αcosθ=aAgsinθα=2(aAgsinθ)Lcosθ\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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