Question 16.SP.3: A pulley weighing 12 lb and having a radius of gyration of 8......

Sense of Motion. Although an arbitrary sense of motion can be assumed (since no friction forces are involved) and later checked by the sign of the answer, we may prefer to determine the actual sense of rotation of the pulley first. The weight of block B required to maintain the equilibrium of the pulley when it is acted upon by the 5-lb block A is first determined. We write

+\uparrow \Sigma M_G=0: \quad W_B(6 \text { in. })-(5\text { lb})(10 \text { in. })=0 \quad W_B=8.33 \text { lb}

Since block B actually weighs 10 lb, the pulley will rotate counterclockwise.
Kinematics of Motion. Assuming α counterclockwise and noting that a_A=r_A \alpha \text { and } a_B=r_B \alpha, we obtain

\mathbf{a}_A=\left(\frac{10}{12}\text{ ft}\right) \alpha \uparrow \quad \mathbf{a}_B=\left(\frac{6}{12} \text{ ft} \right) \alpha \downarrow

Equations of Motion. A single system consisting of the pulley and the two blocks is considered. Forces external to this system consist of the weights of the pulley and the two blocks and of the reaction at G. (The forces exerted by the cables on the pulley and on the blocks are internal to the system considered and cancel out.) Since the motion of the pulley is a centroidal rotation and the motion of each block is a translation, the effective forces reduce to the couple \bar{I} \alpha and the two vectors m \mathbf{a} _A \text { and } m \mathbf{a} _B. The centroidal moment of inertia of the pulley is

\overline{I}=m \overline{k}^2=\frac{W}{\mathrm{g} } \overline{k}^2=\frac{12\text{ lb}}{32.2\text{ ft}/ s ^2}\left(\frac{8}{12} \text{ ft} \right)^2=0.1656 \text{ lb} \cdot \text{ ft} \cdot s ^2

Since the system of the external forces is equipollent to the system of the effective forces, we write