Question 16.SP.4: A pulley weighing 12 lb and having a radius of gyration of 8...
A pulley weighing 12 lb and having a radius of gyration of 8 in. is connected to two blocks as shown. Assuming no axle friction, determine the angular acceleration of the pulley and the acceleration of each block.
STRATEGY: Since you want to determine accelerations and are given the weights, use Newton’s second law.
MODELING: Choose the pulley and the two blocks as a single system. The pulley moves in pure rotation and each block moves in pure translation.
Sense of Motion. Although you can assume an arbitrary sense of motion as shown in Fig. 1 (since no friction forces are involved) and later check it by the sign of the answer, you may prefer to determine the actual sense of rotation of the pulley. First determine the weight of block B, W_B^{\prime}, required to maintain the equilibrium of the pulley when it is acted upon by the 5-lb block A.
+\circlearrowleft\Sigma M_G=0: \quad W_B^{\prime}(6 \text { in. }) – (5 \mathrm{lb})(10 \text { in. })=0 \quad W_B^{\prime}=8.33 \mathrm{lb}
Since block B actually weighs 10 lb, the pulley rotates counterclockwise. The free-body and kinetic diagrams for this system are shown in Fig. 2. The forces external to the system consist of the weights of the pulley and the two blocks and of the reaction at G (Fig. 2). The forces exerted by the cables on the pulley and on the blocks are internal to the system and cancel out. Since the motion of the pulley is a centroidal rotation and the motion of each block is a translation, the inertial terms reduce to the couple \bar{I} \alpha and the two vectors m \mathbf{a}_A and m \mathbf{a}_B.
ANALYSIS:
Kinematics of Motion. Assuming α is counterclockwise and noting that a_A=r_A \alpha and a_B=r_B \alpha, you obtain
\mathbf{a}_A=\left(\frac{10}{12} \mathrm{ft}\right) \alpha \uparrow \quad \mathbf{a}_B=\left(\frac{6}{12} \mathrm{ft}\right) \alpha \downarrow
Equations of Motion. The centroidal moment of inertia of the pulley is
\bar{I}=m \bar{k}^2=\frac{W}{g} \bar{k}^2=\frac{12 \mathrm{lb}}{32.2 \mathrm{ft} / \mathrm{s}^2}\left(\frac{8}{12} \mathrm{ft}\right)^2=0.1656 \mathrm{lb} \cdot \mathrm{ft} \cdot \mathrm{s}^2Since the system of external forces is equivalent to the system of inertial terms, you have
+\circlearrowleft \Sigma M_G=\dot{H}_G: \begin{aligned}(10 \mathrm{lb})\left(\frac{6}{12} \mathrm{ft}\right) – (5 \mathrm{lb})\left(\frac{10}{12} \mathrm{ft}\right) &=+\bar{I} \alpha + m_B a_B\left(\frac{6}{12} \mathrm{ft}\right) + m_A a_A\left(\frac{10}{12} \mathrm{ft}\right) \\(10)\left(\frac{6}{12}\right) – (5)\left(\frac{10}{12}\right) &=0.1656 \alpha + \frac{10}{32.2}\left(\frac{6}{12} \alpha\right)\left(\frac{6}{12}\right) + \frac{5}{32.2}\left(\frac{10}{12} \alpha\right)\left(\frac{10}{12}\right)\end{aligned}\begin{aligned}&\alpha=+2.374 \mathrm{rad} / \mathrm{s}^2 \quad \alpha=2.37 \mathrm{rad} / \mathrm{s}^2\circlearrowleft \\&a_A=r_A \alpha=\left(\frac{10}{12} \mathrm{ft}\right)\left(2.374 \mathrm{rad} / \mathrm{s}^2\right) \quad a_A=1.978 \mathrm{ft} / \mathrm{s}^2 \uparrow\\&a_B=r_B \alpha=\left(\frac{6}{12} \mathrm{ft}\right)\left(2.374 \mathrm{rad} / \mathrm{s}^2\right) \quad \mathrm{a}_B=1.187 \mathrm{ft} / \mathrm{s}^2 \downarrow\end{aligned}
REFLECT and THINK: You could also solve this problem by considering the pulley and each block as separate systems, but you would have more resulting equations. You would have to use this approach if you wanted to know the forces in the cables.