Question 17.SP.1: A 240-lb block is suspended from an inextensible cable which......
A 240-lb block is suspended from an inextensible cable which is wrapped around a drum of 1.25-ft radius rigidly attached to a flywheel. The drum and flywheel have a combined centroidal moment of inertia \bar{I}=10.5 lb · ft · s². At the instant shown, the velocity of the block is 6 ft/s directed downward. Knowing that the bearing at A is poorly lubricated and that the bearing friction is equivalent to a couple M of magnitude 60 lb ? ft, determine the velocity of the block after it has moved 4 ft downward.
We consider the system formed by the flywheel and the block. Since the cable is inextensible, the work done by the internal forces exerted by the cable cancels. The initial and final positions of the system and the external forces acting on the system are as shown.
Kinetic Energy. Position 1.
Block: \bar{v}_1=6 ft/s
Flywheel: w_1=\frac{\bar{v}_1}{r}=\frac{6\text{ ft}/ s }{1.25 \text{ ft} }=4.80\text{ rad}/ s
\begin{aligned}T_1 & =\frac{1}{2} m \bar{v}_1^2+\frac{1}{2} \bar{I} \omega_1^2 \\& =\frac{1}{2} \frac{240\text{ lb}}{32.2 \text{ ft} / s ^2}(6 \text{ ft} / s )^2+\frac{1}{2}\left(10.5\text{ lb} \cdot \text{ft} \cdot s ^2\right)(4.80 \text{ rad} / s )^2 \\& =255 \text{ ft} \cdot \text{ lb}\end{aligned}
Position 2. Noting that \omega_2=\bar{v}_2 / 1.25, we write
\begin{aligned}T_2 & =\frac{1}{2} m \bar{v}_2^2+\frac{1}{2} \bar{I} \omega_2^2 \\& =\frac{1}{2} \frac{240}{32.2}\left(\bar{v}_2\right)^2+\left(\frac{1}{2}\right)(10.5)\left(\frac{\bar{v}_2}{1.25}\right)^2=7.09 \bar{v}_2^2\end{aligned}
Work. During the motion, only the weight W of the block and the friction couple M do work. Noting that W does positive work and that the friction couple M does negative work, we write
\begin{aligned}s_1 & =0 \quad s_2=4 \text{ ft} \\\theta_1 & =0 \quad \theta_2=\frac{s_2}{r}=\frac{4 \text{ ft} }{1.25 \text{ ft} }=3.20 \text{ rad} \\U_{1 \rightarrow 2} & =W\left(s_2-s_1\right)-M\left(\theta_2-\theta_1\right) \\& =(240 \text{ lb} )(4 \text{ ft} )-(60 \text{ lb} \cdot \text{ ft} )(3.20 \text{ rad} ) \\& =768 \text{ ft} \cdot \text{ lb}\end{aligned}
Principle of Work and Energy
\begin{aligned}& T_1+U_{1 \rightarrow 2}=T_2 \\& 255 \text{ ft} \cdot \text{ lb} + 768 \text{ ft}\cdot \text{ lb} =7.09 \bar{v}_2^2 \\& \bar{v}_2=12.01 \text{ ft} / s \quad \overline{ \text{v} }_2=12.01 \text{ ft} / s\downarrow\end{aligned}
