Question 17.SP.1: A 240-lb block is suspended from an inextensible cable that ......
A 240-lb block is suspended from an inextensible cable that is wrapped around a drum with a 1.25-ft radius that is rigidly attached to a flywheel. The drum and flywheel have a combined centroidal moment of inertia of \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 so 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.
STRATEGY: Because you have two positions and are interested in determining the velocity of the block, use the principle of work and energy.
MODELING: Consider the system formed by the flywheel and the block. Because the cable is inextensible, the work done by the internal forces exerted by the cable cancels out to zero. The initial and final positions of the system and the external forces acting on the system are shown in Fig. 1.
ANALYSIS: Apply the principle of work and energy
T_1+U_{1 \rightarrow 2}=T_2 (1)
Kinetic Energy. You need to calculate the initial and final kinetic energy
and the work.
Position 1.
Block: \bar{v}_1 = 6 ft/s
Flywheel: w_1=\frac{\bar{v}_1}{r}=\frac{6~ft / s }{1.25~ft } = 4.80 rad/s
T_1=\frac{1}{2} m \bar{v}_1^2+\frac{1}{2} \bar{I} \omega_1^2
=\frac{1}{2} \frac{240~lb }{32.2~ft / s ^2}\left(6~ft / s ^2\right)+\frac{1}{2}\left(10.5~lb \cdot ft \cdot s ^2\right)(4.80~rad / s )^2
= 255 ft⋅lb
Position 2. Noting that \omega_2=\bar{v}_2 / 1.25, you have
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
Work. During the motion, only the weight W of the block and the friction couple M do work. Note that W does positive work, and the friction couple M does negative work. The total work done is
s_1=0 \quad s_2=4~ft
\theta_1=0 \quad \theta_2=\frac{s_2}{r}=\frac{4~ft }{1.25~ft } = 3.20 rad
U_{1 \rightarrow 2}=W\left(s_2-s_1\right)-M\left(\theta_2-\theta_1\right)
= (240 lb)(4 ft) − (60 lb⋅ft)(3.20 rad)
= 768 ft⋅lb
Substituting these expressions into Eq. (1) gives
T_1+U_{1 \rightarrow 2}=T_2
255~ft \cdot lb +768~ft \cdot lb =7.09 \bar{v}_2^2
\bar{v}_2 = 12.01 ft/s \bar{v}_2 = 12.01 ft/s ↓ ◂
REFLECT and THINK: The speed of the block increases as it falls, but much more slowly than if it were in free fall. This seems like a reasonable result. Rather than calculating the work done by gravity, you could have also treated the effect of the weight using gravitational potential energy, V_g .
