Question 17.SP.4: A 30-lb slender rod AB is 5 ft long and is pivoted about a p......
A 30-lb slender rod AB is 5 ft long and is pivoted about a point O which is 1 ft from end B. The other end is pressed against a spring of constant k = 1800 lb/in. until the spring is compressed 1 in. The rod is then in a horizontal position. If the rod is released from this position, determine its angular velocity and the reaction at the pivot O as the rod passes through a vertical position.
Position 1. Potential Energy. Since the spring is compressed 1 in., we have x_1=1 in.
V_e=\frac{1}{2} k x_1^2=\frac{1}{2}(1800\text{ lb}/ \text{in}.)(1 \text{ in} .)^2=900 \text{ in} \cdot \text{ lb}
Choosing the datum as shown, we have V_\text{g}=0; therefore,
V_1=V_e+V_\text{g}=900 \text { in } \cdot \text {lb} =75\text { ft} \cdot \text {lb}
Kinetic Energy. Since the velocity in position 1 is zero, we have T_1=0.
Position 2. Potential Energy. The elongation of the spring is zero, and we have V_e=0. Since the center of gravity of the rod is now 1.5 ft above the datum,
\begin{aligned}& V_\text{g}=(30 \text { lb} )(+1.5 \text { ft} )=45 \text { ft} \cdot \text {lb} \\& V_2=V_e+V_\text{g}=45 \text { ft} \cdot \text {lb}\end{aligned}
Kinetic Energy. Denoting by \omega _2 the angular velocity of the rod in position
2, we note that the rod rotates about O and write \bar{v}_2=\bar{r} \omega_2=1.5 \omega_2.
\begin{aligned}\bar{I} & =\frac{1}{12} m l^2=\frac{1}{12} \frac{30\text{ lb}}{32.2\text{ ft}/ s ^2}(5 \text{ ft} )^2=1.941 \text{ lb} \cdot \text{ ft} \cdot s ^2 \\T_2 & =\frac{1}{2} m \bar{v}_2^2+\frac{1}{2} \bar{I} \omega_2^2=\frac{1}{2} \frac{30}{32.2}\left(1.5 \omega_2\right)^2+\frac{1}{2}(1.941) \omega_2^2=2.019 \omega_2^2\end{aligned}
Conservation of Energy
\begin{aligned}& T_1+V_1=T_2+V_2 \\& 0+75 \text{ ft} \cdot \text{ lb} =2.019 \omega_2^2+45 \text{ ft} \cdot \text{ lb} \\& &\omega_2=3.86 \text{ rad} / s _{\downarrow} \downarrow\end{aligned}
Reaction in Position 2. Since \omega_2=3.86 rad/s, the components of the acceleration of G as the rod passes through position 2 are
\begin{array}{lll}\bar{a}_n & =\bar{r} \omega_2^2=(1.5\text{ ft})(3.86\text{ rad}/ s )^2=22.3\text{ ft}/ s ^2 & \overline{ \mathbf{a} }_n=22.3\text{ ft}/ s ^2 \downarrow \\\bar{a}_t & =\bar{r} \alpha & \overline{ \mathbf{a} }_t=\bar{r} \alpha \rightarrow\end{array}
We express that the system of external forces is equivalent to the system of effective forces represented by the vector of components m \overline{ \mathbf{a} }_t \text { and } m \overline{ \mathbf{a} }_n attached at G and the couple \bar{I} \alpha.
\begin{aligned}+\downarrow \Sigma M_O=\Sigma\left(M_O\right)_{\text {eff }}: 0&=\bar{I} \alpha+m(\bar{r} \alpha) \bar{r} \quad \alpha=0\\\stackrel{+}{\rightarrow} \Sigma F_x=\Sigma\left(F_x\right)_{\text {eff }}: \quad R_x&=m(\bar{r} \alpha) \quad R_x=0 \\+\uparrow \Sigma F_y=\Sigma\left(F_y\right)_{\text {eff }}: \quad R_y-30 \text{ lb} &=-m \bar{a}_n \\R_y-30 \text{ lb}& =-\frac{30 \text{ lb} }{32.2\text{ ft}/ s ^2}\left(22.3 \text{ ft} / s ^2\right)\end{aligned}
R_y=+9.22 \text{ lb} \quad R =9.22 \text{ lb} \uparrow
