Chapter 17.1, Problem 17.45P

The uniform rods AB and BC are of mass 3 kg and 8 kg, respectively, and collar C has a mass of 4 kg. Knowing that at the instant shown the velocity of collar C is 0.9 m/s downward, determine the velocity of point B after rod AB has rotated through 90°.

Fig. P17.45

To determine

Find the velocity of point B after rod AB has rotated through $90°$.

Answer to Problem 17.45P

The velocity of point B after rod AB has rotated through $90°$.is $\underset{\_}{3.87\text{m/s}}$.

Explanation of Solution

Given information:

The mass $\left(m_{AB}\right)$ of the rod AB is 3 kg.

The mass $\left(m_{BC}\right)$ of the rod BC is 8 kg.

The mass $\left(m_C\right)$ of the collar C is 4 kg.

The velocity $\left(v_C\right)$ of the collar C is 0.9 m/s.

Calculation:

Refer the system shown.

Find the length $\left(l_{BC}\right)$ of rod BC.

$\begin{array}{c}{l}_{BC}=\sqrt{{150}^2+{360}^2}\\ =390\text{mm}\end{array}$

Find the mass moment of inertia $\left({\overline{I}}_{AB}\right)$ of the rod AB using the equation:

${\overline{I}}_{AB}=\frac{1}{12}{m}_{AB}{L}_{AB}^2$

Substitute 3 kg for $m_{AB}$ and 150 mm for $L_{AB}$.

$\begin{array}{c}{\overline{I}}_{AB}=\frac{1}{12}\left(3\right){\left(150\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)}^2\\ =5.625\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}$

Find the mass moment of inertia $\left({\overline{I}}_{BC}\right)$ of the rod BC using the equation:

${\overline{I}}_{BC}=\frac{1}{12}{m}_{BC}{L}_{BC}^2$

Substitute 8 kg for $m_{AB}$ and 390 mm for $L_{AB}$.

$\begin{array}{c}{\overline{I}}_{BC}=\frac{1}{12}\left(8\right){\left(390\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)}^2\\ =101.4\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}$

Consider the position 1 of the system as shown.

Sketch the position 1 as shown in Figure (1).

Refer Figure (1).

Since ${\left(v_B\right)}_1\text{and }{\left(v_C\right)}_1$ are parallel, the instantaneous center of rod BC lies at $\infty$. Therefore the angular velocity of rod BC $\left[{\left({\omega }_{BC}\right)}_1\right]$ at position 1 is 0.

Find the velocity $\left(v_B\right)$ at point B.

$v_B={\overline{v}}_{BC}=v_C$

Here, ${\overline{v}}_{BC}$ is the velocity of rod BC.

Substitute 0.9 m/s for $v_C$.

$v_B=0.9\text{m/s}$

Consider rod AB rotates about point A.

Find the angular velocity of rod AB $\left({\omega }_{AB}\right)$ at position 1 using the kinematics.

$\left({\omega }_{AB}\right)=\frac{v_B}{l_{AB}}$

Substitute 0.9 m/s for $v_B$ and 150 mm for $l_{AB}$.

$\begin{array}{c}{\omega }_{AB}=\frac{0.9}{\left(150\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)}\\ =6\text{rad/s}\end{array}$

Find the velocity of rod AB using the kinematics.

${\overline{v}}_{AB}=\frac{l_{AB}}{2}{\omega }_{AB}$

Substitute 150 mm for $l_{AB}$ and $6\text{rad/s}$ for ${\omega }_{AB}$.

$\begin{array}{c}{\overline{v}}_{AB}=\frac{\left(150\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)}{2}\left(6\right)\\ =0.45\text{m/s}\end{array}$

Find the kinetic energy $\left(T_1\right)$ of the system at position 1 using the equation:

$T_1=\frac{1}{2}{m}_{AB}{\overline{v}}_{AB}^2+\frac{1}{2}{\overline{I}}_{AB}{\omega }_{AB}^2+\frac{1}{2}{m}_{BC}{\overline{v}}_{BC}^2+\frac{1}{2}{\overline{I}}_{BC}{\omega }_{BC}^2+\frac{1}{2}{m}_C{\overline{v}}_C^2$

Substitute 3 kg for $m_{AB}$, 0.45 m/s for ${\overline{v}}_{AB}^2$, $5.625\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for ${\overline{I}}_{AB}$, 6 rad/s for ${\omega }_{AB}$, 8 kg for $m_{BC}$, 0.9 m/s for ${\overline{v}}_{BC}$, $101.4\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for ${\overline{I}}_{BC}$, 0 for ${\omega }_{BC}$, 4 kg for $m_C$, and 0.9 m/s for ${\overline{v}}_C$.

$\begin{array}{c}{T}_1=\frac{1}{2}\left(3\right){\left(0.45\right)}^2+\frac{1}{2}\left(5.625\times {10}^{-3}\right){\left(6\right)}^2+\frac{1}{2}\left(8\right){\left(0.9\right)}^2+\frac{1}{2}\left(101.4\times {10}^{-3}\right)\left(0\right)+\frac{1}{2}\left(4\right){\left(0.9\right)}^2\\ =0.30375+0.10125+3.24+0+1.62\\ =5.265\text{J}\end{array}$

Refer Figure (1),

Consider the datum is a level line through point A.

Find the potential energy $\left(V_1\right)$ of the system at position 1 using the equation:

$V_1=m_{AB}g{\left(h_{AB}\right)}_1+m_{BC}g{\left(h_{BC}\right)}_1+m_{C}g{\left(h_C\right)}_1$

Substitute $\frac{{\left(h_C\right)}_1}{2}$ for ${\left(h_{BC}\right)}_1$.

$V_1=m_{AB}g{\left(h_{AB}\right)}_1+m_{BC}g\frac{{\left(h_C\right)}_1}{2}+m_{C}g{\left(h_C\right)}_1$

Substitute 3 kg for $m_{AB}$, $9.81{\text{m/s}}^2$ for g, 0 for ${\left(h_{AB}\right)}_1$, 8 kg for $m_{BC}$,4 kg for $m_C$ and $-360\text{mm}$ for ${\left(h_C\right)}_1$.

$\begin{array}{c}{V}_1=\left(3\right)\left(9.81\right)\left(0\right)+\left(8\right)\left(9.81\right)\left(\frac{-360\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}}{2}\right)+\left(4\right)\left(9.81\right)\left(-360\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)\\ =0-14.1264-14.1264\\ =-28.2528\text{J}\end{array}$

Consider the position 2 of the system when the rod AB has rotated $90°$.

Sketch the position 2 as shown in Figure (2).

Consider rod AB rotates about point A.

Find the angular velocity of rod AB $\left[{\left({\omega }_{AB}\right)}_2\right]$ at position 2 using the kinematics.

${\left({\omega }_{AB}\right)}_2=\frac{{\left(v_B\right)}_2}{l_{AB}}$

Substitute 150 mm for $l_{AB}$.

$\begin{array}{c}{\left({\omega }_{AB}\right)}_2=\frac{{\left(v_B\right)}_2}{\left(150\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)}\\ =\frac{{\left(v_B\right)}_2}{0.150}\end{array}$

Find the velocity of rod AB using the kinematics.

${\left({\overline{v}}_{AB}\right)}_2=\frac{l_{AB}}{2}{\omega }_{AB}$

Substitute 150 mm for $l_{AB}$ and $\frac{{\left(v_B\right)}_2}{0.150}$ for ${\omega }_{AB}$.

$\begin{array}{c}{\overline{v}}_{AB}=\frac{\left(150\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)}{2}\left[\frac{{\left(v_B\right)}_2}{0.150}\right]\\ =0.5{\left(v_B\right)}_2\end{array}$

Consider the point C is the instantaneous center of rod BC. Therefore the velocity at point C at position 2 ${\left(v_C\right)}_2=0$.

Find the angular velocity of rod BC $\left[{\left({\omega }_{BC}\right)}_2\right]$ at position 2 using the kinematics.

${\left({\omega }_{BC}\right)}_2=\frac{{\left(v_B\right)}_2}{l_{BC}}$

Substitute 390 mm for $l_{BC}$.

$\begin{array}{c}{\left({\omega }_{BC}\right)}_2=\frac{{\left(v_B\right)}_2}{\left(390\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)}\\ =\frac{{\left(v_B\right)}_2}{0.390}\end{array}$

Find the velocity of rod AB using the kinematics.

${\left({\overline{v}}_{BC}\right)}_2=\frac{l_{BC}}{2}{\omega }_{BC}$

Substitute 390 mm for $l_{BC}$ and $\frac{{\left(v_B\right)}_2}{0.390}$ for ${\omega }_{BC}$.

$\begin{array}{c}{\overline{v}}_{BC}=\frac{\left(390\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)}{2}\left[\frac{{\left(v_B\right)}_2}{0.390}\right]\\ =0.5{\left(v_B\right)}_2\end{array}$

Find the kinetic energy $\left(T_2\right)$ of the system at position 2 using the equation:

$T_2=\frac{1}{2}{m}_{AB}{\left({\overline{v}}_{AB}\right)}_2^2+\frac{1}{2}{\overline{I}}_{AB}{\left({\omega }_{AB}\right)}_2^2+\frac{1}{2}{m}_{BC}{\left({\overline{v}}_{BC}\right)}_2^2+\frac{1}{2}{\overline{I}}_{BC}{\left({\omega }_{BC}\right)}_2^2+\frac{1}{2}{m}_C{\left({\overline{v}}_C\right)}_2^2$

Substitute 3 kg for $m_{AB}$, $0.5{\left(v_B\right)}_2$ for ${\overline{v}}_{AB}^2$, $5.625\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for ${\overline{I}}_{AB}$, $\frac{{\left(v_B\right)}_2}{0.150}$ for ${\omega }_{AB}$, 8 kg for $m_{BC}$, $0.5{\left(v_B\right)}_2$ for ${\overline{v}}_{BC}$, $101.4\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for ${\overline{I}}_{BC}$, $\frac{{\left(v_B\right)}_2}{0.390}$ for ${\omega }_{BC}$, 4 kg for $m_C$, and 0 m/s for ${\left({\overline{v}}_C\right)}_2$.

$\begin{array}{c}{T}_2=\left\{\begin{array}{l}\frac{1}{2}\left(3\right){\left[0.5{\left(v_B\right)}_2\right]}^2+\frac{1}{2}\left(5.625\times {10}^{-3}\right){\left[\frac{{\left(v_B\right)}_2}{0.150}\right]}^2\\ +\frac{1}{2}\left(8\right){\left[0.5{\left(v_B\right)}_2\right]}^2+\frac{1}{2}\left(101.4\times {10}^{-3}\right){\left[\frac{{\left(v_B\right)}_2}{0.390}\right]}^2+\frac{1}{2}\left(4\right){\left(0\right)}^2\end{array}\right\}\\ =0.375{\left(v_B\right)}_2^2+0.125{\left(v_B\right)}_2^2+{\left(v_B\right)}_2^2+0.33333{\left(v_B\right)}_2^2+0\\ =1.83333{\left(v_B\right)}_2^2\end{array}$

Refer Figure (1).

Consider the datum is a level line through point A.

Find the potential energy $\left(V_1\right)$ of the system at position 1 using the equation:

$V_1=m_{AB}g{\left(h_{AB}\right)}_1+m_{BC}g{\left(h_{BC}\right)}_1+m_{C}g{\left(h_C\right)}_1$

Substitute $\frac{{\left(h_C\right)}_1}{2}$ for ${\left(h_{BC}\right)}_1$.

$V_1=m_{AB}g\frac{{\left(h_{AB}\right)}_2}{2}+m_{BC}g\left[{\left(h_{AB}\right)}_2+\frac{{\left(h_{BC}\right)}_2}{2}\right]+m_{C}g\left[{\left(h_{AB}\right)}_2+{\left(h_{BC}\right)}_2\right]$

Substitute 3 kg for $m_{AB}$, $9.81{\text{m/s}}^2$ for g, $-150\text{mm}$ for ${\left(h_{AB}\right)}_2$, 8 kg for $m_{BC}$,4 kg for $m_C$ and $-390\text{mm}$ for ${\left(h_{BC}\right)}_2$.

$\begin{array}{c}{V}_1=\left\{\begin{array}{l}\left(3\right)\left(9.81\right)\left(\frac{-0.150\text{m}}{2}\right)+\left(8\right)\left(9.81\right)\left(-0.150\text{m}+\frac{-0.390\text{m}}{2}\right)\\ +\left(4\right)\left(9.81\right)\left[\left(-0.150\text{m}\right)+\left(-0.390\text{m}\right)\right]\end{array}\right\}\\ =-2.20725-27.0756-21.1896\\ =-50.47245\text{J}\end{array}$

Consider the conservation of energy equation:

Find the velocity of point B $\left[{\left(v_B\right)}_2\right]$ after rod AB has rotated through $90°$.

$T_1+V_1=T_2+V_2$

Substitute $5.265\text{J}$ for $T_1$, $-28.2528\text{J}$ for $V_1$, $1.83333{\left(v_B\right)}_2^2$ for $T_2$, and $-50.47245\text{J}$ for $V_2$.

$\begin{array}{l}5.265-28.2528=1.83333{\left(v_B\right)}_2^2-50.47245\\ 1.83333{\left(v_B\right)}_2^2=27.48465\\ {\left(v_B\right)}_2=3.87\text{m/s}\end{array}$

Substitute ${\omega }_1$ for ${\omega }_2$.

$\begin{array}{l}0.3744{\omega }_1^2+13.3185=0.2976{\left({\omega }_1\right)}^2+23.544\\ 0.0768{\omega }_1^2=23.544-13.3185\\ {\omega }_2=11.54\frac{\text{rad}}{\text{s}}\times \frac{1\text{rev}}{2\pi \text{rad}}\times \frac{60\text{s}}{1\text{min}}\\ {\omega }_2\approx 110.8\text{rpm}\end{array}$

Thus, velocity of point B after rod AB has rotated through $90°$.is $\underset{\_}{3.87\text{m/s}}$.