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°.

  

To determine

The velocity of point B after rod AB has rotated through 90°.

Answer to Problem 17.45P

Velocity of point B after $90°$ rotation of rod is $3.871m/s$.

Explanation of Solution

Given information:

Mass of rod AB, $m_{AB}=3kg$

Mass of rod BC, $m_{BC}=8kg$

Mass of collar C, $m_c=4kg$

Velocity of collar, $V_c=0.9m/s$

$\theta =90°$

Concept used:

Energy conservation principle

Calculation:

(a)Initial Position

$\begin{array}{l}{\omega }_{BC}=0\\ v_{BC}=v_C=0.9m/s\\ v_B=v_C=0.9m/s\\ {\omega }_{AB}=\frac{v_B}{L_{AB}}=\frac{0.9}{0.15}=6rad/\sec \\ v_{AB}=(\frac{L_{AB}}{2}){\omega }_{AB}=\frac{0.15}{2}\times 6=0.45rad/\sec \\ L_{BC}=\sqrt{L_{AB}{}^2+L_{AC}{}^2}=0.39m\end{array}$

Moment of inertial of rod BC,

$I_{BC}=\frac{m_{BC}.L_{BC}{}^2}{12}=\frac{{0.39}^2\times 8}{12}=0.1014kg.m^2$

Moment of inertial of rod AB,

$I_{AB}=\frac{m_{AB}.L_{AB}{}^2}{12}=\frac{{0.15}^2\times 3}{12}=5.625\times {10}^{-3}kg.m^2$

Initial kinetic energy,

$\begin{array}{l}{T}_2=\frac{m_{AB}.v_{AB}{}^2}{2}+\frac{I_{AB}.{\omega }_{AB}{}^2}{2}+\frac{m_{BC}.v_{BC}{}^2}{2}+\frac{I_{BC}.{\omega }_{BC}{}^2}{2}+\frac{m_C.v_C{}^2}{2}\\ =0.375v_B{}^2+0.125v_B{}^2+v_B{}^2+0.333v_B+0\\ =1.833v_B{}^2\end{array}$

We know from the figure that,

$\begin{array}{l}{h}_{AB}=0\\ h_{BC}=-\frac{L_{AC}}{2}=-0.18m\\ h_C=-0.36m\end{array}$

Initial Potential energy,

$\begin{array}{l}{v}_1=m_{AB}g{h}_{AB}+m_{BC}g{h}_{BC}+m_{C}g{h}_C\\ =0-14.126-14.126\\ =-28.252J\end{array}$

(b) Final Position

Now, rod AB is rotated through $90°$, the collar is at its maximum position. Hence, the velocity of the collar is zero.

$\begin{array}{l}{v}_C=0\\ {\omega }_{BC}=\frac{v_B}{L_{BC}}=\frac{v_B}{0.39}\\ v_{BC}=\frac{L_{BC}}{2}{\omega }_{BC}=\frac{0.39}{2}\times \frac{v_B}{0.39}=0.5v_B\\ {\omega }_{AB}=\frac{v_B}{L_{AB}}=\frac{v_B}{0.15}\\ v_{AB}=\frac{L_{AB}}{2}{\omega }_{AB}=\frac{0.15}{2}\times \frac{v_B}{0.15}=0.5v_B\end{array}$

Final kinetic energy,

$\begin{array}{l}{T}_2=\frac{m_{AB}.v_{AB}{}^2}{2}+\frac{I_{AB}.{\omega }_{AB}{}^2}{2}+\frac{m_{BC}.v_{BC}{}^2}{2}+\frac{I_{BC}.{\omega }_{BC}{}^2}{2}+\frac{m_C.v_C{}^2}{2}\\ =0.375v_B{}^2+0.125v_B{}^2+v_B{}^2+0.333v_B+0\\ =1.833v_B{}^2\end{array}$

From the figure,

$\begin{array}{l}{h}_{AB}=-\frac{L_{AB}}{2}=-0.075m\\ h_{BC}=-L_{AB}-\frac{L_{BC}}{2}=-0.345m\\ h_C=-L_{AB}-L_{BC}=-0.54m\end{array}$

Final potential energy,

$\begin{array}{l}{v}_2=m_{AB}g{h}_{AB}+m_{BC}g{h}_{BC}+m_{C}g{h}_C\\ =-2.207-27.075-21.189\\ =-50.47J\end{array}$

Using energy conservation principle,

$\begin{array}{l}{T}_1+V_1=T_2+V_2\\ 5.2649-28.252=1.8333v_B{}^2-50.47\\ v_B=3.871m/s\end{array}$

Conclusion:

Therefore, the velocity of point B when rod AB rotates through $90°$ can be determined by using the energy conservation principle.