Chapter 15.4, Problem 15.131P

Knowing that at the instant shown bar AB has a constant angular velocity of 4 rad/s clockwise, determine the angular acceleration (a) of bar BD, (b) of bar DE.

To determine

(a)

The angular acceleration of the bar BD.

Answer to Problem 15.131P

The angular acceleration of the link BD is $-10.752\text{rad}/{\text{s}}^2$ along the counter clockwise direction.

Explanation of Solution

Given information:

The angular velocity of the bar is $4\text{rad}/\text{s}$ in clockwise direction.

The following figure represents the different types of the bar.

Figure-(1)

The unit vector along the $x$ direction is $\text{i}$, the unit vector along the $y$ direction is $\text{j}$ and the unit vector along the angular acceleration is $\text{k}$.

Write the expression for the velocity at point B.

$v_B=r_{B/A}\times {\omega }_{AB}$ ....... (I)

Here, the velocity at point B is $v_B$, the angular velocity at point AB is ${\omega }_{AB}$ and the distance between the point A and the point B is $r_{B/A}$.

Write the expression for the velocity at point D link BD.

$v_D=v_B+r_{D/B}\times {\omega }_{BD}$ ...... (II)

Here, the velocity at point D with link BD is $v_D$, the angular velocity of the link BD is ${\omega }_{BD}$ and the distance between the point D and the point B is $r_{D/B}$.

Write the expression for the velocity at point D with link DE.

$v_D=r_{D/E}\times {\omega }_{DE}$ ...... (III)

Here, the angular velocity of the link DE is ${\omega }_{DE}$ and the distance between the point D and the point E is $r_{D/E}$.

Write the expression for the acceleration of the point B.

$a_B={\alpha }_{AB}\times r_{B/A}-{\left({\omega }_{AB}\right)}^2\times r_{B/A}$ ...... (IV)

Here, the acceleration of the point B is $a_B$ and the angular acceleration of link AB is ${\alpha }_{AB}$.

Write the expression for the acceleration of the point D with a part of the link BD.

$a_D=a_B+{\alpha }_{BD}\times r_{D/B}-{\left({\omega }_{DB}\right)}^2\times r_{D/B}$ ...... (V)

Here, the acceleration at point D with a part of the link BD is $a_D$, the angular acceleration of the link BD is ${\alpha }_{BD}$.

Write the expression for the acceleration of the point D with a part of the link DE.

$a_D={\alpha }_{DE}\times r_{D/E}-{\left({\omega }_{DE}\right)}^2\times r_{D/E}$ ...... (VI)

Here, the acceleration at point D with a part of the link DE is $a_D$, the angular acceleration of the link BD is ${\alpha }_{DE}$.

Write the expression for the position of the unit vector for the link AB with respect to point D.

$r_{B/A}=-q-\left(q+x\right)$ ...... (VII)

Here, the horizontal distance between the point B and the point E is $q$, the horizontal distance between the point E and the point D is $x$ and the unit vector for the link AB with respect to point D is $r_{B/A}$.

Write the expression for the position of the unit vector for the link DE with respect to point D.

$r_{D/E}=-q-o$ ...... (VIII)

Here, the vertical distance between the point D and the point E is and the unit vector for the link DE with respect to point D is $r_{D/E}$.

The angular acceleration of the link AB is zero because it has constant angular velocity.

Calculation:

Substitute $\left(20\text{in}\right)$ for $x$ and $\left(20\text{in}\right)$ for $q$ in Equation (VII).

$\begin{array}{c}{r}_{B/A}=\left(-\left(20\text{in}\right)\right)\text{i}-\left(\left(20\text{in}\right)+\left(20\text{in}\right)\right)\text{j}\\ \text{=}\left(-\left(20\text{in}\right)\right)\text{i}-\left(40\text{in}\right)\text{j}\end{array}$

Substitute $\left(20\text{in}\right)$ for $x$ and $\left(25\text{in}\right)$ for $o$ in Equation (VIII).

$r_{D/E}=-\left(20\text{in}\right)\text{i}-\left(25\text{in}\right)\text{j}$

Substitute $-\left(4\text{rad}/\text{s}\right)\text{k}$ for ${\omega }_{AB}$ and $\left(-\left(20\text{in}\right)\right)\text{i}-\left(40\text{in}\right)\text{j}$ for $r_{B/A}$ in Equation (I).

$\begin{array}{c}{v}_B=\left(-\left(20\text{in}\right)\right)\text{i}-\left(40\text{in}\right)\text{j}\times -\left(4\text{rad}/\text{s}\right)\text{k}\\ \text{=}\left|\begin{array}{l}\begin{array}{ccc}\text{i}& \text{j}& \text{k}\end{array}\\ \begin{array}{ccc}0& 0& -4\end{array}\\ \begin{array}{ccc}-20& -40& 0\end{array}\end{array}\right|\\ =\text{i}\left(0-160\text{in}/\text{s}\right)-\text{j}\left(0-80\text{in}/\text{s}\right)+\text{k}\left(0-0\right)\\ =\left(-160\text{in}/\text{s}\right)\text{i+}\left(80\text{in}/\text{s}\right)\text{j}\end{array}$ ...... (IX)

Substitute $\left(-160\text{in}/\text{s}\right)\text{i+}\left(80\text{in}/\text{s}\right)\text{j}$ for $v_B$, ${\omega }_{BD}\text{k}$ for ${\omega }_{BD}$ and $-\left(40\text{in}\right)\text{i}$ for $r_{D/B}$ in Equation (II).

$\begin{array}{c}{v}_D=\left(-160\text{in}/\text{s}\right)\text{i+}\left(80\text{in}/\text{s}\right)\text{j}-\left(40\text{in}\right)\text{i}\times {\omega }_{BD}\text{k}\\ =\left(-160\text{in}/\text{s}\right)\text{i+}\left(80\text{in}/\text{s}\right)\text{j+}\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 0& 0& {\omega }_{BD}\\ 40& 0& 0\end{array}\right|\\ =\left(-160\text{in}/\text{s}\right)\text{i+}\left(80\text{in}/\text{s}\right)+\left\{\text{i}\left(0-0\right)-\text{j}\left(0-40{\omega }_{BD}\right)+\text{k}\left(0-0\right)\right\}\\ =\left(-160\text{in}/\text{s}\right)\text{i+}\left(80\text{in}/\text{s}+40{\omega }_{BD}\right)\text{j}\end{array}$

Substitute $\left(-160\text{in}/\text{s}\right)\text{i+}\left(80\text{in}/\text{s}+40{\omega }_{BD}\right)\text{j}$ for $v_D$, ${\omega }_{DE}\text{k}$ for ${\omega }_{DE}$ and $-\left(20\text{in}\right)\text{i}-\left(25\text{in}\right)\text{j}$ for $r_{D/E}$ in Equation (III).

$\begin{array}{l}\left(-160\text{in}/\text{s}\right)\text{i+}\left(80\text{in}/\text{s}+40{\omega }_{BD}\right)\text{j}=-\left(20\text{in}\right)\text{i}-\left(25\text{in}\right)\text{j}\times {\omega }_{DE}\text{k}\\ \left(-160\text{in}/\text{s}\right)\text{i+}\left(80\text{in}/\text{s}+40{\omega }_{BD}\right)\text{j}=\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 0& 0& {\omega }_{DE}\\ 20& -25& 0\end{array}\right|\\ \left(-160\text{in}/\text{s}\right)\text{i+}\left(80\text{in}/\text{s}+40{\omega }_{BD}\right)\text{j}=\left[\left\{\begin{array}{l}\text{i}\left(0+25{\omega }_{DE}\right)-\text{j}\left(0-20{\omega }_{DE}\right)\\ +\text{k}\left(0-0\right)\end{array}\right\}\right]\\ \left(-160\text{in}/\text{s}\right)\text{i+}\left(80\text{in}/\text{s}+40{\omega }_{BD}\right)\text{j}=\left(25{\omega }_{DE}\right)\text{i+}\left(20{\omega }_{DE}\right)\text{j}\end{array}$ ...... (X)

Compare $\text{i}$ terms from Equation (X).

$\begin{array}{l}\left(-160\text{in}/\text{s}\right)\text{i}=\left(25{\omega }_{DE}\text{in}\right)\text{i}\\ -160\text{in}/\text{s}=25{\omega }_{DE}\\ {\omega }_{DE}=\frac{-160\text{in}/\text{s}}{25\text{in}}\\ {\omega }_{DE}=-6.4\text{rad}/\text{s}\end{array}$

Compare $\text{j}$ terms from Equation (X).

$\begin{array}{l}\left(80\text{in}/\text{s}+40{\omega }_{BD}\right)\text{j}=\left(20{\omega }_{DE}\text{in}\right)\text{j}\\ 80\text{in}/\text{s}+40{\omega }_{BD}=20{\omega }_{DE}\text{in}\end{array}$ ...... (XI)

Substitute $-6.4\text{rad}/\text{s}$ in for ${\omega }_{DE}$ in Equation (XI).

$\begin{array}{l}80\text{in}/\text{s}+40{\omega }_{BD}\text{in}=20\left(-6.4\text{rad}/\text{s}\right)\text{in}\\ 80\text{in}/\text{s}+40{\omega }_{BD}\text{in}=-128\text{in}/\text{s}\\ 40{\omega }_{BD}\text{in}=-208\text{in}/\text{s}\\ {\omega }_{BD}=-5.2\text{in}/\text{s}\end{array}$

Substitute $0$ in for ${\alpha }_{AB}$, $\left(-\left(20\text{in}\right)\right)\text{i}-\left(40\text{in}\right)\text{j}$ for $r_{B/A}$ and $4\text{rad}/\text{s}$ in for ${\omega }_{AB}$ in Equation (IV).

$\begin{array}{c}{a}_B=0\times \left(-\left(20\text{in}\right)\right)\text{i}-\left(40\text{in}\right)\text{j}-{\left(4\text{rad}/\text{s}\right)}^2\times \left(-\left(20\text{in}\right)\right)\text{i}-\left(40\text{in}\right)\text{j}\\ =-\left(16\text{rad}/{\text{s}}^2\right)\times \left(-\left(20\text{in}\right)\right)\text{i}-\left(40\text{in}\right)\text{j}\\ =-\left(320\text{in}/{\text{s}}^2\right)\text{i+}\left(640\text{in}/{\text{s}}^2\right)\text{j}\end{array}$

Substitute $-\left(320\text{in}/{\text{s}}^2\right)\text{i+}\left(640\text{in}/{\text{s}}^2\right)\text{j}$ for $a_B$, $\left(40\text{in}\right)\text{i}$ for $r_{D/B}$, ${\alpha }_{BD}\text{k}$ in for ${\alpha }_{BD}$ and $-5.2\text{rad}/\text{s}$ in for ${\omega }_{BD}$ in Equation (V).

$\begin{array}{c}{a}_D=\left[\begin{array}{l}-\left(320\text{in}/{\text{s}}^2\right)\text{i+}\left(640\text{in}/{\text{s}}^2\right)\text{j}+\left({\alpha }_{BD}\text{k}\right)\times \left(40\text{in}\right)\text{i}\\ -{\left(-5.2\text{rad}/\text{s}\right)}^2\times \left(40\text{in}\right)\text{i}\end{array}\right]\\ a_D=-\left(320\text{in}/{\text{s}}^2\right)\text{i}+\left(640\text{in}/{\text{s}}^2\right)\text{j}+\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 0& 0& {\alpha }_{BD}\\ 40& 0& 0\end{array}\right|-{\left(-5.2\text{rad}/\text{s}\right)}^2\times \left(40\text{in}\right)\text{i}\\ =\left[\begin{array}{l}-\left(320\text{in}/{\text{s}}^2\right)\text{i}+\left(640\text{in}/{\text{s}}^2\right)\text{j}\\ +\left\{\text{i}\left(0-0\right)-\text{j}\left(0-40{\alpha }_{BD}\text{in}\right)+\text{k}\left(0-0\right)\right\}-1081.6\text{i}\end{array}\right]\\ =-761.6\text{i+}\left(640\text{in}/{\text{s}}^2-40{\alpha }_{BD}\text{in}\right)\text{j}\end{array}$

Substitute $-761.6\text{i+}\left(640\text{in}/{\text{s}}^2-40{\alpha }_{BD}\text{in}\right)\text{j}$ for $a_D$, ${\alpha }_{DE}\text{k}$ in for ${\alpha }_{DE}$, $-\left(20\text{in}\right)\text{i}-\left(25\text{in}\right)\text{j}$ for $r_{D/E}$ and $-6.4\text{rad}/\text{s}$ in for ${\omega }_{DE}$ in Equation (VI).

$\left[\begin{array}{l}-761.6\text{i+}\\ \left(640\text{in}/{\text{s}}^2-40{\alpha }_{BD}\text{in}\right)\text{j}\end{array}\right]=\left[\begin{array}{l}{\alpha }_{DE}\text{k}\times -\left(20\text{in}\right)\text{i}-\left(25\text{in}\right)\text{j}\\ -{\left(-6.4\text{rad}/\text{s}\right)}^2\times -\left(20\text{in}\right)\text{i}-\left(25\text{in}\right)\text{j}\end{array}\right]$

$\left[\begin{array}{l}-761.6\text{i+}\\ \left(640\text{in}/{\text{s}}^2-40{\alpha }_{BD}\text{in}\right)\text{j}\end{array}\right]=\left[\begin{array}{l}\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 0& 0& {\alpha }_{DE}\\ 20& -25& 0\end{array}\right|\\ -{\left(-6.4\text{rad}/\text{s}\right)}^2\times -\left(20\text{in}\right)\text{i}-\left(25\text{in}\right)\text{j}\end{array}\right]$

$\left[\begin{array}{l}-761.6\text{i+}\\ \left(640\text{in}/{\text{s}}^2-40{\alpha }_{BD}\text{in}\right)\text{j}\end{array}\right]=\left[\begin{array}{l}\left\{\begin{array}{l}\text{i}\left(0+25{\alpha }_{DE}\text{in}\right)\\ -\text{j}\left(0-20{\alpha }_{DE}\text{in}\right)+\text{k}\left(0-0\right)\end{array}\right\}\\ -819.2\text{i+1024j}\end{array}\right]$

$\left[\begin{array}{l}-761.6\text{i+}\\ \left(640\text{in}/{\text{s}}^2-40{\alpha }_{BD}\text{in}\right)\text{j}\end{array}\right]=\left[\begin{array}{l}\left(25{\alpha }_{DE}\text{in}-819.2\right)\text{i}\\ \text{+}\left(20{\alpha }_{DE}\text{in+1024}\right)\text{j}\end{array}\right]$ ...... (XII)

Compare $\text{i}$ terms from Equation (XII).

$\begin{array}{l}\left[-761.6\text{i}\right]\text{in}/\text{s}=\left(25{\alpha }_{DE}\text{in}-819.2\text{in}/\text{s}\right)\text{i}\\ 25{\alpha }_{DE}\text{in}=-761.6\text{in}/\text{s}+819.2\text{in}/\text{s}\\ 25{\alpha }_{DE}\text{in}=57.6\text{in}/\text{s}\\ {\alpha }_{DE}=2.304\text{rad}/\text{s}\end{array}$

Compare $\text{j}$ terms from Equation (XII).

$\begin{array}{l}\left(640\text{in}/{\text{s}}^2-40{\alpha }_{BD}\text{in}\right)\text{j}=\left(20{\alpha }_{DE}\text{in}+\text{1024}\text{in}/{\text{s}}^2\right)\text{j}\\ 640\text{in}/{\text{s}}^2-40{\alpha }_{BD}\text{in}=20{\alpha }_{DE}\text{in}+\text{1024}\text{in}/{\text{s}}^2\end{array}$ ...... (XIII)

Substitute $2.304\text{rad}/\text{s}$ for ${\alpha }_{DE}$ in Equation (XIII).

$\begin{array}{l}640\text{in}/{\text{s}}^2-40{\alpha }_{BD}\text{in}=20\left(2.304\text{rad}/{\text{s}}^2\right)\text{in}+\text{1024}\text{in}/{\text{s}}^2\\ -40{\alpha }_{BD}\text{in}=46.08\text{in}/{\text{s}}^2+\text{1024}\text{in}/{\text{s}}^2-640\text{in}/{\text{s}}^2\\ {\alpha }_{BD}=\frac{430.08\text{in}/{\text{s}}^2}{-40\text{in}}\\ {\alpha }_{BD}=-10.752\text{rad}/{\text{s}}^2\end{array}$

Conclusion:

The angular acceleration of the link BD is $-10.752\text{rad}/{\text{s}}^2$ along the counter clockwise direction.

To determine

(b)

The angular acceleration of the link DE.

Answer to Problem 15.131P

The angular acceleration of the link DE is $2.304\text{rad}/{\text{s}}^2$ along counter clockwise direction.

Explanation of Solution

Given information:

The angular velocity of the bar is $4\text{rad}/\text{s}$ in clockwise direction.

Calculation:

Compare $\text{i}$ terms from Equation (XII).

$\begin{array}{l}\left[-761.6\text{i}\right]\text{in}/{\text{s}}^2=\left(25{\alpha }_{DE}\text{in}-819.2\text{in}/{\text{s}}^2\right)\text{i}\\ 25{\alpha }_{DE}\text{in}=-761.6\text{in}/{\text{s}}^2+819.2\text{in}/{\text{s}}^2\\ 25{\alpha }_{DE}\text{in}=57.6\text{in}/{\text{s}}^2\\ {\alpha }_{DE}=2.304\text{rad}/{\text{s}}^2\end{array}$

Conclusion:

The angular acceleration of the link DE is $2.304\text{rad}/{\text{s}}^2$ along counter clockwise direction.