Chapter 15.4, Problem 15.133P

Knowing that at the instant shown bar AB has angular velocity of 4 rad/s and an angular acceleration of 2 rad/s², both clockwise, determine the angular acceleration (a) of bar BD, (b) of bar DE by using the vector approach as is done in Sample Prob. 15.16.

To determine

(a)

The angular acceleration of bar $BD$.

Answer to Problem 15.133P

The angular acceleration of bar $BD$ is $8.152\text{rad}/{\text{s}}^2$ in clockwise direction.

Explanation of Solution

Given information:

The angular velocity of bar $AB$ is $4\text{rad}/\text{s}$ and angular acceleration of bar Y $AB$ is 1 $2{\text{rad}/\text{s}}^2$.

Write the expression of position vector of a point.

YY1 $r=a\text{i}+b\text{j}+c\text{k}$ ...... (I)

Here, the coordinate in 3 $x$ direction is $a$, the coordinate in $y$ direction is $b$, the coordinate in $z$ direction is $c$, the unit vector in $x$ direction is $\text{i}$, the unit vector in $y$ direction is $\text{j}$ and the unit vector in $z$ direction or angular direction is $\text{k}$.

The coordinate of point $B$ with respect to point $A$ are $\left(-20\text{in,}-40\text{in,}\text{0}\right)$ and the position vector is $r_{BA}$.

The coordinate of point $D$ with respect to point $B$ are $\left(40\text{in},0\text{,}\text{0}\right)$ and the position vector is $r_{DB}$.

The coordinate of point $D$ with respect to point $E$ are $\left(20\text{in,}-2\text{5}\text{in,}\text{0}\right)$ and the position vector is $r_{DE}$.

Write the expression of angular velocity of $AB$ in vector form.

${\omega }_{AB}=-{{\omega }^{\prime }}_{AB}\text{k}$ ...... (II)

Here, the angular velocity of $AB$ is ${{\omega }^{\prime }}_{AB}$ and the unit vector in $z$ direction or angular direction is $\text{k}$.

Write the expression of angular acceleration of $AB$ in vector form.

$a_{AB}=-{a^{\prime }}_{AB}\text{k}$ ...... (III)

Here, the angular acceleration of $AB$ is ${a^{\prime }}_{AB}$ and the unit vector in $z$ direction or angular direction is $\text{k}$.

Write the expression of velocity of point $B$.

$V_B={\omega }_{AB}\times r_{BA}$ ...... (IV)

Here, the angular velocity of $AB$ in vector form is ${\omega }_{AB}$ and the position vector is $r_{BA}$.

Write the expression of velocity of point $D$ when it is considered as a part of link $BD$.

$V_D=V_B+\left({\omega }_{DB}\times r_{DB}\right)$ ...... (V)

Here, the angular velocity of $BD$ in vector form is ${\omega }_{DB}$ and the position vector is $r_{DB}$.

Write the expression of velocity of point $D$ when it is considered as a part of a link $DE$.

$V_D={\omega }_{DE}{r}_{DE}$ ...... (VI)

Here, the angular velocity of $DE$ in vector form is ${\omega }_{DE}$ and the position vector is $r_{DE}$.

Write the expression of acceleration of point $B$ in vector form.

$a_B=a_{BA}\times r_{BA}-{{\omega }^{\prime }}_{AB}^2{r}_{BA}$ ...... (VII)

Here, acceleration of point $B$ is $a_{BA}$, the angular velocity of $AB$ in vector form is ${{\omega }^{\prime }}_{AB}$ and position vector is $r_{BA}$.

Write the expression of acceleration of point $D$ when it is considered as a part of the link $BD$.

$a_D=a_B+a_{DB}\times r_{DB}-{\omega }_{DB}^2\times r_{DB}$ ...... (VIII)

Here, the acceleration of point $B$ in vector form is $a_B$, the angular acceleration of link $BD$ is $a_{BD}$, the position vector is $r_{DE}$ and angular velocity of $DB$ in vector form is ${\omega }_{DB}$.

Write the expression of acceleration of point $D$ when it is considered as a part of the link $DE$.

$a_D=a_{DE}\times r_{DE}-{\omega }_{DE}^2{r}_{DE}$ ...... (IX)

Here, the angular acceleration of link $DE$ is $a_{DE}$ position vector is $r_{DE}$ and, the angular velocity of $DE$ in vector form is ${\omega }_{DE}$.

Calculation:

Substitute $r_{BA}$ for $r$, $-20\text{in}$ for $a$, $-40\text{in}$ for $b$ and $0$ for $c$ in Equation (I).

$\begin{array}{c}{r}_{BA}=\left(-20\text{in}\right)\text{i}+\left(-40\text{in}\right)\text{j}+\left(0\right)\text{k}\\ \text{=}-\left(20\text{in}\right)\text{i}-\left(40\text{in}\right)\text{j}\end{array}$

Substitute $r_{DB}$ for $r$, $40\text{in}$ for $a$, $0$ for $b$ and $0$ for $c$ in Equation (I).

$\begin{array}{c}{r}_{DB}=\left(40\text{in}\right)\text{i}+\left(0\right)\text{j}+\left(0\right)\text{k}\\ \text{=}\left(40\text{in}\right)\text{i}\end{array}$

Substitute $r_{DE}$ for $r$, $-275\text{mm}$ for $a$, $\text{75}\text{mm}$ for $b$ and $0$ for $c$ in Equation (I).

$\begin{array}{c}{r}_{DE}=\left(20\text{in}\right)\text{i}+\left(-2\text{5}\text{in}\right)\text{j}+\left(0\right)\text{k}\\ \text{=}\left(20\text{in}\right)\text{i}-\left(2\text{5}\text{in}\right)\text{j}\end{array}$

Substitute $4\text{rad}/\text{s}$ for ${{\omega }^{\prime }}_{AB}$ in Equation (II).

${\omega }_{AB}=-\left(4\text{rad}/\text{s}\right)\text{k}$

Substitute $2{\text{rad}/\text{s}}^2$ for ${a^{\prime }}_{AB}$ in Equation (III).

$a_{AB}=-\left(2\text{rad}/{\text{s}}^2\right)\text{k}$

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

$V_B=\left\{-\left(4\text{rad}/\text{s}\right)\text{k}\right\}\times \left\{-\left(20\text{in}\right)\text{i}-\left(40\text{in}\right)\text{j}\right\}$ ...... (X)

Here, the Equation (X) is the cross product of two vector, now change it in determinant form.

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

Substitute $-\left(160\text{in}/\text{s}\right)\text{i}+\left(80\text{in}/\text{s}\right)\text{j}$ for $V_B$, ${\omega }_{DB}\text{k}$ for ${\omega }_{DB}$ and $\left(40\text{in}\right)\text{i}$ for $r_{DB}$ in Equation (V).

$\begin{array}{l}{V}_D=\left\{-\left(160\text{in}/\text{s}\right)\text{i}+\left(80\text{in}/\text{s}\right)\text{j}\right\}+\left\{{\omega }_{DB}\text{k}\times \left(40\text{in}\right)\text{i}\right\}\\ V_D=-\left(160\text{in}/\text{s}\right)\text{i}+\left(80\text{in}/\text{s}\right)\text{j}+\left\{{\omega }_{DB}\text{k}\times \left(40\text{in}\right)\text{i}\right\}\end{array}$ ...... (XI)

Here, the Equation (XI) is the cross product of two vector, now change it in determinant form.

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

Substitute $-\left(160\text{in}/\text{s}\right)\text{i}+\left\{\left(80\text{in}/\text{s}\right)+{\omega }_{DB}\left(40\text{in}\right)\right\}\text{j}$ for $V_D$, $\left(20\text{in}\right)\text{i}-\left(2\text{5}\text{in}\right)\text{j}$ for $r_{DE}$ and ${\omega }_{DE}\text{k}$ for ${\omega }_{DE}$ in Equation (VI).

$-\left(160\text{in}/\text{s}\right)\text{i}+\left\{\left(80\text{in}/\text{s}\right)+{\omega }_{DB}\left(40\text{in}\right)\right\}\text{j}=\left\{{\omega }_{DE}\text{k}\right\}\times \left\{\left(20\text{in}\right)\text{i}-\left(2\text{5}\text{in}\right)\text{j}\right\}$ ...... (XII)

Here, the Equation (XII) is the cross product of two vector, now change it in determinant form.

$\begin{array}{l}\left[\begin{array}{l}-\left(160\text{in}/\text{s}\right)\text{i}+\\ \left\{\left(80\text{in}/\text{s}\right)+{\omega }_{DB}\left(40\text{in}\right)\right\}\text{j}\end{array}\right]=\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 0& 0& {\omega }_{DE}\\ 20\text{in}& -2\text{5}\text{in}& 0\end{array}\right|\\ \left[\begin{array}{l}-\left(160\text{in}/\text{s}\right)\text{i}+\\ \left\{\left(80\text{in}/\text{s}\right)+{\omega }_{DB}\left(40\text{in}\right)\right\}\text{j}\end{array}\right]=\left[\begin{array}{l}\left\{0-\left(-2\text{5}\text{in}\right)\times {\omega }_{DE}\right\}\text{i}-\\ \left\{0-\left({\omega }_{DE}20\text{in}\right)\right\}\text{j}-\left\{0-0\right\}\text{k}\end{array}\right]\\ \left[\begin{array}{l}-\left(160\text{in}/\text{s}\right)\text{i}+\\ \left\{\left(80\text{in}/\text{s}\right)+{\omega }_{DB}\left(40\text{in}\right)\right\}\text{j}\end{array}\right]={\omega }_{DE}\left(2\text{5}\text{in}\right)\text{i+}{\omega }_{DE}\left\{20\text{in}\right\}\text{j}-0\text{k}\\ \left[\begin{array}{l}-\left(160\text{in}/\text{s}\right)\text{i}+\\ \left\{\left(80\text{in}/\text{s}\right)+{\omega }_{DB}\left(40\text{in}\right)\right\}\text{j}\end{array}\right]={\omega }_{DE}\left(2\text{5}\text{in}\right)\text{i}+{\omega }_{DE}\left\{20\text{in}\right\}\text{j}\end{array}$ ...... (XIII)

Compare the term of $\text{i}$ on both sides of Equation (XIII).

$\begin{array}{l}-\left(160\text{in}/\text{s}\right)={\omega }_{DE}\left(2\text{5}\text{in}\right)\\ {\omega }_{DE}=\frac{-\left(160\text{in}/\text{s}\right)}{\left(2\text{5}\text{in}\right)}\\ {\omega }_{DE}=-6.4\text{rad}/\text{s}\end{array}$

Compare the term of $\text{j}$ on both sides of Equation (XIII) and substitute $-6.4\text{rad}/\text{s}$ for ${\omega }_{DE}$.

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

Substitute $-\left(2\text{rad}/{\text{s}}^2\right)\text{k}$ for $a_{AB}$, $-\left(20\text{in}\right)\text{i}-\left(40\text{in}\right)\text{j}$ for $r_{BA}$ and $\left(4\text{rad}/\text{s}\right)$ for ${{\omega }^{\prime }}_{AB}$ in Equation (VI).

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

Here, the Equation (XIV) is the cross product of two vector, now change it in determinant form.

$\begin{array}{c}{a}_B=\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 0& 0& -2\text{rad}/{\text{s}}^2\\ -20\text{in}& -40\text{in}& 0\end{array}\right|+\left(320{\text{in}/\text{s}}^2\right)\text{i}+\left(640{\text{in}/\text{s}}^2\right)\text{j}\\ \text{=}\left[\begin{array}{l}\left\{0-\left(-2\text{rad}/{\text{s}}^2\right)\times \left(-40\text{in}\right)\right\}\text{i}-\left\{0-\left(-2\text{rad}/{\text{s}}^2\right)\times \left(-20\text{in}\right)\right\}\text{j}-\\ \left\{0-0\right\}\text{k}+\left(320{\text{in}/\text{s}}^2\right)\text{i}+\left(640{\text{in}/\text{s}}^2\right)\text{j}\end{array}\right]\\ =\left(-80\text{in}\cdot \text{rad}/{\text{s}}^2\right)\text{i}-\left(-40\text{in}\cdot \text{rad}/{\text{s}}^2\right)\text{j}+\left(320{\text{in}/\text{s}}^2\right)\text{i}+\left(640{\text{in}/\text{s}}^2\right)\text{j}\\ =\left(240\text{in}/{\text{s}}^2\right)\text{i}+\left(680\text{in}/{\text{s}}^2\right)\text{j}\end{array}$

Substitute $\left(240\text{in}/{\text{s}}^2\right)\text{i}+\left(680\text{in}/{\text{s}}^2\right)\text{j}$ for $a_B$, $\left(a_{DB}\right)\text{k}$ for $a_{DB}$

$\left(40\text{in}\right)\text{i}$ for $r_{DB}$ and $-5.2\text{rad}/\text{s}$ for ${\omega }_{DB}$ in Equation (VIII).

$\begin{array}{c}{a}_D=\left[\begin{array}{l}\left\{\left(240\text{in}/{\text{s}}^2\right)\text{i}+\left(680\text{in}/{\text{s}}^2\right)\text{j}\right\}+\left\{\left(a_{DB}\right)\text{k}\right\}\times \left\{\left(40\text{in}\right)\text{i}\right\}\\ -{\left(-5.2\text{rad}/\text{s}\right)}^2\times \left\{\left(40\text{in}\right)\text{i}\right\}\end{array}\right]\\ =\left(240\text{in}/{\text{s}}^2\right)\text{i}+\left(680\text{in}/{\text{s}}^2\right)\text{j}+\left\{\left(a_{DB}\right)\text{k}\times \left(40\text{in}\right)\text{i}\right\}-\left(1081.6\text{in}/{\text{s}}^2\right)\text{i}\end{array}$ ...... (XV)

Here, the Equation (XV) is the cross product of two vector, now change it in determinant form.

$\begin{array}{c}{a}_D=\left(240\text{in}/{\text{s}}^2\right)\text{i}+\left(680\text{in}/{\text{s}}^2\right)\text{j}+\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 0& 0& a_{DB}\\ 40\text{in}& 0& 0\end{array}\right|-\left(1081.6\text{in}/{\text{s}}^2\right)\text{i}\\ =\left[\begin{array}{l}\left(240\text{in}/{\text{s}}^2\right)\text{i}+\left(680\text{in}/{\text{s}}^2\right)\text{j}+\left\{0-0\right\}\text{i}-\\ \left\{0-\left(a_{DB}\times 40\text{in}\right)\right\}\text{j}-\left\{0-0\right\}\text{k}-\left(1081.6\text{in}/{\text{s}}^2\right)\text{i}\end{array}\right]\\ =\left(240\text{in}/{\text{s}}^2\right)\text{i}+\left(680\text{in}/{\text{s}}^2\right)\text{j}+\left(a_{DB}\times 40\text{in}\right)\text{j}-\left(1081.6\text{in}/{\text{s}}^2\right)\text{i}\\ =-\left(841.6\text{in}/{\text{s}}^2\right)\text{i}+\left\{\left(680\text{in}/{\text{s}}^2\right)+\left(a_{DB}\times 40\text{in}\right)\right\}\text{j}\end{array}$

Substitute $-\left(841.6\text{in}/{\text{s}}^2\right)\text{i}+\left\{\left(680\text{in}/{\text{s}}^2\right)+\left(a_{DB}\times 40\text{in}\right)\right\}\text{j}$ for $a_D$, $a_{DE}\text{k}$ for $a_{DE}$

$\left(20\text{in}\right)\text{i}-\left(2\text{5}\text{in}\right)\text{j}$ for $r_{DE}$ and $-6.4\text{rad}/\text{s}$ for ${\omega }_{DE}$ in Equation (IX).

$\begin{array}{l}\left[-\left(841.6\text{in}/{\text{s}}^2\right)\text{i}+\left\{\begin{array}{l}\left(680\text{in}/{\text{s}}^2\right)\\ +\left(a_{DB}\times 40\text{in}\right)\end{array}\right\}\text{j}\right]=\left[\begin{array}{l}\left(a_{DE}\text{k}\right)\times \left\{\left(20\text{in}\right)\text{i}-\left(2\text{5}\text{in}\right)\text{j}\right\}\\ -{\left(-6.4\text{rad}/\text{s}\right)}^2\times \left\{\begin{array}{l}\left(20\text{in}\right)\text{i}\\ -\left(2\text{5}\text{in}\right)\text{j}\end{array}\right\}\end{array}\right]\\ \left[-\left(841.6\text{in}/{\text{s}}^2\right)\text{i}+\left\{\begin{array}{l}\left(680\text{in}/{\text{s}}^2\right)\\ +\left(a_{DB}\times 40\text{in}\right)\end{array}\right\}\text{j}\right]=\left[\begin{array}{l}\left(a_{DE}\text{k}\right)\times \left\{\left(20\text{in}\right)\text{i}-\left(2\text{5}\text{in}\right)\text{j}\right\}\\ -\left(819.2\text{in}/{\text{s}}^2\right)\text{i}+\left(1024\text{in}/{\text{s}}^2\right)\text{j}\end{array}\right]\end{array}$ ...... (XVI)

Here, the Equation (XVI) is the cross product of two vector, now change it in determinant form

$\left[\begin{array}{l}-\left(841.6\text{in}/{\text{s}}^2\right)\text{i}+\\ \left\{\left(680\text{in}/{\text{s}}^2\right)+\left(a_{DB}\times 40\text{in}\right)\right\}\text{j}\end{array}\right]=\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 0& 0& a_{DE}\\ 20\text{in}& -25\text{in}& 0\end{array}\right|-\left\{\begin{array}{l}\left(819.2\text{in}/{\text{s}}^2\right)\text{i}\\ +\left(1024\text{in}/{\text{s}}^2\right)\text{j}\end{array}\right\}$

$\left[-\left(841.6\text{in}/{\text{s}}^2\right)\text{i}+\left\{\begin{array}{l}\left(680\text{in}/{\text{s}}^2\right)\\ +\left(a_{DB}\times 40\text{in}\right)\end{array}\right\}\text{j}\right]=\left[\begin{array}{l}\left\{0-\left(-25\text{in}\right)\times a_{DE}\right\}\text{i}-\\ \left\{0-\left(20\text{in}\right)\times a_{DE}\right\}\text{j}\\ -\left\{0-0\right\}\text{k}-\left(819.2\text{in}/{\text{s}}^2\right)\text{i}\\ +\left(1024\text{in}/{\text{s}}^2\right)\text{j}\end{array}\right]$

$\left[-\left(841.6\text{in}/{\text{s}}^2\right)\text{i}+\left\{\begin{array}{l}\left(680\text{in}/{\text{s}}^2\right)\\ +\left(a_{DB}\times 40\text{in}\right)\end{array}\right\}\text{j}\right]=\left[\begin{array}{l}{a}_{DE}\left(25\text{in}\right)\text{i}+a_{DE}\left(20\text{in}\right)\text{j}\\ -\left(819.2\text{in}/{\text{s}}^2\right)\text{i}+\left(1024\text{in}/{\text{s}}^2\right)\text{j}\end{array}\right]$

$\left[-\left(841.6\text{in}/{\text{s}}^2\right)\text{i}+\left\{\begin{array}{l}\left(680\text{in}/{\text{s}}^2\right)\\ +\left(a_{DB}\times 40\text{in}\right)\end{array}\right\}\text{j}\right]=\left[\begin{array}{l}\left\{a_{DE}\left(25\text{in}\right)-\left(819.2\text{in}/{\text{s}}^2\right)\right\}\text{i}+\\ \left\{a_{DE}\left(20\text{in}\right)+\left(1024\text{in}/{\text{s}}^2\right)\right\}\text{j}\end{array}\right]$ ....... (XVII)

Compare the term of $\text{i}$ on both sides of Equation (XVII).

$\begin{array}{l}-\left(841.6\text{in}/{\text{s}}^2\right)=a_{DE}\left(25\text{in}\right)-\left(819.2\text{in}/{\text{s}}^2\right)\\ a_{DE}\left(25\text{in}\right)=-\left(841.6\text{in}/{\text{s}}^2\right)+\left(819.2\text{in}/{\text{s}}^2\right)\\ a_{DE}=\frac{-22.4\text{in}/{\text{s}}^2}{25\text{in}}\\ a_{DE}=-0.896\text{rad}/{\text{s}}^2\end{array}$

Compare the term of $\text{j}$ on both sides of Equation (XVII) and substitute $-0.896\text{rad}/{\text{s}}^2$ for $a_{DE}$.

$\begin{array}{l}\left(680\text{in}/{\text{s}}^2\right)+\left(a_{DB}\times 40\text{in}\right)=\left(-0.896\text{rad}/{\text{s}}^2\right)\left(20\text{in}\right)+\left(1024\text{in}/{\text{s}}^2\right)\\ \left(a_{DB}\times 40\text{in}\right)=\left(-17.92\text{in}/{\text{s}}^2\right)+\left(1024\text{in}/{\text{s}}^2\right)-\left(680\text{in}/{\text{s}}^2\right)\\ a_{DB}=\frac{326.08\text{in}/{\text{s}}^2}{40\text{in}}\\ a_{DB}=8.152\text{rad}/{\text{s}}^2\end{array}$

Conclusion:

The angular acceleration of bar $BD$ is $8.152\text{rad}/{\text{s}}^2$ in clockwise direction.

To determine

(b)

The angular acceleration of bar $DE$ by using the vector approach.

Answer to Problem 15.133P

The angular acceleration of bar $DE$ is $0.896\text{rad}/{\text{s}}^2$ in clockwise direction.

Explanation of Solution

Calculation:

Compare the term of $\text{i}$ on both sides of Equation (XVII).

$\begin{array}{l}-\left(841.6\text{in}/{\text{s}}^2\right)=a_{DE}\left(25\text{in}\right)-\left(819.2\text{in}/{\text{s}}^2\right)\\ a_{DE}\left(25\text{in}\right)=-\left(841.6\text{in}/{\text{s}}^2\right)+\left(819.2\text{in}/{\text{s}}^2\right)\\ a_{DE}=\frac{-22.4\text{in}/{\text{s}}^2}{25\text{in}}\\ a_{DE}=-0.896\text{rad}/{\text{s}}^2\end{array}$

Conclusion:

The angular acceleration of bar $DE$ is $0.896\text{rad}/{\text{s}}^2$ in clockwise direction.