Chapter 15.4, Problem 15.133P

(a)

To determine

The angular acceleration of bar BD using vector approach.

Answer to Problem 15.133P

The angular acceleration of bar BD $\left({\alpha }_{BD}\right)$ is $\underset{\_}{8.15{\text{rad/s}}^2\left(\text{Anticlockwise}\right)}$.

Explanation of Solution

Given information:

The angular velocity of the bar AB $\left({\omega }_{AB}\right)$ is 4 rad/s.

The angular acceleration of the bar AB $\left({\alpha }_{AB}\right)$ is $2{\text{rad/s}}^{\text{2}}$.

Calculation:

Write the relative position vector for point B with respect to A.

${\text{r}}_{B/A}=-\left(20\text{in}\text{.}\right)\text{i}-\left(40\text{in}\text{.}\right)\text{j}$

Write the relative position vector for point D with respect to B.

${\text{r}}_{D/B}=\left(40\text{in}\text{.}\right)\text{i}$

Write the relative position vector for point D with respect to E.

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

Bar AB (Rotation about A):

Determine the angular velocity at point B $\left({\text{v}}_B\right)$ using the relation.

${\text{v}}_B={\text{ω}}_{AB}\times {\text{r}}_{B/A}$

Substitute –4k rad/s for ${\text{ω}}_{AB}$ and $\left[-\left(20\text{in}\text{.}\right)\text{i}-\left(40\text{in}\text{.}\right)\text{j}\right]$ for ${\text{r}}_{B/A}$.

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

Bar BD $\left[\text{Plane}\text{motion}=\text{Translation}\text{with}B+\text{Rotation}\text{about}B\right]$.

Determine the angular velocity at point D $\left({\text{v}}_D\right)$ using the relation.

${\text{v}}_D={\text{v}}_B+{\text{ω}}_{BD}\times {\text{r}}_{D/B}$

Substitute $\left[-160\text{i}+80\text{j}\right]\text{in}\text{./s}$ for ${\text{v}}_B$, ${\omega }_{BD}\text{k}$ for ${\text{ω}}_{BD}$, and $\left(40\text{in}\text{.}\right)\text{i}$ for ${\text{r}}_{D/B}$.

$\begin{array}{c}{\text{v}}_D=\left[-160\text{i}+80\text{j}\right]+\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 0& 0& {\omega }_{BD}\\ 40& 0& 0\end{array}\right|\\ =\left[-160\text{i}+80\text{j}\right]+\left(0\right)\text{i}-\left(0-40{\omega }_{BD}\right)\text{j}+0\text{k}\\ =\left[-160\text{i}+\left(80+40{\omega }_{BD}\right)\text{j}\right]\text{in}\text{./s}\end{array}$ (1)

Bar DE $\left[\text{Rotation}\text{about}E\right]$.

Determine the angular velocity at point D $\left(v_D\right)$ using the relation.

${\text{v}}_D={\text{ω}}_{DE}\times {\text{r}}_{D/E}$

Substitute ${\omega }_{DE}\text{k}$ for ${\text{ω}}_{DE}$ and $\left(20\text{in}\text{.}\right)\text{i}-\left(25\text{in}\text{.}\right)\text{j}$ for ${\text{r}}_{D/E}$.

$\begin{array}{c}{\text{v}}_D=\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 0& 0& {\omega }_{DE}\\ 20& -25& 0\end{array}\right|\\ =\left(0+25{\omega }_{DE}\right)\text{i}-\left(0-20{\omega }_{DE}\right)\text{j}+0\text{k}\\ =\left[25{\omega }_{DE}\text{i}+20{\omega }_{DE}\text{j}\right]\text{in}\text{./s}\end{array}$ (2)

Equate the i components in Equations (1) and (2).

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

Equate the j components in Equations (1) and (2).

$80+40{\omega }_{BD}=20{\omega }_{DE}$

Substitute –6.4 rad/s for ${\omega }_{DE}$.

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

Angular acceleration Analysis:

Bar AB (Rotation about A):

Determine the acceleration at point B using the relation.

$\begin{array}{c}{\text{a}}_B={\text{α}}_{AB}\times {\text{r}}_{B/A}-{\omega }_{AB}^2{\text{r}}_{B/A}\\ ={\alpha }_{AB}\text{k}\times {\text{r}}_{B/A}-{\omega }_{AB}^2{\text{r}}_{B/A}\end{array}$

Substitute $-2{\text{rad/s}}^{\text{2}}$ for ${\text{α}}_{AB}$, $\left[-\left(20\text{in}\text{.}\right)\text{i}-\left(40\text{in}\text{.}\right)\text{j}\right]$ for ${\text{r}}_{B/A}$, and 4k rad/s for ${\text{ω}}_{AB}$.

$\begin{array}{c}{\text{a}}_B=\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 0& 0& -2\\ -20& -40& 0\end{array}\right|-{\left(4\right)}^2\left[-\left(20\right)\text{i}-\left(40\right)\text{j}\right]\\ =\left[\text{i}\left(0-80\right)-\text{j}\left(0-40\right)+\text{k}\left(0\right)\right]+\left[\left(320\right)\text{i}+\left(640\right)\text{j}\right]\\ =\left[240\text{i}+680\text{j}\right]\text{in}{\text{./s}}^{\text{2}}\end{array}$

Bar BD $\left[\text{Plane}\text{motion}=\text{Translation}\text{with}B+\text{Rotation}\text{about}B\right]$.

Determine the acceleration at point D using the relation.

${\text{a}}_D={\text{a}}_B+{\text{α}}_{BD}\times {\text{r}}_{D/B}-{\omega }_{BD}^2{\text{r}}_{D/B}$

Substitute $\left[\left(240\text{in}{\text{./s}}^2\right)\text{i}+\left(680\text{in}{\text{./s}}^2\right)\text{j}\right]$ for ${\text{a}}_B$, ${\alpha }_{BD}\text{k}$ for ${\text{α}}_{BD}$, $\left[\left(40\text{in}\text{.}\right)\text{i}\right]$ for ${\text{r}}_{D/B}$, and $5.2\text{rad/s}$ for ${\text{ω}}_{BD}$.

$\begin{array}{c}{\text{a}}_D=\left[\left(240\right)\text{i}+\left(680\right)\text{j}\right]+\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 0& 0& {\alpha }_{BD}\\ 40& 0& 0\end{array}\right|-{\left(5.2\right)}^2\left[\left(40\right)\text{i}\right]\\ =\left[\left(240\right)\text{i}+\left(680\right)\text{j}\right]+\left[\text{i}\left(0\right)-\text{j}\left(0-40{\alpha }_{BD}+\text{k}\left(0\right)\right)\right]-1,081.6\text{i}\\ =-841.6\text{i}+\left[680+40{\alpha }_{BD}\text{j}\right]\end{array}$ (3)

Bar DE $\left[\text{Rotation}\text{about}E\right]$.

Determine the acceleration at point D using the relation.

${\text{a}}_D={\text{α}}_{DE}\times {\text{r}}_{D/E}-{\omega }_{DE}^2{\text{r}}_{D/E}$

Substitute ${\alpha }_{DE}\text{k}$ for ${\text{α}}_{DE}$, $\left(20\text{in}\text{.}\right)\text{i}-\left(25\text{in}\text{.}\right)\text{j}$ for ${\text{r}}_{D/E}$, and $6.4\text{rad/s}$ for ${\text{ω}}_{DE}$.

$\begin{array}{c}{\text{a}}_D=\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 0& 0& {\alpha }_{DE}\\ 20& -25& 0\end{array}\right|-{\left(6.4\right)}^2\left[20\text{i}-25\text{j}\right]\\ =\left[\text{i}\left(0+25{\alpha }_{DE}\right)-\text{j}\left(0-20{\alpha }_{DE}+\text{k}\left(0\right)\right)\right]-\left[819.2\text{i}-1,024\text{j}\right]\\ =\left[25{\alpha }_{DE}-819.2\right]\text{i}+\left[\left(20{\alpha }_{DE}+1,024\right)\text{j}\right]\end{array}$ (4)

Equate the i components in Equations (3) and (4).

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

Equate the j components in Equations (3) and (4).

$680+40{\alpha }_{BD}=\left(20{\alpha }_{DE}+1,024\right)$

Substitute $-0.896{\text{rad/s}}^2$ for ${\alpha }_{DE}$.

$\begin{array}{l}680+40{\alpha }_{BD}=\left(-20\times 0.896+1,024\right)\\ 40{\alpha }_{BD}=1,006.08-680\\ {\alpha }_{BD}=\frac{326.08}{40}\\ {\alpha }_{BD}=8.15{\text{rad/s}}^{\text{2}}\left(\text{Anticlockwise}\right)\end{array}$

Therefore, the angular acceleration of bar BD $\left({\alpha }_{BD}\right)$ is $\underset{\_}{8.15{\text{rad/s}}^2\left(\text{Anticlockwise}\right)}$.

(b)

To determine

The angular acceleration of the bar DE.

Answer to Problem 15.133P

The angular acceleration of the bar DE is $\underset{\_}{0.896{\text{rad/s}}^2\left(\text{Clockwise}\right)}$.

Explanation of Solution

Given information:

The angular velocity of the bar AB $\left({\omega }_{AB}\right)$ is 4 rad/s.

The angular acceleration of the bar AB $\left({\alpha }_{AB}\right)$ is $2{\text{rad/s}}^{\text{2}}$.

Calculation:

Determine the angular acceleration of the bar DE using the relation.

Refer Part (a).

The angular acceleration of the bar DE is $0.896{\text{rad/s}}^2\left(\text{Clockwise}\right)$.

Therefore, the angular acceleration of the bar DE $\left({\alpha }_{DE}\right)$ is $\underset{\_}{0.896{\text{rad/s}}^2\left(\text{Clockwise}\right)}$.