Chapter 15.4, Problem 15.133P

15.133 and 15.134    Knowing that at the instant shown bar AB has an angular velocity of 10 rad/s and an angular acceleration of 4 rad/s², both clockwise, determine the angular acceleration of (a) bar BD, (b) bar DE by using the vector approach as is done in Sample Prob. 15.16.

Fig. P15.131 and P15.133

To determine

Find the magnitude of angular acceleration of bar BD.

Answer to Problem 15.133P

The angular acceleration of the bar BD is $\underset{\_}{62{\text{rad/s}}^2\left(\text{Counterclockwise}\right)}$.

Explanation of Solution

Given information:

The angular velocity of the bar AB is ${\text{ω}}_{AB}=10\text{rad/s}\left(\text{Clockwise}\right)=-10\text{k}$.

Consider the bar AB.

Consider the position of the point B with respect to the point A is denoted by ${\text{r}}_{B/A}=\left(-20\text{in}\text{.}\right)\text{i}-\left(40\text{in}\text{.}\right)\text{j}$.

Calculate the velocity at B using the relation:

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

Substitute $\left(-20\text{in}\text{.}\right)\text{i}-\left(40\text{in}\text{.}\right)\text{j}$ for ${\text{r}}_{B/A}$ and $-10\text{k}$ for ${\text{ω}}_{AB}$.

$\begin{array}{c}{\text{v}}_B=-10\text{k}\times \left[\left(-20\text{in}\text{.}\right)\text{i}-\left(40\text{in}\text{.}\right)\text{j}\right]\\ \text{=}\left(200\text{in}\text{.}\right)\text{j}-\left(400\text{in}\text{.}\right)\text{i}\\ \text{=}\left[-\left(400\text{in}\text{.}\right)\text{i}+\left(200\text{in}\text{.}\right)\text{j}\right]\text{in}\text{./s}\end{array}$

Consider the bar BD.

Consider the position of the point D with respect to the point B is ${\text{r}}_{D/B}=\left(40\text{in}\text{.}\right)\text{i}$.

Consider the angular acceleration of the bar BD is ${\text{ω}}_{BD}={\omega }_{BD}\text{k}$.

Calculate the velocity of the point D using the relation:

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

Substitute $\left[-\left(400\text{in}\text{.}\right)\text{i}+\left(200\text{in}\text{.}\right)\text{j}\right]\text{in}\text{./s}$ for ${\text{v}}_B$, ${\omega }_{BD}\text{k}$ for ${\text{ω}}_{BD}$, and $\left(40\text{in}\text{.}\right)\text{i}$ for ${\text{r}}_{D/B}$.

$\begin{array}{c}{\text{v}}_D=\left[-\left(400\text{in}\text{.}\right)\text{i}+\left(200\text{in}\text{.}\right)\text{j}\right]+{\omega }_{BD}\text{k}\times \left(40\text{in}\text{.}\right)\text{i}\\ =-400\text{i}+200\text{j}+\left(40{\omega }_{BD}\right)\text{j}\\ \text{=}-400\text{i}+\left(200+40{\omega }_{BD}\right)\text{j}\end{array}$ (1)

Consider the bar DE.

Consider the position of the point D with respect to the point E is ${\text{r}}_{D/E}=\left(20\text{in}\text{.}\right)\text{i}-\left(25\text{in}\text{.}\right)\text{j}$.

Consider the angular acceleration of the bar DE is ${\text{ω}}_{DE}={\omega }_{DE}\text{k}$.

Calculate the velocity of the point D 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 ${\text{r}}_{D/E}$.

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

Equate the Equation (1) and (2).

$-400\text{i}+\left(200+40{\omega }_{BD}\right)\text{j}=25{\omega }_{DE}\text{i}+20{\omega }_{DE}\text{j}$ (3)

Equate j component of the Equation (3).

$\begin{array}{l}-400=25{\omega }_{DE}\\ {\omega }_{DE}=-\frac{\text{400}}{25}\\ {\omega }_{DE}=-16\text{rad/s}\\ {\omega }_{DE}=16\text{rad/s}\left(\text{Clockwise}\right)\end{array}$

Equate i component of the Equation (3).

$\begin{array}{l}\left(200+40{\omega }_{BD}\right)=20{\omega }_{DE}\\ 40{\omega }_{BD}=20{\omega }_{DE}-200\\ {\omega }_{BD}=\left(\frac{20{\omega }_{DE}-200}{40}\right)\end{array}$

Substitute $-16\text{rad/s}$ for ${\omega }_{DE}$.

$\begin{array}{c}{\omega }_{BD}=\left(\frac{20\times \left(-16\right)-200}{40}\right)\\ =-13\text{rad/s}\end{array}$

Consider the bar AB.

The angular velocity of the bar AB is ${\text{ω}}_{AB}=10\text{rad/s}\left(\text{Clockwise}\right)$.

The angular velocity of the bar BD is ${\text{ω}}_{BD}=13\text{rad/s}\left(\text{Clockwise}\right)$.

The angular velocity of the bar DE is ${\text{ω}}_{DE}=16\text{rad/s}\left(\text{Clockwise}\right)$.

The angular acceleration of the bar AB is ${\text{α}}_{AB}=-4\text{k}$.

The angular acceleration of the bar BD is ${\text{α}}_{BD}={\alpha }_{BD}\text{k}$.

The angular acceleration of the bar DE is ${\text{α}}_{DE}={\alpha }_{DE}\text{k}$.

Consider the bar AB.

Calculate the acceleration $\left({\text{a}}_B\right)$ of the point B using the relation:

${\text{a}}_B={\text{α}}_{AB}\times {\text{r}}_{B/A}-{\omega }_{AB}^2{\text{r}}_{B/A}$

Substitute $10\text{rad/s}$ for ${\omega }_{AB}$, $\left(-20\text{in}\text{.}\right)\text{i}-\left(40\text{in}\text{.}\right)\text{j}$ for ${\text{r}}_{B/A}$, and $-4\text{k}$ for ${\text{α}}_{AB}$.

$\begin{array}{l}{\text{a}}_B=-4\text{k}\times \left[\left(-20\text{in}\text{.}\right)\text{i}-\left(40\text{in}\text{.}\right)\text{j}\right]-{10}^2\times \left[\left(-20\text{in}\text{.}\right)\text{i}-\left(40\text{in}\text{.}\right)\text{j}\right]\\ {\text{a}}_B=\left[\text{80j}-160\text{i}\right]-\left[-2000\text{i}-4000\text{j}\right]\\ {\text{a}}_B=\text{80j}-160\text{i}+2000\text{i}+4000\text{j}\\ {\text{a}}_B=184\text{0i}+4080\text{j}\end{array}$

Consider the bar BD.

Calculate the acceleration of the point D using the relation:

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

Substitute $\left(184\text{0i}+4080\text{j}\right)$ for ${\text{a}}_B$, ${\alpha }_{BD}\text{k}$ for ${\text{α}}_{BD}$, $13\text{rad/s}$ for ${\omega }_{BD}$, and $\left(40\text{in}\text{.}\right)\text{i}$ for ${\text{r}}_{D/B}$.

$\begin{array}{c}{\text{a}}_D=\left(184\text{0i}+4080\text{j}\right)+{\alpha }_{BD}\text{k}\times \left(40\text{in}\text{.}\right)\text{i}-{13}^2\times \left(40\text{in}\text{.}\right)\text{i}\\ {\text{a}}_D=\left(184\text{0i}+4080\text{j}\right)+40{\alpha }_{BD}\text{j}-6,760\text{i}\\ {\text{a}}_D\text{=}-4920\text{i}+\left(40{\alpha }_{BD}+4080\right)\text{j}\end{array}$ (4)

Consider the bar DE.

Consider the angular acceleration of the bar DE is ${\text{α}}_{DE}={\alpha }_{DE}\text{k}$.

Calculate the acceleration of the point D using the relation:

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

Substitute ${\alpha }_{DE}\text{k}$ for ${\text{α}}_{DE}$, $16\text{rad/s}$ for ${\omega }_{DE}$, and $\left(20\text{in}\text{.}\right)\text{i}-\left(25\text{in}\text{.}\right)\text{j}$ for ${\text{r}}_{D/E}$.

$\begin{array}{c}{\text{a}}_D={\alpha }_{DE}\text{k}\times \left[\left(20\text{in}\text{.}\right)\text{i}-\left(25\text{in}\text{.}\right)\text{j}\right]-{16}^2\times \left[\left(20\text{in}\text{.}\right)\text{i}-\left(25\text{in}\text{.}\right)\text{j}\right]\\ {\text{a}}_D=\left[20{\alpha }_{DE}\text{j}+25{\alpha }_{DE}\text{i}\right]-\left[\text{5,120i}-6,400\text{j}\right]\\ {\text{a}}_D=\left(25{\alpha }_{DE}-5,120\right)\text{i}+\left(20{\alpha }_{DE}+6,400\right)\text{j}\end{array}$ (5)

Equate Equation (4) and (5)

$-4920\text{i}+\left(40{\alpha }_{BD}+4080\right)\text{j}=\left(25{\alpha }_{DE}-5,120\right)\text{i}+\left(20{\alpha }_{DE}+6,400\right)\text{j}$ (6)

Equate i component of the Equation (6).

$\begin{array}{l}-4920=\left(25{\alpha }_{DE}-5,120\right)\\ {\alpha }_{DE}=\frac{-4920+5,120}{25}\\ {\alpha }_{DE}=8{\text{rad/s}}^2\left(\text{Counterclockwise}\right)\end{array}$

Equate j component of the Equation (6).

$\begin{array}{l}\left(40{\alpha }_{BD}+4,080\right)=\left(20{\alpha }_{DE}+6,400\right)\\ 40{\alpha }_{BD}=20{\alpha }_{DE}+6,400-4,080\\ {\alpha }_{BD}=\left(\frac{20{\alpha }_{DE}+6,400-4,080}{40}\right)\end{array}$

Substitute $8{\text{rad/s}}^2$ for ${\alpha }_{DE}$.

$\begin{array}{c}{\alpha }_{BD}=\left(\frac{20\times 8+6,400-4,080}{40}\right)\\ =62{\text{rad/s}}^2\end{array}$

Thus, the angular acceleration of the bar BD is $\underset{\_}{62{\text{rad/s}}^2\left(\text{Counterlockwise}\right)}$.

To determine

Find the magnitude of angular acceleration of bar DE.

Answer to Problem 15.133P

The magnitude of the angular acceleration of bar DE is $\underset{\_}{8{\text{rad/s}}^2\left(\text{Counterclockwise}\right)}$.

Explanation of Solution

Given information:

Calculation:

Refer Part (a).

The angular acceleration of the bar DE is ${\alpha }_{DE}=8{\text{rad/s}}^2\left(\text{Counterclockwise}\right)$.

Thus, The magnitude of the angular acceleration of bar DE is $\underset{\_}{{\alpha }_{DE}=8{\text{rad/s}}^2\left(\text{Counterclockwise}\right)}$.