Chapter 15, Problem 15.252RP

Knowing that at the instant shown bar AB has an angular velocity of 10 rad/s clockwise and it is slowing down at a rate of 2 rad/s², determine the angular accelerations of bar BD and bar DE.

    

To determine

The angular acceleration of bar $BD$.

The angular acceleration of bar $DE$.

Answer to Problem 15.252RP

The angular acceleration of bar $BD$ is $306\text{rad}/{\text{s}}^{\text{2}}$counter clockwise.

The angular acceleration of bar $DE$ is $737\text{rad}/{\text{s}}^{\text{2}}$counter clockwise.

Explanation of Solution

Given information:

The angular velocity of bar $AB$ is $10\text{rad}/\text{s}$ and angular acceleration of bar $AB$ is $2\text{rad}/{\text{s}}^2$

Figure-(1) represents the geometry of mechanism.

Figure-(1)

Figure-(2) represents the instantaneous center of rotation $C$.

Figure-(2)

The instantaneous centre of rotation is the point of intersection perpendiculars drawn from the velocity at point $B$ and at point $D$ on the bar $BD$

Write the expression for velocity at point $B$.

$v_B=\left(l_{AB}\right){\omega }_{AB}$ ...... (I)

Here, angular velocity of bar $AB$ is ${\omega }_{AB}$ and length of bar $AB$ is $l_{AB}$.

Write the expression for angular velocity of bar $BD$.

${\omega }_{BD}=\frac{v_B}{BC}$ ...... (II)

Here, the distance between point $B$ and $C$ is $BC$.

Write the expression for velocity at point $D$.

$v_D=\left(DC\right){\omega }_{BD}$ ...... (III)

Here, distance between point $D$ and instantaneous centre is $DC$.

Write the expression for angular velocity of bar $DE$.

${\omega }_{DE}=\frac{v_D}{DE}$ ...... (IV)

Here, the distance between point $D$ and $E$ is $DE$.

Write the expression for position vector of point $B$ with respect to point $A$.

${\overrightarrow{r}}_{B/A}={\overrightarrow{r}}_B-{\overrightarrow{r}}_A$ ...... (V)

Here, position vector of point $B$ is ${\overrightarrow{r}}_B$ and position vector of point $A$ is ${\overrightarrow{r}}_A$.

Write the expression for position vector of point $D$ with respect to point $B$.

${\overrightarrow{r}}_{D/B}={\overrightarrow{r}}_D-{\overrightarrow{r}}_B$ ...... (VI)

Here, position vector of point $D$ is ${\overrightarrow{r}}_D$.

Write the expression for position vector of point $D$ with respect to point $E$.

${\overrightarrow{r}}_{D/E}={\overrightarrow{r}}_D-{\overrightarrow{r}}_E$ ...... (VII)

Write the expression for acceleration of point $B$.

${\overrightarrow{a}}_B={\overrightarrow{a}}_A+{\overrightarrow{\alpha }}_{AB}\times {\overrightarrow{r}}_{B/A}-{\omega }_{AB}^2{\overrightarrow{r}}_{B/A}$ ...... (VIII)

Here, acceleration of point $A$ is $a_A$, angular acceleration of bar $AB$ is ${\alpha }_{AB}$, angular velocity ob bar $AB$ is ${\omega }_{AB}$.

Write the expression for acceleration of point $D$ for point $D$ lie on bar $BD$.

${\overrightarrow{a}}_D={\overrightarrow{a}}_B+{\overrightarrow{\alpha }}_{BD}\times {\overrightarrow{r}}_{D/B}-{\omega }_{BD}^2{\overrightarrow{r}}_{D/B}$ ...... (IX)

Here, angular acceleration of bar $BD$ is ${\alpha }_{BD}$.

Write the expression for acceleration of point $D$ for point $D$ lie on bar $DE$.

${\overrightarrow{a}}_D={\overrightarrow{a}}_E+{\overrightarrow{\alpha }}_{DE}\times {\overrightarrow{r}}_{D/E}-{\omega }_{DE}^2{\overrightarrow{r}}_{D/E}$ ...... (X)

Here, angular acceleration of bar $DE$ is ${\alpha }_{DE}$.

Calculation:

Consider point $A$ as origin, the coordinate of point $B$ is $\left(0,-0.2\text{m}\right)$, coordinate of point $D$ is $\left(-0.6\text{m},-0.45\text{m}\right)$, coordinate of point $E$ is $\left(-0.4\text{m},-0.45\text{m}\right)$.

Consider clockwise direction as negative and counter clockwise direction as positive.

Substitute $0.2\text{m}$ for $l_{AB}$ and $10\text{rad}/\text{s}$ for ${\omega }_{AB}$ in Equation (I).

$\begin{array}{c}{v}_B=\left(0.2\text{m}\right)10\text{rad}/\text{s}\\ =2\text{m}/\text{s}\end{array}$

Substitute $2\text{m}/\text{s}$ for $v_B$ and $0.25\text{m}$ for $BC$ in Equation (II).

$\begin{array}{c}{\omega }_{BD}=\frac{2\text{m}/\text{s}}{0.25\text{m}}\\ =8\text{rad}/\text{s}\end{array}$

Substitute $8\text{rad}/\text{s}$ for ${\omega }_{BD}$ and $0.6\text{m}$ for $DC$ in Equation (III).

$\begin{array}{c}{v}_D=\left(0.6\text{m}\right)8\text{rad}/\text{s}\\ =4.8\text{mm}/\text{s}\end{array}$

Substitute $4.8\text{mm}/\text{s}$ for $v_D$ and $0.2\text{m}$ for $DE$ in Equation (IV).

$\begin{array}{c}{\omega }_{DE}=\frac{4.8\text{mm}/\text{s}}{0.2\text{m}}\\ =24\text{rad}/\text{s}\end{array}$

Substitute $\left(0\text{m}\right)\text{i}-\left(0.2\text{m}\right)\text{j}$ for ${\overrightarrow{r}}_B$ and $0\text{i}+0\text{j}$ for ${\overrightarrow{r}}_A$ in Equation (V).

$\begin{array}{c}{\overrightarrow{r}}_{B/A}=\left(0\text{m}\right)\text{i}-\left(0.2\text{m}\right)\text{j}-0\text{i}+0\text{j}\\ =-\left(0.2\text{m}\right)\text{j}\end{array}$

Substitute $0$ for ${\overrightarrow{a}}_A$, $\left(2\text{rad}/{\text{s}}^2\right)\text{k}$ for ${\alpha }_{AB}$, $-\left(0.2\text{m}\right)\text{j}$ for ${\overrightarrow{r}}_{B/A}$ and $10\text{rad}/\text{s}$ for ${\omega }_{AB}$ in Equation (VIII)

$\begin{array}{c}{\overrightarrow{a}}_B=0+\left(2\text{rad}/{\text{s}}^2\right)\text{k}\times \left(-0.2\text{m}\right)\text{j}-{\left(10\text{rad}/\text{s}\right)}^2\left(-\left(0.2\text{m}\right)\text{j}\right)\\ =0.4\text{m}/{\text{s}}^{\text{2}}\text{i}+20\text{m}/{\text{s}}^{\text{2}}\text{j}\end{array}$

Substitute $\left(-0.6\text{m}\right)\text{i}-\left(0.45\text{m}\right)\text{j}$ for ${\overrightarrow{r}}_D$ and $\left(0\text{m}\right)\text{i}-\left(0.2\text{m}\right)\text{j}$ for ${\overrightarrow{r}}_B$ in Equation (VI).

$\begin{array}{c}{\overrightarrow{r}}_{D/B}=\left(-0.6\text{m}\right)\text{i}-\left(0.45\text{m}\right)\text{j}-\left\{\left(0\text{m}\right)\text{i}-\left(0.2\text{m}\right)\text{j}\right\}\\ =\left(-0.6\text{m}\right)\text{i}-\left(0.25\text{m}\right)\text{j}\end{array}$

Substitute $0.4\text{m}/{\text{s}}^{\text{2}}\text{i}+20\text{m}/{\text{s}}^{\text{2}}\text{j}$ for ${\overrightarrow{a}}_B$, ${\alpha }_{BD}\text{k}$ for ${\alpha }_{BD}$, $-\left(0.6\text{m}\right)\text{i}-\left(0.25\text{m}\right)\text{j}$ for ${\overrightarrow{r}}_{D/B}$ and $8\text{rad}/\text{s}$ for ${\omega }_{BD}$ .in Equation (IX).

$\begin{array}{c}{\overrightarrow{a}}_D=\left[\begin{array}{l}0.4\text{m}/{\text{s}}^{\text{2}}\text{i}+20\text{m}/{\text{s}}^{\text{2}}\text{j}+\\ {\alpha }_{BD}\text{k}\times \left\{-\left(0.6\text{m}\right)\text{i}-\left(0.25\text{m}\right)\text{j}\right\}-\\ {\left(8\text{rad}/\text{s}\right)}^2\left\{-\left(0.6\text{m}\right)\text{i}-\left(0.25\text{m}\right)\text{j}\right\}\end{array}\right]\\ =\left[\begin{array}{c}0.4\text{m}/{\text{s}}^{\text{2}}\text{i}+20\text{m}/{\text{s}}^{\text{2}}\text{j}+\\ \left\{\left(0-\left(-0.25\text{m}{\alpha }_{BD}\right)\right)\text{i}-\left(0-\left(0.6\text{m}{\alpha }_{BD}\right)\right)\text{j}+0\text{k}\right\}\\ +38.4\text{mm}/{\text{s}}^{\text{2}}\text{i}+16\text{mm}/{\text{s}}^{\text{2}}\text{j}\end{array}\right]\\ =\left[\begin{array}{c}\left(38.8\text{mm}/{\text{s}}^{\text{2}}+0.25\text{m}{\alpha }_{BD}\right)\text{i}+\\ \left(36\text{mm}/{\text{s}}^{\text{2}}-\left(0.6\text{m}{\alpha }_{BD}\right)\right)\text{j}\end{array}\right]\end{array}$ ...... (XI)

Substitute $\left(-0.6\text{m}\right)\text{i}-\left(0.45\text{m}\right)\text{j}$ for ${\overrightarrow{r}}_D$ and $\left(-0.4\text{m}\right)\text{i}-\left(0.45\text{m}\right)\text{j}$ for ${\overrightarrow{r}}_E$ in Equation (VII).

$\begin{array}{c}{\overrightarrow{r}}_{D/E}=\left(-0.6\text{m}\right)\text{i}-\left(0.45\text{m}\right)\text{j}-\left\{\left(-0.4\text{m}\right)\text{i}-\left(0.45\text{m}\right)\text{j}\right\}\\ =\left(-0.2\text{m}\right)\text{i}\end{array}$

Substitute $0$ for ${\overrightarrow{a}}_E$, $-\left(0.2\text{m}\right)\text{i}$ for ${\overrightarrow{r}}_{D/E}$, $24\text{rad}/\text{s}$ for ${\omega }_{DE}$ in Equation (X).

$\begin{array}{c}{\overrightarrow{a}}_D=0+{\alpha }_{DE}\text{k}\times \left\{-\left(0.2\text{m}\right)\text{i}\right\}-{\left(24\text{rad}/\text{s}\right)}^2\left\{-\left(0.2\text{m}\right)\text{i}\right\}\\ =-\left(0-\left(-0.2\text{m}\times {\alpha }_{DE}\right)\right)\text{j}+115.2\text{mm}/{\text{s}}^{\text{2}}\text{i}\\ \text{=}115.2\text{i}-0.2\text{m}\times {\alpha }_{DE}\text{j}\end{array}$ ...... (XII)

Substitute $115.2$ for coefficient of $\text{i}$ in Equation (XI).

$\begin{array}{l}38.8\text{m}/{\text{s}}^{\text{2}}+0.25\text{m}{\alpha }_{BD}=115.2\text{m}/{\text{s}}^{\text{2}}\\ 0.25\text{m}{\alpha }_{BD}=115.2\text{m}/{\text{s}}^{\text{2}}-38.8\text{m}/{\text{s}}^{\text{2}}\\ {\alpha }_{BD}=\frac{115.2\text{m}/{\text{s}}^{\text{2}}-38.8\text{m}/{\text{s}}^{\text{2}}}{0.25\text{m}}\\ {\alpha }_{BD}=306\text{rad}/{\text{s}}^{\text{2}}\end{array}$

Substitute $0.2\text{m}\times {\alpha }_{DE}$ for the coefficient of $\text{j}$ in Equation (XI).

$\begin{array}{l}36\text{m}/{\text{s}}^{\text{2}}-\left(0.6\text{m}\times 306\text{rad}/{\text{s}}^{\text{2}}\right)=0.2\text{m}\times {\alpha }_{DE}\\ {\alpha }_{DE}=\frac{36\text{m}/{\text{s}}^{\text{2}}-\left(0.6\text{m}\times 306\text{rad}/{\text{s}}^{\text{2}}\right)}{0.2\text{m}}\\ {\alpha }_{DE}=737\text{rad}/{\text{s}}^{\text{2}}\end{array}$

Here, angular acceleration is positive, so direction of angular acceleration is counter clockwise.

Conclusion:

The angular acceleration of bar $BD$ is $306\text{rad}/{\text{s}}^{\text{2}}$counter clockwise.

The angular acceleration of bar $DE$ is $737\text{rad}/{\text{s}}^{\text{2}}$counter clockwise.