Chapter 15.4, Problem 15.134P

To determine

(a)

Angular acceleration of bar BD.

Answer to Problem 15.134P

The angular acceleration of bar BD is $25.162rad/s^2$ clockwise.

Explanation of Solution

Given information:

Angular velocity of bar AB is $10rad/s$ clockwise.

Angular acceleration of bar AB is $4rad/s^2$ clockwise.

The absolute value of point A

$v_A=v_B+v_{A/B}$

The relative velocity of A with respect to B is defined as

$v_{A/B}=\omega k\times r_{A/B}$

$r_{A/B}$ - Position vector of A with respect to B

$\omega$ - Angular velocity

The absolute acceleration of point B is defined as:

$a_B=a_A+a_{B/A}$

$a_B=a_A+{\left(a_{B/A}\right)}_t+{\left(a_{B/A}\right)}_n$

The tangential acceleration is defined as:

${\left(a_{B/A}\right)}_t=\alpha k\times r_{B/A}$

The normal acceleration is defined as:

${\left(a_{B/A}\right)}_n=-r_{B/A}{\omega }_{AB}{}^2$

In above equations

$\alpha$ - Angular acceleration

$\omega$ - Angular velocity

$r$ - Distance from A to B

Calculation:

Position vector of B relative to A:

$r_{B/A}=-175i$

Position vector of D relative to B:

$r_{D/B}=-200j$

Position vector of D relative to E:

$r_{D/E}=-275i+75j$

For bar AB

Velocity of point B

$\begin{array}{l}{v}_B={\omega }_{AB}\times r_{B/A}=-\left(10k\right)\times \left(-175i\right)\\ v_B=\left(1750mm\right)j\end{array}$

For bar BD:

Velocity of point D:

$\begin{array}{l}{v}_D=v_B+{\omega }_{BD}\times r_{D/B}\\ v_D=\left(1750mm\right)j+{\omega }_{BD}k\times \left(-200j\right)\\ v_D=\left(1750mm\right)j+\left(200{\omega }_{BD}\right)i\to \left(1\right)\end{array}$

For bar DE:

Velocity of point D:

$\begin{array}{l}{v}_D=v_E+{\omega }_{ED}\times r_{D/E}\\ v_D=0+{\omega }_{ED}k\times \left(-275i+75j\right)\\ v_D=-275{\omega }_{ED}j-75{\omega }_{ED}i\to \left(2\right)\end{array}$

Equate above equations:

$\left(1750mm\right)j+\left(200{\omega }_{BD}\right)i=-275{\omega }_{ED}j-75{\omega }_{ED}i$

Equate components:

$\begin{array}{l}i:200{\omega }_{BD}=-75{\omega }_{ED}\\ j:1750=-275{\omega }_{ED}\end{array}$

Therefore:

$\begin{array}{l}{\omega }_{ED}=-6.364rad/s\\ {\omega }_{BD}=2.3865rad/s\end{array}$

For bar AB:

Acceleration of point B:

$a_B=a_A+{\alpha }_{AB}k\times r_{B/A}-{\omega }_{AB}{}^2{r}_{B/A}$

We know that:

$a_A=0$

Therefore:

$\begin{array}{l}{a}_B={\alpha }_{AB}k\times r_{B/A}-{\omega }_{AB}{}^2{r}_{B/A}\\ a_B=\left(-4k\right)\times \left(-175i\right)-{\left(10rad/s\right)}^2\left(-175i\right)\\ a_B=17500i+700j\end{array}$

For bar BD:

Acceleration of point D:

$a_D=a_B+{\alpha }_{BD}k\times r_{D/B}-{\omega }_{BD}{}^2{r}_{D/B}$

Substitute:

$\begin{array}{l}{a}_D=17500i+700j+{\alpha }_{BD}k\times \left(-200j\right)-{\left(2.3865\right)}^2\left(-200j\right)\\ a_D=17500i+700j+200{\alpha }_{BD}i+1139.076j\end{array}$

Therefore:

$a_D=\left(17500+200{\alpha }_{BD}\right)i+\left(700+1139.076\right)j$

For bar DE:

Acceleration of point D:

$a_D=a_E+{\alpha }_{DE}k\times r_{D/E}-{\omega }_{DE}{}^2{r}_{D/E}$

Substitute:

$\begin{array}{l}{a}_D=0+{\alpha }_{DE}k\times \left(-275i+75j\right)-{\left(-6.364\right)}^2\left(-275i+75j\right)\\ a_D=-275{\alpha }_{DE}j-75{\alpha }_{DE}i+11137.64i-3037.54j\end{array}$

Therefore:

$a_D=\left(-75{\alpha }_{DE}+11137.64\right)i-\left(275{\alpha }_{DE}+3037.54\right)j$

Equate above equations:

$\left(17500+200{\alpha }_{BD}\right)i+\left(7000+1139.076\right)j=\left(-75{\alpha }_{DE}+11137.64\right)i-\left(275{\alpha }_{DE}+3037.54\right)j$

Equate components:

$\begin{array}{l}i:\left(17500+200{\alpha }_{BD}\right)=\left(-75{\alpha }_{DE}+11137.64\right)\\ j:\left(7000+1139.076\right)=-\left(275{\alpha }_{DE}+3037.54\right)\end{array}$

Solve above equations:

$\begin{array}{l}{\alpha }_{DE}=-17.733rad/s^2\\ {\alpha }_{BD}=-25.162rad/s^2\end{array}$

Conclusion:

The angular acceleration of bar BD is $25.162rad/s^2$ clockwise.

To determine

(b)

Angular acceleration of bar DE

Answer to Problem 15.134P

The angular acceleration of bar DE is $17.733rad/s^2$ clockwise.

Explanation of Solution

Given information:

Angular velocity of bar AB is $10rad/s$ clockwise.

Angular acceleration of bar AB is $4rad/s^2$ clockwise.

The absolute value of point A

$v_A=v_B+v_{A/B}$

The relative velocity of A with respect to B is defined as

$v_{A/B}=\omega k\times r_{A/B}$

$r_{A/B}$ - Position vector of A with respect to B

$\omega$ - Angular velocity

The absolute acceleration of point B is defined as

$a_B=a_A+a_{B/A}$

$a_B=a_A+{\left(a_{B/A}\right)}_t+{\left(a_{B/A}\right)}_n$

The tangential acceleration is defined as

${\left(a_{B/A}\right)}_t=\alpha k\times r_{B/A}$

The normal acceleration is defined as

${\left(a_{B/A}\right)}_n=-r_{B/A}{\omega }_{AB}{}^2$

In above equations

$\alpha$ - Angular acceleration

$\omega$ - Angular velocity

$r$ - Distance from A to B

Calculation:

According to sub part a

Acceleration of point D for bar BD

$a_D=\left(17500+200{\alpha }_{BD}\right)i+\left(700+1139.076\right)j$

Acceleration of point D for bar DE

$a_D=\left(-75{\alpha }_{DE}+11137.64\right)i-\left(275{\alpha }_{DE}+3037.54\right)j$

Equate above equations

$\left(17500+200{\alpha }_{BD}\right)i+\left(700+1139.076\right)j=\left(-75{\alpha }_{DE}+11137.64\right)i-\left(275{\alpha }_{DE}+3037.54\right)j$

Equate components

$j:\left(700+1139.076\right)=-\left(275{\alpha }_{DE}+3037.54\right)$

Therefore

${\alpha }_{DE}=-17.733rad/s^2$

Conclusion:

The angular acceleration of bar DE is $17.733rad/s^2$ clockwise.