Chapter 15.4, Problem 15.132P

To determine

(a)

Angular acceleration of bar BD.

Answer to Problem 15.132P

The angular acceleration of bar BD is $62rad/s^2$ counter 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:

$r_{B/A}=-20i-40j$

Position vector of D relative to B:

$r_{D/B}=40i$

Position vector of D relative to E:

$r_{D/E}=20i-25j$

For bar AB:

Velocity of point B:

$\begin{array}{l}{v}_B={\omega }_{AB}\times r_{B/A}=-\left(10k\right)\times \left(-20i-40j\right)\\ v_B=-\left(400in/s\right)i+\left(200in/s\right)j\end{array}$

For bar BD:

Velocity of point D:

$\begin{array}{l}{v}_D=v_B+{\omega }_{BD}\times r_{D/B}\\ v_D=-\left(400in/s\right)i+\left(200in/s\right)j+{\omega }_{BD}k\times 40i\\ v_D=-\left(400in/s\right)i+\left(200in/s+40{\omega }_{BD}\right)j\to \left(1\right)\end{array}$

For bar DE:

Velocity of point D:

$\begin{array}{l}{v}_D=v_E+{\omega }_{ED}\times r_{D/E}\\ v_D=0+{\omega }_{ED}k\times \left(20i-25j\right)\\ v_D=25{\omega }_{ED}i+20{\omega }_{ED}j\to \left(2\right)\end{array}$

Equate above equations

$-\left(400in/s\right)i+\left(200in/s+40{\omega }_{BD}\right)j=25{\omega }_{ED}i+20{\omega }_{ED}j$

Equate components:

$\begin{array}{l}i:-400=25{\omega }_{ED}\\ j:200+40{\omega }_{BD}=20{\omega }_{ED}\end{array}$

Therefore:

$\begin{array}{l}{\omega }_{ED}=-16rad/s\\ {\omega }_{BD}=-13rad/s\end{array}$

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:

$\begin{array}{l}{a}_B={\alpha }_{AB}k\times r_{B/A}-{\omega }_{AB}{}^2{r}_{B/A}\\ a_B=-\left(4k\right)\times \left(-20i-40j\right)-{\left(10rad/s\right)}^2\left(-20i-40j\right)\\ a_B=80j-160i+2000i+4000j\\ a_B=1840i+4080j\end{array}$

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:

$\begin{array}{l}{a}_D=1840i+4080j+{\alpha }_{BD}k\times 40i-{\left(13\right)}^{2}40i\\ a_D=1840i+4080j+40{\alpha }_{BD}j-6760i\end{array}$

Therefore:

$a_D=-4920i+\left(4080+40{\alpha }_{BD}\right)j$

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:

$\begin{array}{l}{a}_D=0+{\alpha }_{DE}k\times \left(20i-25j\right)-{\left(16\right)}^2\left(20i-25j\right)\\ a_D=20{\alpha }_{DE}j+25{\alpha }_{DE}i-5120i+6400j\end{array}$

Therefore:

$a_D=\left(25{\alpha }_{DE}-5120\right)i+\left(20{\alpha }_{DE}+6400\right)j$

Equate above equations

$-4920i+\left(4080+40{\alpha }_{BD}\right)j=\left(25{\alpha }_{DE}-5120\right)i+\left(20{\alpha }_{DE}+6400\right)j$

Equate components:

$\begin{array}{l}i:-4920=25{\alpha }_{DE}-5120\\ j:4080+40{\alpha }_{BD}=20{\alpha }_{DE}+6400\end{array}$

Solve above equations:

$\begin{array}{l}{\alpha }_{DE}=8rad/s^2\\ {\alpha }_{BD}=62rad/s^2\end{array}$

Conclusion:

The angular acceleration of bar BD is $62rad/s^2$ counter clockwise.

To determine

(b)

Angular acceleration of bar DE

Answer to Problem 15.132P

The angular acceleration of bar DE is $8rad/s^2$ counter 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

$a_D=-4920i+\left(4080+40{\alpha }_{BD}\right)j$

Acceleration of point D for bar DE

$a_D=\left(25{\alpha }_{DE}-5120\right)i+\left(20{\alpha }_{DE}+6400\right)j$

Equate above equations

$-4920i+\left(4080+40{\alpha }_{BD}\right)j=\left(25{\alpha }_{DE}-5120\right)i+\left(20{\alpha }_{DE}+6400\right)j$

Equate components

$i:-4920=25{\alpha }_{DE}-5120$

Therefore

${\alpha }_{DE}=8rad/s^2$

Conclusion:

The angular acceleration of bar DE is $8rad/s^2$ counter clockwise.