Chapter 15.6, Problem 15.219P

In Prob. 15.209, determine the acceleration of collar A.

To determine

Acceleration of collar A when $c=120mm$

Answer to Problem 15.219P

The acceleration $a_A$ of collar A is $-\left(8.764mm/s\right)j$.

Explanation of Solution

Given information:

Length of rod AB is $300mm$.

Constant velocity of collar B is $50mm/s$ towards point D.

The vector product of angular velocity $\omega$ and of the position vector r of the particle:

$v=\frac{dr}{dt}=\omega \times r$

The velocity $v_A$ of point A can be defined as

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

Acceleration of particle B can be defined as

$\begin{array}{l}{a}_B=a_A+a_{B/A}\\ a_B=a_A+\left(\alpha \times r_{B/A}+\omega \times \left(\omega \times r_{B/A}\right)\right)\end{array}$

Calculation:

The position vector $r_{D/C}$ of point D with respect to point C

$\begin{array}{l}{r}_{D/C}=r_D-r_C\\ =\left(90mm\right)i-\left(180mm\right)k\end{array}$

The length $l_{CD}$ of rod CD

$l_{CD}=\sqrt{{90}^2+{180}^2}=201.246mm$

The position vector $r_{B/C}$ of point B with respect to point C

$\begin{array}{l}{r}_{B/C}=\frac{c\left(r_{D/C}\right)}{180}\\ =\frac{120\left(90i-180k\right)}{180}\\ =\left(60mm\right)i-\left(120mm\right)k\end{array}$

The position vector $r_B$ of point B

$\begin{array}{l}{r}_B=r_C+r_{B/C}\\ =\left(180mm\right)k+\left(60mm\right)i-\left(120mm\right)k\\ =\left(60mm\right)i+\left(60mm\right)k\end{array}$

The position vector $r_{A/B}$ of point A with respect to point B

$\begin{array}{l}{r}_{A/B}=r_A-r_B\\ =-\left(60mm\right)i+yj-\left(60mm\right)k\end{array}$

The length of AB

$l_{AB}{}^2=x_{AB}{}^2+y_{AB}{}^2+z_{AB}{}^2$

Substitute

$\begin{array}{l}{300}^2={\left(60\right)}^2+y_{AB}{}^2+{\left(60\right)}^2\\ y_{AB}=287.75mm\end{array}$

Therefore

$r_{A/B}=-\left(60mm\right)i+\left(287.75mm\right)j-\left(60mm\right)k$

Velocity $v_B$ of point B

$\begin{array}{l}{v}_B=\left(50\right)\frac{r_D{}_{/C}}{l_{CD}}\\ =50\left(\frac{90i-180k}{201.246}\right)\\ =\left(22.3607mm/s\right)i-\left(44.7214mm/s\right)k\end{array}$

Velocity $v_A$ of point A

$\begin{array}{l}{v}_A=v_{A}j\\ v_A=v_B+v_{A/B}\to \left(1\right)\end{array}$

Where

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

$v_{A/B}$ is perpendicular to $r_{A/B}$.

Therefore

$r_{A/B}.v_{A/B}=0$

Multiply equation 1 by $r_A{{}_{/}}_B$.

$\begin{array}{l}{r}_A{{}_{/}}_B.v_A=r_A{{}_{/}}_B\cdot (v_B+v_A{{}_{/}}_B)\\ r_A{{}_{/}}_B.v_A=r_A{{}_{/}}_B.v_B+r_A{{}_{/}}_B.v_A{{}_{/}}_B\\ r_A{{}_{/}}_B.v_A=r_A{{}_{/}}_B.v_B\end{array}$

Substitute and solve

$\begin{array}{l}\left(-60i+287.75j-60k\right).\left(v_{A}j\right)=\left(-60i+287.75j-60k\right).\left(22.3607i-44.7214k\right)\\ 287.75v_A=-1341.642+2683.284\\ v_A=\left(4.6625mm/s\right)j\end{array}$

The relative velocity $v_{A/B}$ of point A with respect to point B

$\begin{array}{l}{v}_{A/B}=v_A-v_B\\ =\left(4.6625mm/s\right)j-\left(22.3607mm/s\right)i+\left(44.7214mm/s\right)k\\ v_{A/B}{}^2={\left(4.6625\right)}^2+{\left(22.3607\right)}^2+{\left(44.7214\right)}^2\\ =2521.74\end{array}$

Acceleration $a_B$ of collar B

$a_B=0$

Acceleration $a_A$ of collar A

$a_A=a_{A}j$

$a_A=a_B+a_{A/B}\to \left(1\right)$

The relative acceleration $a_{A/B}$ of point A with respect to point B

$a_{A/B}={\alpha }_{AB}\times r_{A/B}+{\omega }_{AB}\times v_{A/B}$

${\alpha }_{AB}\times r_{A/B}$ is perpendicular to $r_{A/B}$.

Therefore

$r_{A/B}.\left({\alpha }_{AB}\times r_{A/B}\right)=0$

Hence

$\begin{array}{l}{r}_{A/B}.\left({\alpha }_{AB}\times r_{A/B}\right)=-r_{A/B}.\left({\omega }_{AB}\times v_{A/B}\right)\\ 0=-v_{A/B}{}^2\\ r_{A/B}.a_{A/B}=-v_{A/B}{}^2=0\end{array}$

Multiply equation 1 by $r_{A/B}$

$\begin{array}{l}{r}_{A/B}.a_A=r_{A/B}.a_B-v_{A/B}{}^2\\ \left(-60i+287.75j-60k\right).\left(a_{A}j\right)=0-2521.74\end{array}$

Solve

$\begin{array}{l}287.75a_A=-2521.74\\ a_A=-\left(8.764mm/s\right)j\end{array}$

Conclusion:

The acceleration $a_A$ of collar A is $-\left(8.764mm/s\right)j$.