Chapter 15.1, Problem 15.11P

In Prob. 15.10, determine the velocity and acceleration of corner B, assuming that the angular velocity is 9 rad/s and increases at the rate of 45 rad/s².

To determine

The velocity and acceleration of corner3 $B$.

Answer to Problem 15.11P

The velocity of corner $B$ is $\left(0.75\text{m}/\text{s}\right)\text{i}+\left(1.5\text{m}/\text{s}\right)\text{k}$.

The acceleration of corner $B$ is $12.75\text{i+11}\text{.25j+3k}$.

Explanation of Solution

Given information:

The angular velocity is $9\text{rad}/\text{s}$ and increase at the rate of $45{\text{rad}/\text{s}}^2$.

Expression of position vector of point $B$ with respect to point $E$.

$r_{BE}=\left(p\right)\text{i}+\left(q\right)\text{j+}\left(r\right)k$ ...... (I)

Here, the distance in $x$ direction is $p$, the direction in $y$ distance is $q$ and the distance in $z$ direction is $r$, the unit vector of $x$ direction is $\text{i}$, the unit vector of $y$ direction is $\text{j}$ and the unit vector of $z$ direction is $\text{k}$.

Expression of position vector of point $A$ with respect to point $E$.

$r_{AE}=\left(a\right)\text{i}+\left(b\right)\text{j}+\left(c\right)\text{k}$ ...... (II)

Here, the distance in $x$ direction is $a$, the direction in $y$ distance is $b$ and the distance in $z$ direction is $c$, the unit vector of $x$ direction is $\text{i}$, the unit vector of $y$ direction is $\text{j}$ and the unit vector of $z$ direction is $\text{k}$.

Expression of rotation vector along the line $AE$ and in clockwise direction when seen from point $E$.

$\omega ={\omega }_{BE}{\lambda }_{AE}$ ...... (III)

Here, the angular velocity is ${\omega }_{BE}$ and the unit vector is ${\lambda }_{AE}$.

Expression of unit vector of point $A$ along with point $E$

${\lambda }_{AE}=\frac{r_{AE}}{AE}$ ...... (IV)

Here, the magnitude of position vector of point $A$ along with point $C$ is $AE$.

Expression of magnitude of position vector of point $A$ along with point $C$

$AE=\sqrt{a^2+b^2+c^2}$ ...... (V)

Here, the distance in $x$ direction is $a$, the direction in $y$ distance is $b$ and the distance in $z$ direction is $c$.

Expression of velocity of point $B$.

$V_B=\omega \times r_{BE}$ ...... (VI)

Here, the of rotation vector along the line $AE$ is $\omega$ and the position vector of point $B$ with respect to point $E$ is $r_{BE}$.

Expression of angular acceleration vector

$\alpha ={\alpha }^{\prime }\times {\lambda }_{AE}$ ...... (VII)

Here, the angular acceleration is ${\alpha }^{\prime }$ and the unit vector is ${\lambda }_{AE}$.

Expression of acceleration of corner $B$.

$a_B=\alpha \times r_{BE}+\omega \times V_B$ ...... (VIII)

Calculation:

Substitute $-400\text{mm}$ for $p$, $150\text{mm}$ for $q$ and $200\text{mm}$ for $r$ in Equation (I).

$\begin{array}{c}{r}_{BE}=\left(-400\text{mm}\right)\text{i}+\left(150\text{mm}\right)\text{j+}\left(200\text{mm}\right)k\\ =\left[\begin{array}{l}-\left\{\left(400\text{mm}\right)\times \left(\frac{1\text{m}}{1000\text{mm}}\right)\right\}\text{i}+\left\{\left(150\text{mm}\right)\times \left(\frac{1\text{m}}{1000\text{mm}}\right)\right\}\text{j}\\ \text{+}\left\{\left(200\text{mm}\right)\times \left(\frac{1\text{m}}{1000\text{mm}}\right)\right\}\text{k}\end{array}\right]\\ \text{=}-\left(0.4\text{m}\right)\text{i}+\left(0.15\text{m}\right)\text{j+}\left(0.2\text{m}\right)\text{k}\end{array}$

Substitute $-400\text{mm}$ for $a$, $400\text{mm}$ for $b$ and $200\text{mm}$ for $c$ in Equation (II).

$\begin{array}{c}{r}_{EA}=\left(-400\text{mm}\right)\text{i}+\left(400\text{mm}\right)\text{j}+\left(200\text{mm}\right)\text{k}\\ =\left[\begin{array}{l}-\left\{\left(400\text{mm}\right)\times \left(\frac{1\text{m}}{1000\text{mm}}\right)\right\}\text{i}+\left\{\left(400\text{mm}\right)\times \left(\frac{1\text{m}}{1000\text{mm}}\right)\right\}\text{j}\\ +\left\{\left(200\text{mm}\right)\times \left(\frac{1\text{m}}{1000\text{mm}}\right)\right\}\text{k}\end{array}\right]\\ \text{=}-\left(0.4\text{m}\right)\text{i}+\left(0.4\text{m}\right)\text{j}+\left(0.2\text{m}\right)\text{k}\end{array}$

Substitute $-400\text{mm}$ for $a$, $400\text{mm}$ for $b$ and $200\text{mm}$ for $c$ in Equation (V).

$\begin{array}{c}AE=\sqrt{{\left(-400\text{mm}\right)}^2+{\left(400\text{mm}\right)}^2+{\left(200\text{mm}\right)}^2}\\ =\left[\sqrt{\begin{array}{l}{\left\{\left(-400\text{mm}\right)\times \left(\frac{1\text{m}}{1000\text{mm}}\right)\right\}}^2+{\left\{\left(400\text{mm}\right)\times \left(\frac{1\text{m}}{1000\text{mm}}\right)\right\}}^2\\ +{\left\{\left(200\text{mm}\right)\times \left(\frac{1\text{m}}{1000\text{mm}}\right)\right\}}^2\end{array}}\right]\\ =\sqrt{{\left(-0.4\text{m}\right)}^2+{\left(0.4\text{m}\right)}^2+{\left(0.2\text{m}\right)}^2}\\ =0.6\text{m}\end{array}$

Substitute $0.6\text{m}$ for $AE$ and $-\left(0.4\text{m}\right)\text{i}+\left(0.4\text{m}\right)\text{j}+\left(0.2\text{m}\right)\text{k}$ for $r_{EA}$ in Equation (IV).

$\begin{array}{c}{\lambda }_{AE}=\frac{\left(-\left(0.4\text{m}\right)\text{i}+\left(0.4\text{m}\right)\text{j}+\left(0.2\text{m}\right)\text{k}\right)}{0.6\text{m}}\\ =\frac{\left(\frac{2\text{m}}{10}\right)\left(-2\text{i}+2\text{j}+\text{k}\right)}{\left(\frac{6\text{m}}{10}\right)}\\ =\left(\frac{2\text{m}}{10}\right)\left(-2\text{i}+2\text{j}+\text{k}\right)\left(\frac{10}{6\text{m}}\right)\\ =\frac{1}{3}\left(-2\text{i}+2\text{j}+\text{k}\right)\end{array}$

Substitute $\frac{1}{3}\left(-2\text{i}+2\text{j}+\text{k}\right)$ for ${\lambda }_{AE}$ and $9\text{rad}/\text{s}$ for ${\omega }_{AE}$ in Equation (III).

$\begin{array}{c}\omega =\left(9\text{rad}/\text{s}\right)\left\{\frac{1}{3}\left(-2\text{i}+2\text{j}+\text{k}\right)\right\}\\ =\left(3\text{rad}/\text{s}\right)\left(-2\text{i}+2\text{j}+\text{k}\right)\\ =-\left(6\text{rad}/\text{s}\right)\text{i}+\left(6\text{rad}/\text{s}\right)\text{j}+\left(3\text{rad}/\text{s}\right)\text{k}\end{array}$

Substitute $-\left(6\text{rad}/\text{s}\right)\text{i}+\left(6\text{rad}/\text{s}\right)\text{j}+\left(3\text{rad}/\text{s}\right)\text{k}$ for $\omega$ and $-\left(0.4\text{m}\right)\text{i}+\left(0.15\text{m}\right)\text{j+}\left(0.2\text{m}\right)\text{k}$ for $r_{BE}$ in Equation (VI).

$V_B=\left[\begin{array}{l}\left\{-\left(6\text{rad}/\text{s}\right)\text{i}+\left(6\text{rad}/\text{s}\right)\text{j}+\left(3\text{rad}/\text{s}\right)\text{k}\right\}\times \\ \left\{-\left(0.4\text{m}\right)\text{i}+\left(0.15\text{m}\right)\text{j+}\left(0.2\text{m}\right)\text{k}\right\}\end{array}\right]$ ...... (IX)

Here, the Equation (IX) is a cross product of two vector now change it in determinant form.

$\begin{array}{c}{V}_B=\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ -6\text{rad}/\text{s}& 6\text{rad}/\text{s}& 3\text{rad}/\text{s}\\ -0.4\text{m}& 0.15\text{m}& 0.2\text{m}\end{array}\right|\\ =\left[\begin{array}{l}\left(\left(6\times 0.2\text{rad}\cdot \text{m}/\text{s}\right)-\left(3\times 0.15\text{rad}\cdot \text{m}/\text{s}\right)\right)\text{i}-\left(\left(-6\times 0.2\text{rad}\cdot \text{m}/\text{s}\right)-\left(-0.4\times 3\text{rad}\cdot \text{m}/\text{s}\right)\right)\text{j}\\ -\left(\left(-6\times 0.15\text{rad}\cdot \text{m}/\text{s}\right)-\left(-0.4\times 6\text{rad}\cdot \text{m}/\text{s}\right)\right)\text{k}\end{array}\right]\\ =\left(0.75\text{rad}\cdot \text{m}/\text{s}\right)\text{i}-\left(0\text{rad}\cdot \text{m}/\text{s}\right)\text{j}+\left(1.5\text{rad}\cdot \text{m}/\text{s}\right)\text{k}\\ =\left(0.75\text{m}/\text{s}\right)\text{i}+\left(1.5\text{m}/\text{s}\right)\text{k}\end{array}$

Substitute $45{\text{rad}/\text{s}}^2$ for ${\alpha }^{\prime }$ and $\frac{1}{3}\left(-2\text{i}+2\text{j}+\text{k}\right)$ for ${\lambda }_{AE}$ in Equation (VII).

$\begin{array}{c}\alpha =\left(45{\text{rad}/\text{s}}^2\right)\times \left(\frac{1}{3}\left(-2\text{i}+2\text{j}+\text{k}\right)\right)\\ =\left(15{\text{rad}/\text{s}}^2\right)\left(-2\text{i}+2\text{j}+\text{k}\right)\\ =-\left(30{\text{rad}/\text{s}}^2\right)\text{i}+\left(30{\text{rad}/\text{s}}^2\right)\text{j}+\left(15\right)\text{k}\end{array}$

Substitute $-\left(30{\text{rad}/\text{s}}^2\right)\text{i}+\left(30{\text{rad}/\text{s}}^2\right)\text{j}+\left(15\right)\text{k}$ for $\alpha$, $-\left(0.4\text{m}\right)\text{i}+\left(0.15\text{m}\right)\text{j+}\left(0.2\text{m}\right)\text{k}$ for $r_{BE}$, $-\left(6\text{rad}/\text{s}\right)\text{i}+\left(6\text{rad}/\text{s}\right)\text{j}+\left(3\text{rad}/\text{s}\right)\text{k}$ for $\omega$ and $\left(0.75\text{m}/\text{s}\right)\text{i}+\left(1.5\text{m}/\text{s}\right)\text{k}$ for $V_B$ in Equation (VIII).

$a_B=\left[\begin{array}{l}\left\{-\left(30{\text{rad}/\text{s}}^2\right)\text{i}+\left(30{\text{rad}/\text{s}}^2\right)\text{j}+\left(15\right)\text{k}\right\}\times \left\{-\left(0.4\text{m}\right)\text{i}+\left(0.15\text{m}\right)\text{j+}\left(0.2\text{m}\right)\text{k}\right\}\\ +\left\{-\left(6\text{rad}/\text{s}\right)\text{i}+\left(6\text{rad}/\text{s}\right)\text{j}+\left(3\text{rad}/\text{s}\right)\text{k}\right\}\times \left\{\left(0.75\text{m}/\text{s}\right)\text{i}+\left(1.5\text{m}/\text{s}\right)\text{k}\right\}\end{array}\right]$

Here, the cross product of two vector, now change it in determinant form.

${\text{a}}_{\text{B}}=\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ -\text{30}{\text{rad}/\text{s}}^{\text{2}}& \text{30}{\text{rad}/\text{s}}^{\text{2}}& \text{15}{\text{rad}/\text{s}}^{\text{2}}\\ -\text{0}\text{.4}\text{m}& \text{0}\text{.15}\text{m}& \text{0}\text{.2}\text{m}\end{array}\right|+\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ -\text{6}\text{rad}\cdot \text{m}/\text{s}& \text{6}\text{rad}\cdot \text{m}/\text{s}& \text{3}\text{rad}\cdot \text{m}/\text{s}\\ \text{0}\text{.75}\text{rad}/\text{s}& \text{0}\text{rad}/\text{s}& \text{1}\text{.5}\text{rad}/\text{s}\end{array}\right|$

$=\left[\begin{array}{l}\left\{\left(6{\text{rad}\cdot \text{m}/\text{s}}^{\text{2}}\right)-\left(2.25{\text{rad}\cdot \text{m}/\text{s}}^{\text{2}}\right)\right\}\text{i}-\left\{\left(-6{\text{rad}\cdot \text{m}/\text{s}}^{\text{2}}\right)+\left(6{\text{rad}\cdot \text{m}/\text{s}}^{\text{2}}\right)\right\}\text{j}\\ -\left\{\left(-4.5{\text{rad}\cdot \text{m}/\text{s}}^{\text{2}}\right)+\left(12{\text{rad}\cdot \text{m}/\text{s}}^{\text{2}}\right)\right\}\text{k+}\left\{\left(9{\text{rad}\cdot \text{m}/\text{s}}^{\text{2}}\right)-\left(0{\text{rad}\cdot \text{m}/\text{s}}^{\text{2}}\right)\right\}\text{i}\\ -\left\{\left(-9{\text{rad}\cdot \text{m}/\text{s}}^{\text{2}}\right)-\left(2.25{\text{rad}\cdot \text{m}/\text{s}}^{\text{2}}\right)\right\}\text{j}-\left\{\left(0{\text{rad}\cdot \text{m}/\text{s}}^{\text{2}}\right)-\left(4.5{\text{rad}\cdot \text{m}/\text{s}}^{\text{2}}\right)\right\}\text{k}\end{array}\right]$

$=\left(3.75{\text{m}/\text{s}}^{\text{2}}\right)\text{i}+\left(7.5{\text{m}/\text{s}}^{\text{2}}\right)\text{k}+\left(9{\text{m}/\text{s}}^{\text{2}}\right)\text{i}-\left(11.25{\text{m}/\text{s}}^{\text{2}}\right)\text{j}-\left(4.5{\text{m}/\text{s}}^{\text{2}}\right)\text{k}$

$=\left(12.75{\text{m}/\text{s}}^{\text{2}}\right)\text{i+}\left(\text{11}\text{.25}{\text{m}/\text{s}}^{\text{2}}\right)\text{j+}\left(\text{3}{\text{m}/\text{s}}^{\text{2}}\right)\text{k}$

Conclusion:

The velocity of corner $B$ is $\left(0.75\text{m}/\text{s}\right)\text{i}+\left(1.5\text{m}/\text{s}\right)\text{k}$.

The acceleration of corner $B$ is $\left(12.75{\text{m}/\text{s}}^{\text{2}}\right)\text{i+}\left(\text{11}\text{.25}{\text{m}/\text{s}}^{\text{2}}\right)\text{j+}\left(\text{3}{\text{m}/\text{s}}^{\text{2}}\right)\text{k}$.