Chapter 15.4, Problem 15.128P

The elliptical exercise machine has fixed axes of rotation at points A and E. Knowing that at the instant shown the flywheel AB has a constant angular velocity of 6 rad/s clockwise, determine (a) the angular acceleration of bar DEF, (b) the acceleration of point F.

Fig. P15.127 and P15.128

To determine

Find the angular acceleration of the bar DEF.

Answer to Problem 15.128P

The angular acceleration of the bar DEF is $\underset{\_}{{\alpha }_{DE}=1.47\text{rad/s}}$.

Explanation of Solution

Given information:

Consider the angular acceleration of the bar DEF is denoted by ${\omega }_{DE}$.

The constant angular acceleration of the flywheel AB is ${\omega }_{AB}=6\text{rad/s}\left(\text{Clockwise}\right)=-6\text{k}$.

The angular acceleration of the flywheel AB is ${\alpha }_{AB}=0$.

Consider the position of the point B with respect to point A is denoted by ${\text{r}}_{B/A}=0.2\text{j}$.

Consider the position of the point D with respect to point B is denoted by ${\text{r}}_{D/B}=1.2\text{i}-0.2\text{j}$.

Consider the position of the point D with respect to point E is denoted by ${\text{r}}_{D/E}=-0.12\text{i}-0.8\text{j}$.

The axis of rotation points A and E are fixed. Then,

The velocity at the point A and E are ${\text{v}}_A=0$ and ${\text{v}}_E=0$.

Show the relation between the velocity of the point B and A as follows:

$\begin{array}{c}{\text{v}}_B={\text{v}}_A+{\omega }_{AB}\text{k}\times {\text{r}}_{B/A}\\ =0+{\omega }_{AB}\text{k}\times {\text{r}}_{B/A}\\ ={\omega }_{AB}\text{k}\times {\text{r}}_{B/A}\end{array}$ (1)

Show the relation between the velocity of the point B and D as follows:

${\text{v}}_D={\text{v}}_B+{\omega }_{BD}\text{k}\times {\text{r}}_{D/B}$ (2)

Modify Equation (2) using Equation (1).

${\text{v}}_D={\omega }_{AB}\text{k}\times {\text{r}}_{B/A}+{\omega }_{BD}\text{k}\times {\text{r}}_{D/B}$ (3)

Show the relation between the velocity of the point E and D as follows:

$\begin{array}{l}{\text{v}}_D={\text{v}}_E+{\omega }_{DE}\text{k}\times {\text{r}}_{D/E}\\ {\text{v}}_D=0+{\omega }_{DE}\text{k}\times {\text{r}}_{D/E}\\ {\text{v}}_D={\omega }_{DE}\text{k}\times {\text{r}}_{D/E}\end{array}$ (4)

Equate Equation (3) and (4).

${\omega }_{DE}\text{k}\times {\text{r}}_{D/E}={\omega }_{AB}\text{k}\times {\text{r}}_{B/A}+{\omega }_{BD}\text{k}\times {\text{r}}_{D/B}$

Substitute $\left(-6\text{rad/s}\right)$ for ${\omega }_{AB}$, $0.2\text{j}$ for ${\text{r}}_{B/A}$, $\left(1.2\text{i}-0.2\text{j}\right)$ for ${\text{r}}_{D/B}$, and $\left(-0.12\text{i}-0.8\text{j}\right)$ for ${\text{r}}_{D/E}$.

$\begin{array}{l}{\omega }_{DE}\text{k}\times \left(-0.12\text{i}-0.8\text{j}\right)=-6\text{k}\times 0.2\text{j}+{\omega }_{BD}\text{k}\times \left(1.2\text{i}-0.2\text{j}\right)\\ -0.12{\omega }_{DE}\text{j}+0.8{\omega }_{DE}\text{i}=1.2\text{i}+1.2{\omega }_{BD}\text{j}+0.2{\omega }_{BD}\text{i}\end{array}$ (5)

Equate j component of the Equation (5).

$\begin{array}{l}-0.12{\omega }_{DE}=1.2{\omega }_{BD}\\ {\omega }_{DE}=-\frac{1.2{\omega }_{BD}}{0.12}\\ {\omega }_{DE}=-10{\omega }_{BD}\end{array}$ (6)

Equate i component of the Equation (5).

$\begin{array}{l}0.8{\omega }_{DE}=1.2+0.2{\omega }_{BD}\\ 0.2{\omega }_{BD}=0.8{\omega }_{DE}-1.2\end{array}$

Substitute $-\left(10{\omega }_{BD}\right)$ for ${\omega }_{DE}$.

$\begin{array}{l}0.2{\omega }_{BD}=0.8\left(-10{\omega }_{BD}\right)-1.2\\ 0.2{\omega }_{BD}=-8{\omega }_{BD}-1.2\\ 8.2{\omega }_{BD}=-1.2\\ {\omega }_{BD}=-\frac{1.2}{8.2}\end{array}$

${\omega }_{BD}=-0.1463\text{rad/s}$

Substitute $-0.1463\text{rad/s}$ for ${\omega }_{BD}$.

$\begin{array}{c}{\omega }_{DE}=-10\left(-0.1463\text{rad/s}\right)\\ =1.463\text{rad/s}\end{array}$

The axis of rotation points A and E are fixed. Then,

The acceleration at the point A and E are ${\text{a}}_A=0$ and ${\text{a}}_E=0$.

Show the relation between the acceleration of the point B and A as follows:

$\begin{array}{c}{\text{a}}_B={\text{a}}_A+\left(0\right)\times {\text{r}}_{B/A}-{\omega }_{AB}^2{\text{r}}_{B/A}\\ =-{\omega }_{AB}^2{\text{r}}_{B/A}\end{array}$ (7)

Show the relation between the acceleration of the point B and D as follows:

${\text{a}}_D={\text{a}}_B+{\alpha }_{BD}\text{k}\times {\text{r}}_{D/B}-{\omega }_{BD}^2{\text{r}}_{D/B}$ (8)

Modify Equation (2) using Equation (1).

${\text{a}}_D=-{\omega }_{AB}^2{\text{r}}_{B/A}+{\alpha }_{BD}\text{k}\times {\text{r}}_{D/B}-{\omega }_{BD}^2{\text{r}}_{D/B}$ (9)

Show the relation between the acceleration of the point E and D as follows:

$\begin{array}{c}{\text{a}}_D={\text{a}}_E+{\alpha }_{DE}\text{k}\times {\text{r}}_{D/E}-{\omega }_{DE}^2{\text{r}}_{D/E}\\ =0+{\alpha }_{DE}\text{k}\times {\text{r}}_{D/E}-{\omega }_{DE}^2{\text{r}}_{D/E}\\ ={\alpha }_{DE}\text{k}\times {\text{r}}_{D/E}-{\omega }_{DE}^2{\text{r}}_{D/E}\end{array}$ (10)

Equate Equation (9) and (10).

$-{\omega }_{AB}^2{\text{r}}_{B/A}+{\alpha }_{BD}\text{k}\times {\text{r}}_{D/B}-{\omega }_{BD}^2{\text{r}}_{D/B}={\alpha }_{DE}\text{k}\times {\text{r}}_{D/E}-{\omega }_{DE}^2{\text{r}}_{D/E}$

Substitute $\left(-6\text{rad/s}\right)$ for ${\omega }_{AB}$, $0.2\text{j}$ for ${\text{r}}_{B/A}$, $\left(1.2\text{i}-0.2\text{j}\right)$ for ${\text{r}}_{D/B}$, and $\left(-0.12\text{i}-0.8\text{j}\right)$ for ${\text{r}}_{D/E}$, $-0.1463\text{rad/s}$ for ${\omega }_{BD}$, and $1.463\text{rad/s}$ for ${\omega }_{DE}$.

$\begin{array}{l}\left\{\begin{array}{l}-{\left(-6\right)}^2\times \left(0.2\text{j}\right)\\ +{\alpha }_{BD}\text{k}\times \left(1.2\text{i}-0.2\text{j}\right)\\ -{\left(-0.1463\right)}^2\times \left(1.2\text{i}-0.2\text{j}\right)\end{array}\right\}=\left\{\begin{array}{l}{\alpha }_{DE}\text{k}\times \left(-0.12\text{i}-0.8\text{j}\right)\\ -{\left(1.463\right)}^2\times \left(-0.12\text{i}-0.8\text{j}\right)\end{array}\right\}\\ \left\{\begin{array}{l}1.2{\alpha }_{BD}\text{j+}0.2{\alpha }_{BD}\text{i}\\ -\text{0}\text{.02568i+}0.00428\text{j}\end{array}\right\}=\left\{\begin{array}{l}-0.12{\alpha }_{DE}\text{j+}0.8{\alpha }_{DE}\text{i}\\ +0.2568\text{i}+1.712\text{j}\end{array}\right\}\end{array}$ (11)

Equate j component of the Equation (5).

$\begin{array}{l}-7.2+1.2{\alpha }_{BD}\text{+}0.00428=-0.12{\alpha }_{DE}+1.712\\ 1.2{\alpha }_{BD}=-0.12{\alpha }_{DE}+8.907\\ {\alpha }_{BD}=\frac{-0.12{\alpha }_{DE}+8.907}{1.2}\\ {\alpha }_{BD}=-0.1{\alpha }_{DE}+7.4225\end{array}$ (12)

Equate i component of the Equation (5).

$\begin{array}{l}0.2{\alpha }_{BD}-\text{0}\text{.02568}=0.8{\alpha }_{DE}+0.2568\\ 0.2{\alpha }_{BD}=0.8{\alpha }_{DE}+0.28248\end{array}$

Substitute $\left(-0.1{\alpha }_{DE}+7.4225\right)$ for ${\alpha }_{BD}$.

$\begin{array}{l}0.2\left(-0.1{\alpha }_{DE}+7.4225\right)=0.8{\alpha }_{DE}+0.28248\\ -0.02{\alpha }_{DE}+1.4845=0.8{\alpha }_{DE}+0.2824\\ {\alpha }_{DE}=-1.2021\\ {\alpha }_{DE}=\frac{-1.2021}{-0.82}\end{array}$

$\begin{array}{c}{\alpha }_{DE}=1.466\text{rad/s}\\ \text{=1}\text{.47}\text{rad/s}\end{array}$

Thus, the angular acceleration of the rod DEF is $\underset{\_}{{\alpha }_{DE}=1.47\text{rad/s}}$.

To determine

Find the acceleration at point F.

Answer to Problem 15.128P

The acceleration at F is $\underset{\_}{1.575{\text{m/s}}^2\left(\text{Acting at }47.1°\text{counterclockwise below horizontal}\right)}$.

Explanation of Solution

Given information:

Calculation:

Consider the position of the point F with respect to the point E is ${\text{r}}_{F/E}=0.09\text{i}+0.6\text{j}$.

Calculate the acceleration at F using the relation:

${\text{a}}_F={\alpha }_{DE}\text{k}\times {\text{r}}_{F/E}-{\omega }_{DE}^2\times {\text{r}}_{F/E}$

Substitute $1.466\text{rad/s}$ for ${\alpha }_{DE}$, $\left(0.09\text{i}+0.6\text{j}\right)$ for ${\text{r}}_{F/E}$, and $1.463\text{rad/s}$ for ${\omega }_{DE}$.

$\begin{array}{c}{\text{a}}_F=1.466\text{k}\times \left(0.09\text{i}+0.6\text{j}\right)-{\left(1.463\right)}^2\times \left(0.09\text{i}+0.6\text{j}\right)\\ =0.13194\text{j}-0.8796\text{i}-0.19263\text{i}-1.2842\text{j}\\ \text{=}-\text{1}\text{.0723i}-\text{1}\text{.153j}\end{array}$

Calculate the magnitude of the acceleration at D as follows:

$\begin{array}{c}{\text{a}}_D\text{=}\sqrt{{\left(-\text{1}\text{.0723}\right)}^2+{\left(-\text{1}\text{.153}\right)}^2}\\ =1.575{\text{m/s}}^2\end{array}$

Calculate the direction of the acceleration at D as follows:

$\begin{array}{c}\tan \theta \text{=}\frac{-\text{1}\text{.153}}{-\text{1}\text{.0723}}\\ \theta ={\tan }^{-1}\left(\frac{\text{1}\text{.153}}{\text{1}\text{.0723}}\right)\\ \approx 47.1°\left(\text{Counterclockwise below the horizontal}\right)\end{array}$

Thus, the acceleration at F is $\underset{\_}{1.575{\text{m/s}}^2\left(\text{Acting at }47.1°\text{counterclockwise below horizontal}\right)}$.