Chapter 15.6, Problem 15.200P

(a)

To determine

The common angular acceleration of unit AB.

Answer to Problem 15.200P

The common angular acceleration of unit AB $\left(\text{α}\right)$ is $\underset{\_}{\left(135.1{\text{rad/s}}^2\right)\text{k}}$.

Explanation of Solution

Given information:

The constant angular velocities of gears C and D is 30 rad/s and 20 rad/s respectively.

Calculation:

Draw the free body diagram of the planetary gear system as in Figure (1).

Place origin at F.

Write the relative position vector $\left({\text{r}}_1\right)$.

${\text{r}}_1=-\left(80\text{mm}\right)\text{i}+\left(260\text{mm}\right)\text{j}$

Write the relative position vector $\left({\text{r}}_2\right)$.

${\text{r}}_2=\left(80\text{mm}\right)\text{i}+\left(50\text{mm}\right)\text{j}$

Determine the velocity value $\left({\text{v}}_1\right)$.

${\text{v}}_1={\text{ω}}_E\times {\text{r}}_1$

Here, ${\text{ω}}_E$ is the angular velocity of E.

Substitute $\left(30\text{rad/s}\right)\text{i}$ for ${\text{ω}}_E$ and $\left[-\left(80\text{mm}\right)\text{i}+\left(260\text{mm}\right)\text{j}\right]$ for ${\text{r}}_1$.

$\begin{array}{c}{\text{v}}_1=\left(30\right)\text{i}\times \left[-\left(80\right)\text{i}+\left(260\right)\text{j}\right]\\ =0+\left(7,800\text{mm/s}\right)\text{k}\\ =\left(7,800\text{mm/s}\right)\text{k}\end{array}$

Determine the velocity value $\left({\text{v}}_2\right)$.

${\text{v}}_2={\text{ω}}_G\times {\text{r}}_2$

Here, ${\text{ω}}_G$ is the angular velocity of G.

Substitute $\left(20\text{rad/s}\right)\text{i}$ for ${\text{ω}}_G$ and $\left[\left(80\text{mm}\right)\text{i}+\left(50\text{mm}\right)\text{j}\right]$ for ${\text{r}}_2$.

$\begin{array}{c}{\text{v}}_2=\left(20\right)\text{i}\times \left[\left(80\right)\text{i}+\left(50\right)\text{j}\right]\\ =0+\left(1,000\text{mm/s}\right)\text{k}\\ =\left(1,000\text{mm/s}\right)\end{array}$

The motion of gear unit AB is $\text{ω}={\omega }_x\text{i}+{\omega }_y\text{j}+{\omega }_z\text{k}$.

Determine the velocity vector $\left({\text{v}}_1\right)$ with common angular velocity.

${\text{v}}_1=\text{ω}\times {\text{r}}_1$

Substitute $\left(7,800\text{mm/s}\right)\text{k}$ for ${\text{v}}_1$, ${\omega }_x\text{i}+{\omega }_y\text{j}+{\omega }_z\text{k}$ for $\text{ω}$ and $\left[-\left(80\text{mm}\right)\text{i}+\left(260\text{mm}\right)\text{j}\right]$ for ${\text{r}}_1$.

$\begin{array}{c}7,800\text{k}=\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ {\omega }_x& {\omega }_y& {\omega }_z\\ -80& 260& 0\end{array}\right|\\ =\text{i}\left(0-260{\omega }_z\right)-\text{j}\left(0+80{\omega }_z\right)+\text{k}\left(260{\omega }_x+80{\omega }_y\right)\\ =-\left(260{\omega }_z\right)\text{i}-\left(80{\omega }_z\right)\text{j}+\left(260{\omega }_x+80{\omega }_y\right)\text{k}\end{array}$

Substitute 0 for ${\omega }_z$.

$\begin{array}{c}7,800\text{k}=-\left(260\left(0\right)\right)\text{i}-\left(80\left(0\right)\right)\text{j}+\left(260{\omega }_x+80{\omega }_y\right)\text{k}\\ =\left(260{\omega }_x+80{\omega }_y\right)\text{k}\end{array}$ (1)

Equate the k component in Equation (1).

$7,800=\left(260{\omega }_x+80{\omega }_y\right)$ (2)

Determine the velocity vector $\left({\text{v}}_2\right)$ with common angular velocity.

${\text{v}}_2=\text{ω}\times {\text{r}}_2$

Substitute $\left(1,000\text{mm/s}\right)\text{k}$ for ${\text{v}}_2$, ${\omega }_x\text{i}+{\omega }_y\text{j}+{\omega }_z\text{k}$ for $\text{ω}$, 0 for ${\omega }_z$, and $\left[\left(80\text{mm}\right)\text{i}+\left(50\text{mm}\right)\text{j}\right]$ for ${\text{r}}_2$.

$\begin{array}{c}1,000\text{k}=\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ {\omega }_x& {\omega }_y& 0\\ 80& 50& 0\end{array}\right|\\ =\text{i}\left(0-0\right)-\text{j}\left(0-0\right)+\text{k}\left(50{\omega }_x-80{\omega }_y\right)\\ =\left(50{\omega }_x-80{\omega }_y\right)\text{k}\end{array}$

Equate the k component.

$1,000=\left(50{\omega }_x-80{\omega }_y\right)$ (3)

Add the Equation (2) and (3).

$\begin{array}{l}7,800+1,000=\left(260{\omega }_x+80{\omega }_y\right)+\left(50{\omega }_x-80{\omega }_y\right)\\ 8,800=310{\omega }_x\\ {\omega }_x=\frac{8,800}{310}\\ {\omega }_x=28.387\text{rad/s}\end{array}$

Determine the value of ${\omega }_y$.

Substitute $28.387\text{rad/s}$ for ${\omega }_x$ in Equation (3).

$\begin{array}{l}1,000=\left(50\times 28.387-80{\omega }_y\right)\\ 1,000-1,419.35=-80{\omega }_y\\ {\omega }_y=\frac{-419.35}{-80}\\ {\omega }_y=5.242\text{rad/s}\end{array}$

Determine the common angular velocity of gears A and B $\left(\text{ω}\right)$ using the relation.

$\text{ω}={\omega }_x\text{i}+{\omega }_y\text{j}+{\omega }_z\text{k}$

Substitute $28.387\text{rad/s}$ for ${\omega }_x$, $5.242\text{rad/s}$ for ${\omega }_y$, and 0 for ${\omega }_z$.

$\text{ω}=\left(28.387\text{rad/s}\right)\text{i}+\left(5.242\text{rad/s}\right)\text{j}$

Draw the free body diagram of the shaft system with FH as in Figure (2).

The point N is at nut, which is a part of unit AB and also is a part of shaft GH.

Write the distance of point N along x axis.

$x_N=\frac{1}{2}{y}_N$

Write the relative position vector $\left({\text{r}}_N\right)$.

${\text{r}}_N=x_N\text{i}+y_N\text{j}$

Substitute $\frac{1}{2}{y}_N$ for $x_N$.

${\text{r}}_N=\frac{1}{2}{y}_N\text{i}+y_N\text{j}$

The nut N as a part of unit AB.

Determine the velocity vector $\left({\text{v}}_N\right)$ using the relation.

${\text{v}}_N=\text{ω}\times {\text{r}}_N$

Substitute $\left(28.4\text{rad/s}\right)\text{i}+\left(5.24\text{rad/s}\right)\text{j}$ for $\text{ω}$ and $\left[\frac{1}{2}{y}_N\text{i}+y_N\text{j}\right]$ for ${\text{r}}_N$.

$\begin{array}{c}{v}_N=\left[\left(28.4\right)\text{i}+\left(5.24\right)\text{j}\right]\times \left[\frac{1}{2}{y}_N\text{i}+y_N\text{j}\right]\\ =\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 28.4& 5.24& 0\\ \frac{1}{2}{y}_N& y_N& 0\end{array}\right|\\ =\text{i}\left(0-0\right)-\text{j}\left(0-0\right)+\text{k}\left(28.4y_N-2.62y_N\right)\\ =\text{k}\left(25.78y_N\right)\end{array}$ (4)

The nut N as a part of shaft FH.

Determine the velocity vector $\left({\text{v}}_N\right)$ using the relation.

${\text{v}}_N={\text{ω}}_{FH}\times {\text{r}}_N$

Substitute ${\omega }_{FH}\text{i}$ for ${\text{ω}}_{FH}$ and $\left[\frac{1}{2}{y}_N\text{i}+y_N\text{j}\right]$ for ${\text{r}}_N$.

$\begin{array}{c}{v}_N={\omega }_{FH}\text{i}\times \left[\frac{1}{2}{y}_N\text{i}+y_N\text{j}\right]\\ =\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ {\omega }_{FH}& 0& 0\\ \frac{1}{2}{y}_N& y_N& 0\end{array}\right|\\ =\text{i}\left(0-0\right)-\text{j}\left(0-0\right)+\text{k}\left({\omega }_{FH}{y}_N-0\right)\\ =\text{k}\left({\omega }_{FH}{y}_N\right)\end{array}$ (5)

Equate the Equations (4) and (5).

$\begin{array}{l}\left(25.78y_N\right)=\left({\omega }_{FH}{y}_N\right)\\ {\omega }_{FH}=\left(25.8\text{rad/s}\right)\text{i}\end{array}$

The angular velocity vector $\omega$ rotates about the x-axis with angular velocity ${\text{ω}}_{FH}$.

Determine the common angular acceleration of unit AB.

$\alpha ={\text{ω}}_{FH}\times \text{ω}$

Substitute $\left(25.8\text{rad/s}\right)\text{i}$ for ${\text{ω}}_{FH}$ and $\left(28.4\text{rad/s}\right)\text{i}+\left(5.24\text{rad/s}\right)\text{j}$ for $\text{ω}$.

$\begin{array}{c}\alpha =\left(25.8\right)\text{i}\times \left[\left(28.4\right)\text{i}+\left(5.24\right)\text{j}\right]\\ =0+135.1\text{k}\\ =\left(135.1{\text{rad/s}}^{\text{2}}\right)\text{k}\end{array}$

Therefore, the common angular acceleration of unit AB $\left(\text{α}\right)$ is $\underset{\_}{\left(135.1{\text{rad/s}}^2\right)\text{k}}$.

(b)

To determine

The acceleration of the tooth of gear A which is in contact with gear C at point 1.

Answer to Problem 15.200P

The acceleration of the tooth of gear A which is in contact with gear C at point 1

$\left({\text{a}}_1\right)$ is $\underset{\_}{\left(5.8{\text{m/s}}^2\right)\text{i}-\left(232{\text{m/s}}^2\right)\text{j}}$.

Explanation of Solution

Given information:

The constant angular velocities of gears C and D is 30 rad/s and 20 rad/s respectively.

Calculation:

Determine the acceleration of the tooth of gear A which is in contact with gear C at point 1.

${\text{a}}_1=\text{α}\times {\text{r}}_1+\text{ω}\times {\text{v}}_1$

Substitute $\left(135.1{\text{rad/s}}^{\text{2}}\right)\text{k}$ for $\text{α}$, $\left[-\left(80\text{mm}\right)\text{i}+\left(260\text{mm}\right)\text{j}\right]$ for ${\text{r}}_1$, $\left(28.4\text{rad/s}\right)\text{i}+\left(5.24\text{rad/s}\right)\text{j}$ for $\text{ω}$, and $\left(7,800\text{mm/s}\right)\text{k}$ for ${\text{v}}_1$.

$\begin{array}{c}{\text{a}}_1=\text{α}\times {\text{r}}_1+\text{ω}\times {\text{v}}_1\\ =\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 0& 0& 135.1\\ -80& 260& 0\end{array}\right|+\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 28.4& 5.24& 0\\ 0& 0& 7,800\end{array}\right|\\ =\left\{\begin{array}{l}\left[\text{i}\left(0-35,126\right)-\text{j}\left(0+10,808\right)+\text{k}\left(0-0\right)\right]\times \frac{1\text{m}}{1,000\text{mm}}+\\ \left[\text{i}\left(40,872-0\right)-\text{j}\left(221,520-0\right)+\text{k}\left(0-0\right)\right]\times \frac{1\text{m}}{1,000\text{mm}}\end{array}\right\}\\ =\left\{\left[-35.13\text{i}-10.81\text{j}\right]+\left[40.9\text{i}-221.52\text{j}\right]\right\}\\ =\left[\left(5.8{\text{m/s}}^{\text{2}}\right)\text{i}-\left(232{\text{m/s}}^{\text{2}}\right)\text{j}\right]\end{array}$

Therefore, the acceleration of the tooth of gear A which is in contact with gear C at point 1

$\left({\text{a}}_1\right)$ is $\underset{\_}{\left(5.8{\text{m/s}}^2\right)\text{i}-\left(232{\text{m/s}}^2\right)\text{j}}$.