Chapter 15.6, Problem 15.200P

In Prob. 15.199, determine (a) the common angular acceleration of gears A and B, (b) the acceleration of the tooth of gear A that is in contact with gear C at point 1.

    

To determine

(a)

The common angular acceleration of the gears $A$ and $B$.

Answer to Problem 15.200P

The common angular acceleration of the gears $A$ and $B$ is $\left(135.1\text{rad}/{\text{s}}^2\right)\text{k}$.

Explanation of Solution

Given Information:

The angular velocity of the gear $C$ is $30\text{rad}/\text{s}$ and the angular velocity of the gear $D$ is $20\text{rad}/\text{s}$.

Draw the schematic diagram of the given system.

Figure-(1)

Write the expression for the velocity of point 1.

$v_1={\omega }_E\times r_1$ ...... (I)

Here, the angular velocity of the gear $E$ is ${\omega }_E$ and the position vector at point 1 is $r_1$.

Write the expression for the velocity of point 1 when gear $C$ is considered to be driven by the gear unit $A$ and $B$.

$v_1=\omega \times r_1$ ...... (II)

Here, the angular velocity of the shaft unit carries gear $A$ and $B$ is $\omega$.

Write the expression for the velocity of point 2.

$v_2={\omega }_G\times r_2$ ...... (III)

Here, the angular velocity of the gear $G$ is ${\omega }_G$ and the position vector at point 2 is $r_2$.

Write the expression for the velocity of point 2 when gear $D$ is considered to be driven by the gear unit $A$ and $B$.

$v_2=\omega \times r_2$ ...... (IV)

Write the expression for the angular velocity of the inclined shaft unit which carries gear $A$ and $B$ in the vector form.

$\omega ={\omega }_x\text{i}+{\omega }_y\text{j}+{\omega }_z\text{k}$ ...... (V)

Here, the angular velocity of the shaft in x- direction is ${\omega }_x$, the angular velocity of the shaft in y- direction is ${\omega }_y$ and the angular velocity of the shaft in z- direction is ${\omega }_z$.

Draw the diagram to show the motion of the shaft is $FH$.

Figure-(2)

Write the expression for the velocity of point $N$.

$v_N=\omega \times r_N$ ...... (VI)

Here, the position vector at point $N$ is $r_N$.

Write the expression for the velocity at point $N$ when it is considered as a part of the shaft $FH$.

$v_N={\omega }_{FH}\times r_N$ ...... (VII)

Here, the angular velocity of shaft $FH$ is ${\omega }_{FH}$.

Write the expression for the common angular acceleration of the gear $A$ and $B$.

$\alpha ={\omega }_{FH}\times \omega$ ...... (VIII)

Calculation:

Consider the unit vector along $X$, $Y$, $Z$ axis as $\text{i}$, $\text{j}$ and $\text{k}$.

From Figure-(1) the coordinate of the point 1 at the intersection of the gears $A$ and $C$ are $\left(-80\text{mm},260\text{mm}\right)$. Therefore, the position vector at point 1 is $r_1=-\left(80\text{mm}\right)\text{i}+\left(260\text{mm}\right)\text{j}$.

From Figure-(1) the coordinate of the point 2 at the intersection of the gears $B$ and $D$ are $\left(80\text{mm},50\text{mm}\right)$. Therefore, the position vector at point 2 is $r_2=\left(80\text{mm}\right)\text{i}+\left(50\text{mm}\right)\text{j}$.

Since, the shaft $E$ rotate with the angular velocity of the gear $C$ hence the velocity of the gear $E$ in vector form is $\left(30\text{rad}/\text{s}\right)\text{i}$.

Substitute $\left(30\text{rad}/\text{s}\right)\text{i}$ for ${\omega }_E$ and $-\left(80\text{mm}\right)\text{i}+\left(260\text{mm}\right)\text{j}$ for $r_1$ in Equation (I).

$\begin{array}{c}{v}_1=\left(30\text{rad}/\text{s}\right)\text{i}\left[-\left(80\text{mm}\right)\text{i}+\left(260\text{mm}\right)\text{j}\right]\\ =\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 30\text{rad}/\text{s}& 0& 0\\ -80\text{mm}& 260\text{mm}& 0\end{array}\right|\\ =\left(0-0\right)\text{i}-\left(0-0\right)\text{j}+\left(7800\text{mm}/\text{s}-0\right)\text{k}\\ \text{=}\left(7800\text{mm}/\text{s}\right)\text{k}\end{array}$

Substitute ${\omega }_x\text{i}+{\omega }_y\text{j}+{\omega }_z\text{k}$ for $\omega$ and $-\left(80\text{mm}\right)\text{i}+\left(260\text{mm}\right)\text{j}$ for $r_1$ in Equation (II).

$\begin{array}{c}{v}_1={\omega }_x\text{i}+{\omega }_y\text{j}+{\omega }_z\text{k}\left[-\left(80\text{mm}\right)\text{i}+\left(260\text{mm}\right)\text{j}\right]\\ =\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ {\omega }_x& {\omega }_y& {\omega }_z\\ -80\text{mm}& 260\text{mm}& 0\end{array}\right|\\ =\left(0-260\text{mm}{\omega }_z\right)\text{i}-\left(0-\left(-80\text{mm}{\omega }_z\right)\right)\text{j}+\left(260\text{mm}{\omega }_x-\left(-80\text{mm}{\omega }_y\right)\right)\text{k}\\ =-\left(260\text{mm}{\omega }_z\right)\text{i}-\left(80\text{mm}{\omega }_z\right)\text{j}+\left(260\text{mm}{\omega }_x+80\text{mm}{\omega }_y\right)\text{k}\end{array}$ ... (IX)

Substitute $\left(7800\text{mm}/\text{s}\right)\text{k}$ for $v_1$ in equation (IX).

$\left(7800\text{mm}/\text{s}\right)\text{k}=\left[\begin{array}{l}-\left(260\text{mm}{\omega }_z\right)\text{i}-\left(80\text{mm}{\omega }_z\right)\text{j}+\\ \left(260\text{mm}{\omega }_x+80\text{mm}{\omega }_y\right)\text{k}\end{array}\right]$ ...... (X)

Compare the terms along the x- direction in Equation (X).

$\begin{array}{l}\left(260\text{mm}\right){\omega }_z=0\\ {\omega }_z=0\end{array}$

Compare the terms along the z- direction in Equation (X).

$\left(7\text{800}\text{mm}/\text{s}\right)=260\text{mm}{\omega }_x+80\text{mm}{\omega }_y$ ...... (XI)

Since, the shaft $G$ rotate with the angular velocity of the gear $D$ hence the velocity of the gear $G$ in vector form is $\left(20\text{rad}/\text{s}\right)\text{i}$.

Substitute $\left(20\text{rad}/\text{s}\right)\text{i}$ for ${\omega }_G$ and $\left(80\text{mm}\right)\text{i}+\left(50\text{mm}\right)\text{j}$ for $r_2$ in Equation (III).

$\begin{array}{c}{v}_2=\left(20\text{rad}/\text{s}\right)\text{i}\left[\left(80\text{mm}\right)\text{i}+\left(50\text{mm}\right)\text{j}\right]\\ =\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 20\text{rad}/\text{s}& 0& 0\\ 80\text{mm}& 50\text{mm}& 0\end{array}\right|\\ =\left(0-0\right)\text{i}-\left(0-0\right)\text{j}+\left(1000\text{mm}/\text{s}-0\right)\text{k}\\ \text{=}\left(1000\text{mm}/\text{s}\right)\text{k}\end{array}$

Substitute ${\omega }_x\text{i}+{\omega }_y\text{j}+{\omega }_z\text{k}$ for $\omega$ and $\left(80\text{mm}\right)\text{i}+\left(50\text{mm}\right)\text{j}$ for $r_2$ in Equation (IV).

$\begin{array}{c}{v}_2={\omega }_x\text{i}+{\omega }_y\text{j}+{\omega }_z\text{k}\left[\left(80\text{mm}\right)\text{i}+\left(50\text{mm}\right)\text{j}\right]\\ =\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ {\omega }_x& {\omega }_y& {\omega }_z\\ 80\text{mm}& 50\text{mm}& 0\end{array}\right|\\ =\left(0-50\text{mm}{\omega }_z\right)\text{i}-\left(0-\left(80\text{mm}{\omega }_z\right)\right)\text{j}+\left(50\text{mm}{\omega }_x-\left(80\text{mm}{\omega }_y\right)\right)\text{k}\\ =-\left(50\text{mm}{\omega }_z\right)\text{i}+\left(80\text{mm}{\omega }_z\right)\text{j}+\left(50\text{mm}{\omega }_x-80\text{mm}{\omega }_y\right)\text{k}\end{array}$ ... (XII)

Substitute $\left(1000\text{mm}/\text{s}\right)\text{k}$ for $v_2$ in equation (XII).

$\left(1000\text{mm}/\text{s}\right)\text{k}=\left[\begin{array}{l}-\left(50\text{mm}{\omega }_z\right)\text{i}+\left(80\text{mm}{\omega }_z\right)\text{j}+\\ \left(50\text{mm}{\omega }_x-80\text{mm}{\omega }_y\right)\text{k}\end{array}\right]$ ...... (XIII)

Compare the terms along the z- direction in Equation (X).

$\begin{array}{l}\left(\text{1000}\text{mm}/\text{s}\right)=50\text{mm}{\omega }_x-80\text{mm}{\omega }_y\\ {\omega }_x=\frac{\left(\text{1000}\text{mm}/\text{s}\right)+80\text{mm}{\omega }_y}{\left(50\text{mm}\right)}\end{array}$ ...... (XIV)

Substitute $\frac{\left(\text{1000}\text{mm}/\text{s}\right)+80\text{mm}{\omega }_y}{\left(50\text{mm}\right)}$ for ${\omega }_x$ in Equation (XII).

$\begin{array}{l}\left(7\text{800}\text{mm}/\text{s}\right)=260\text{mm}\left(\frac{\left(\text{1000}\text{mm}/\text{s}\right)+80\text{mm}{\omega }_y}{\left(50\text{mm}\right)}\right)+80\text{mm}{\omega }_y\\ \left(7\text{800}\text{mm}/\text{s}\right)-\left(5200\text{mm}/\text{s}\right)=496\text{mm}{\omega }_y\\ {\omega }_y=\frac{\left(\text{2600}\text{mm}/\text{s}\right)}{496\text{mm}}\\ {\omega }_y=5.24\text{rad}/\text{s}\end{array}$

Substitute $5.24\text{rad}/\text{s}$ for ${\omega }_y$ in Equation (XI).

$\begin{array}{c}{\omega }_x=\frac{\left(\text{1000}\text{mm}/\text{s}\right)+80\text{mm}\left(5.24\text{rad}/\text{s}\right)}{\left(50\text{mm}\right)}\\ =\frac{\left(\text{1000}\text{mm}/\text{s}\right)}{\left(50\text{mm}\right)}+\frac{80\text{mm}\left(5.24\text{rad}/\text{s}\right)}{\left(50\text{mm}\right)}\\ =20\text{rad}/\text{s}+8.384\text{rad}/\text{s}\\ =28.38\text{rad}/\text{s}\end{array}$

Substitute $28.38\text{rad}/\text{s}$ for ${\omega }_x$, $5.24\text{rad}/\text{s}$ for ${\omega }_y$ and $0$ for ${\omega }_z$ in Equation (V).

$\begin{array}{c}\omega =\left(28.38\text{rad}/\text{s}\right)\text{i}+\left(5.24\text{rad}/\text{s}\right)\text{j}+\left(0\right)\text{k}\\ =\left(28.38\text{rad}/\text{s}\right)\text{i}+\left(5.24\text{rad}/\text{s}\right)\text{j}\end{array}$

From Figure-(2) the coordinate of the point $N$ is the half of the coordinate of $y$ that is $\left(\frac{1}{2}{y}_N,y_N\right)$. Therefore, the position vector at point $N$ is $r_N=\left(1/2y_N\right)\text{i}+\left(y_N\right)\text{j}$.

Substitute $\left(28.38\text{rad}/\text{s}\right)\text{i}+\left(5.24\text{rad}/\text{s}\right)\text{j}$ for $\omega$ and $\left(1/2y_N\right)\text{i}+\left(y_N\right)\text{j}$ for $r_N$ in Equation (VI).

$\begin{array}{c}{v}_N=\left[\left(28.38\text{rad}/\text{s}\right)\text{i}+\left(5.24\text{rad}/\text{s}\right)\text{j}\right]\left(\left(1/2y_N\right)\text{i}+\left(y_N\right)\text{j}\right)\\ =\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 28.38\text{rad}/\text{s}& 5.24\text{rad}/\text{s}& 0\\ 1/2y_N& y_N& 0\end{array}\right|\\ =\left(0-0\right)\text{i}-\left(0-0\right)\text{j}+\left(\left(28.387\text{rad}/\text{s}\right)y_N-\left(2.621\text{rad}/\text{s}\right)y_N\right)\text{k}\\ \text{=}\left(25.766\text{rad}/\text{s}\right)y_N\text{k}\end{array}$

Substitute $\left({\omega }_{FH}\right)\text{i}$ for ${\omega }_{FH}$ and $\left(1/2y_N\right)\text{i}+\left(y_N\right)\text{j}$ for $r_N$ in Equation (VII).

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

Substitute $\left(25.766\text{rad}/\text{s}\right)y_N\text{k}$ for $v_N$ in Equation (XV).

$\begin{array}{l}\left(25.766\text{rad}/\text{s}\right)y_N\text{k}=\left({\omega }_{FH}\right)y_N\text{k}\\ {\omega }_{FH}=\frac{\left(25.766\text{rad}/\text{s}\right)y_N\text{k}}{y_N\text{k}}\\ {\omega }_{FH}=\left(25.766\text{rad}/\text{s}\right)\text{i}\end{array}$

Substitute $\left(25.766\text{rad}/\text{s}\right)\text{i}$ for ${\omega }_{FH}$ and $\left(28.38\text{rad}/\text{s}\right)\text{i}+\left(5.24\text{rad}/\text{s}\right)\text{j}$ for $\omega$ in equation (VIII).

$\begin{array}{c}\alpha =\left[\left(25.766\text{rad}/\text{s}\right)\text{i}\right]\left[\left(28.38\text{rad}/\text{s}\right)\text{i}+\left(5.24\text{rad}/\text{s}\right)\text{j}\right]\\ =\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 25.766\text{rad}/\text{s}& 0& 0\\ 28.38\text{rad}/\text{s}& 5.24\text{rad}/\text{s}& 0\end{array}\right|\\ =\left(0-0\right)\text{i}-\left(0-0\right)\text{j}+\left(135.1\text{rad}/{\text{s}}^2-0\right)\text{k}\\ \text{=}\left(135.1\text{rad}/{\text{s}}^2\right)\text{k}\end{array}$

Conclusion:

The common angular acceleration of the gear $A$ and $B$ is $\left(135.1\text{rad}/{\text{s}}^2\right)\text{k}$.

To determine

(b)

The acceleration of the tooth of the gear $A$ at point 1.

Answer to Problem 15.200P

The acceleration of the tooth of the gear $A$ at point 1 is $\left(5761.06\text{mm}/{\text{s}}^2\right)\text{i}-\left(232226.6\text{mm}/{\text{s}}^2\right)\text{j}$.

Explanation of Solution

Write the expression for the acceleration of the gear $A$ at point 1.

$a_1=\left(\alpha \times r_1\right)+\left(\omega \times v_1\right)$ ...... (XVI)

Calculation:

Substitute $\left(135.1\text{rad}/{\text{s}}^2\right)\text{k}$ for $\alpha$, $-\left(80\text{mm}\right)\text{i}+\left(260\text{mm}\right)\text{j}$ for $r_1$, $\left(7800\text{mm}/\text{s}\right)\text{k}$ for $v_1$, $\left(28.38\text{rad}/\text{s}\right)\text{i}+\left(5.24\text{rad}/\text{s}\right)\text{j}$ for $\omega$ in Equation (XVI).

$\begin{array}{c}{a}_1=\left[\begin{array}{l}\left\{\left(\left(135.1\text{rad}/{\text{s}}^2\right)\text{k}\right)\left[-\left(80\text{mm}\right)\text{i}+\left(260\text{mm}\right)\text{j}\right]\right\}+\\ \left\{\left[\left(28.38\text{rad}/\text{s}\right)\text{i}+\left(5.24\text{rad}/\text{s}\right)\text{j}\right]\left(\left(7800\text{mm}/\text{s}\right)\text{k}\right)\right\}\end{array}\right]\\ =\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 0& 0& 135.1\text{rad}/{\text{s}}^2\\ 80\text{mm}& 260\text{mm}& 0\end{array}\right|+\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 28.38\text{rad}/\text{s}& 5.24\text{rad}/\text{s}& 0\\ 0& 0& 7800\text{mm}/\text{s}\end{array}\right|\\ =\left\{\begin{array}{l}\left(0-35126\text{mm}/{\text{s}}^2\right)\text{i}-\left(0-\left(-10808\text{mm}/{\text{s}}^2\right)\right)\text{j}+\left(0-0\right)\text{k}+\\ \left(40887.6\text{mm}/{\text{s}}^2-0\right)\text{i}-\left(221418.6\text{mm}/{\text{s}}^2-0\right)\text{j}+\left(0-0\right)\text{k}\end{array}\right\}\\ =\left(5761.06\text{mm}/{\text{s}}^2\right)\text{i}-\left(232226.6\text{mm}/{\text{s}}^2\right)\text{j}\end{array}$

Conclusion:

The acceleration of the tooth of the gear $A$ at point 1 is $\left(5761.06\text{mm}/{\text{s}}^2\right)\text{i}-\left(232226.6\text{mm}/{\text{s}}^2\right)\text{j}$.