Chapter 15.3, Problem 15.80P

The arm ABC rotates with an angular velocity of 4 rad/s counterclockwise. Knowing that the angular velocity of the intermediate gear B is 8 rad/s counterclockwise, determine (a) the instantaneous centers of rotation of gears A and C, (b) the angular velocities of gears A and C.

Fig. P15.80

To determine

Find the instantaneous center of rotation of gear A and C.

Answer to Problem 15.80P

The instantaneous center of rotation of the gear A is $\underset{\_}{300\text{mm}\text{left of }A}$.

The instantaneous center of rotation of the gear C is $\underset{\_}{600\text{mm}\text{left of }C}$.

Explanation of Solution

Given information:

The angular velocity of the arm ABC is ${\omega }_{ABC}=4\text{rad/s}\left(\text{Counterclockwise}\right)$.

The angular velocity of the gear B is ${\omega }_B=8\text{rad/s}\left(\text{Counterclockwise}\right)$.

Consider the contact point between the gear A and B is denoted by 1.

Consider the contact point between the gear B and C is denoted by 2.

Consider the velocity at point A, C, 1, and 2 are denoted by $v_A,v_C,v_1,\text{and }{v}_2$.

Calculation:

Show the velocity at point A and C as shown in Figure 1.

Refer to Figure 1.

Consider the arm ABC.

Calculate the velocity at A and C as follows:

Take the value of ${\omega }_{ABC}=4\text{rad/s}\left(\text{Counterclockwise}\right)$.

$\begin{array}{c}{v}_A=\left(AB\right){\omega }_{ABC}\\ =300\text{mm}\times \left(\frac{1\text{m}}{1000\text{mm}}\right)\times 4\text{rad/s}\\ =1.2\text{m/s}(\downarrow )\end{array}$

$\begin{array}{c}{v}_B=\left(BC\right){\omega }_{ABC}\\ =300\text{mm}\times \left(\frac{1\text{m}}{1000\text{mm}}\right)\times 4\text{rad/s}\\ =1.2\text{m/s}(\downarrow )\end{array}$

Show the gear B as shown in Figure 2.

Refer Figure 2.

Calculate the velocity at point 1 and 2 as follows:

Take the angular velocity of the gear B is ${\omega }_B=8\text{rad/s}\left(\text{Counterclockwise}\right)$.

$\begin{array}{c}{v}_1=\left(B1\right){\omega }_B\\ =200\text{mm}\times \left(\frac{1\text{m}}{1000\text{mm}}\right)\times 8\text{rad/s}\\ =1.6\text{m/s}(\downarrow )\end{array}$

$\begin{array}{c}{v}_2=\left(B2\right){\omega }_B\\ =100\text{mm}\times \left(\frac{1\text{m}}{1000\text{mm}}\right)\times 8\text{rad/s}\\ =0.8\text{m/s}(\uparrow )\end{array}$

Show the gear A as shown in Figure 3.

Refer Figure 3.

Consider the instantaneous center of rotation of the gear A is denoted by $I_A$.

Calculate the angular acceleration $\left({\omega }_A\right)$ of the gear A using the relation:

$\begin{array}{c}{\omega }_A=\frac{v_1-v_A}{100\text{mm}\times \left(\frac{1\text{m}}{1000\text{mm}}\right)}\\ =\frac{v_1-v_A}{0.100}\end{array}$

Substitute $1.6\text{m/s}$ for $v_1$ and $1.2\text{m/s}$ for $v_A$.

$\begin{array}{c}{\omega }_A=\frac{1.6-1.2}{0.1}\\ =4\text{rad/s}\left(\text{Clockwise}\right)\end{array}$ (1)

Calculate the distance $l_A$ using the relation:

$l_A=\frac{v_A}{{\omega }_A}$

Substitute $4\text{rad/s}\left(\text{Clockwise}\right)$ for ${\omega }_A$ and $1.2\text{m/s}$ for $v_A$.

$\begin{array}{c}{l}_A=\frac{1.2}{4}\\ =0.3\text{m}\times \left(\frac{1000\text{mm}}{1\text{m}}\right)\\ =300\text{mm}\end{array}$

Thus, the instantaneous center of rotation of the gear A is $\underset{\_}{300\text{mm}\text{left of }A}$.

Show the gear C as shown in Figure 4.

Refer Figure 4.

Consider the instantaneous center of rotation of the gear C is denoted by $I_C$.

Calculate the angular acceleration $\left({\omega }_C\right)$ of the gear C using the relation:

$\begin{array}{c}{\omega }_C=\frac{v_C-v_2}{200\text{mm}\times \left(\frac{1\text{m}}{1000\text{mm}}\right)}\\ =\frac{v_C-v_2}{0.200}\end{array}$

Substitute $0.8\text{m/s}$ for $v_2$ and $1.2\text{m/s}$ for $v_C$.

$\begin{array}{c}{\omega }_C=\frac{0.8-1.2}{0.2}\\ =-2\text{rad/s}\\ =2\text{rad/s}\left(\text{Counterclockwise}\right)\end{array}$ (2)

Calculate the distance $l_C$ using the relation:

$l_C=\frac{v_C}{{\omega }_C}$

Substitute $2\text{rad/s}\left(\text{Counterclockwise}\right)$ for ${\omega }_C$ and $1.2\text{m/s}$ for $v_C$.

$\begin{array}{c}{l}_C=\frac{1.2}{2}\\ =0.6\text{m}\times \left(\frac{1000\text{mm}}{1\text{m}}\right)\\ =600\text{mm}\end{array}$

Thus, the instantaneous center of rotation of the gear C is $\underset{\_}{600\text{mm}\text{left of }C}$.

To determine

Find the angular velocity of gear A and C.

Answer to Problem 15.80P

The angular velocity of the gear A and C are $\underset{\_}{4\text{rad/s}\left(\text{Clockwise}\right)}$ and $\underset{\_}{2\text{rad/s}\left(\text{Clockwise}\right)}$.

Explanation of Solution

Given information:

Calculation:

Refer Part (a).

Refer Equation (1).

The angular velocity of the gear A is $4\text{rad/s}\left(\text{Clockwise}\right)$.

Refer Equation (2).

The angular velocity of the gear C is $2\text{rad/s}\left(\text{Clockwise}\right)$

Thus, the angular velocity of the gear A and C are $\underset{\_}{4\text{rad/s}\left(\text{Clockwise}\right)}$ and $\underset{\_}{2\text{rad/s}\left(\text{Clockwise}\right)}$.