Chapter 15, Problem 15.251RP

(a)

To determine

The angular velocity of arm AB.

Answer to Problem 15.251RP

The angular velocity of arm AB $\left({\text{ω}}_{AB}\right)$ is $\underset{\_}{11.11\text{rad/s}\left(\text{Clockwise}\right)}$.

Explanation of Solution

Given information:

The constant angular acceleration of gear C $\left({\alpha }_C\right)$ is $4{\text{rad/s}}^{\text{2}}$.

The given time t is 5 s.

Calculation:

Determine the angular velocity of gear C at the given time.

${\omega }_C={\alpha }_{C}t$

Substitute $4{\text{rad/s}}^{\text{2}}$ for ${\alpha }_C$ and 5 s for t.

$\begin{array}{c}{\omega }_C=5\times 4\\ =20\text{rad/s}\left(\text{Clockwise}\right)\end{array}$

Consider point 1 be the contact point between gears A and B. Let point 2 be the contact point between gears B and C. Points A, B, and C are the centers respectively, of gears A, B, C.

Take the position of x axis along the straight line $\text{A}1\text{B}2$

The distance of point A and point C is $x_A=x_C=0$, the distance of point 1 is $x_1=80\text{mm}$, the distance of point B is $x_B=\left(80+40\right)\text{mm}=120\text{mm}$, and the distance of point 2 is $x_2=\left(120+80\right)\text{mm}=200\text{mm}$.

Velocity at point 1.

The gear A stationary, the velocity at point 1 is $v_1=0$.

Velocity at point 2.

Determine the velocity at point 2 using the relation.

$v_2=x_2{\omega }_C$

Substitute 200 mm for $x_2$ and 20 rad/s for ${\omega }_C$.

$\begin{array}{c}{v}_2=200\times 20\\ =4,000\text{mm/s}(\downarrow )\end{array}$

Point 1 is the instantaneous center of gear B.

Determine the angular velocity at point B using the relation.

$v_2=\left(x_2-x_1\right){\omega }_B$

Substitute 4,000 mm/s for $v_2$, 200 mm for $x_2$, 80 mm for $x_1$, and 20 rad/s for ${\omega }_B$.

$\begin{array}{l}4,000=\left(200-80\right)\times {\omega }_B\\ \frac{4,000}{120}={\omega }_B\\ {\omega }_B=33.33\text{rad/s}\end{array}$

Determine the velocity at point B using the relation.

$v_B=\left(x_B-x_1\right){\omega }_B$

Substitute 120 mm for $x_B$, 80 mm for $x_1$, and 33.33 rad/s for ${\omega }_B$.

$\begin{array}{c}{v}_B=\left(120-80\right)\times 33.33\\ =1,333.2\text{mm/s}\end{array}$

Determine the angular velocity of arm AB using the relation.

${\omega }_{AB}=\frac{v_B}{x_B}$

Substitute 1,333.2 mm/s for $v_B$ and 120 mm for $x_B$.

$\begin{array}{c}{\omega }_{AB}=\frac{1,333.2}{120}\\ =11.11\text{rad/s}\end{array}$

Therefore, the angular velocity of arm AB $\left({\text{ω}}_{AB}\right)$ is $\underset{\_}{11.11\text{rad/s}\left(\text{Clockwise}\right)}$.

(b)

To determine

The angular velocity of gear B.

Answer to Problem 15.251RP

The angular velocity of gear B $\left({\text{ω}}_B\right)$ is $\underset{\_}{33.33\text{rad/s}\left(\text{Clockwise}\right)}$.

Explanation of Solution

Given information:

The constant angular acceleration of gear C $\left({\alpha }_C\right)$ is $4{\text{rad/s}}^{\text{2}}$.

The given time t is 5 s.

Calculation:

Refer part (a).

The angular velocity of gear B is 33.33 rad/s.

Therefore, the angular velocity of gear B $\left({\text{ω}}_B\right)$ is $\underset{\_}{33.33\text{rad/s}\left(\text{Clockwise}\right)}$.

(c)

To determine

The acceleration of the point on gear B that is in contact with gear A.

Answer to Problem 15.251RP

The acceleration of the point on gear B that is in contact with gear A $\left({\text{a}}_1\right)$ is $\underset{\_}{29.6{\text{m/s}}^{\text{2}}(\to )}$.

Explanation of Solution

Given information:

The constant angular acceleration of gear C $\left({\alpha }_C\right)$ is $4{\text{rad/s}}^{\text{2}}$.

The given time t is 5 s.

Calculation:

The tangential component of acceleration at point 1 is zero ${\left(a_1\right)}_t=0$.

Determine the tangential component of acceleration at point 2.

$\begin{array}{c}{\left(a_2\right)}_t={\left(a_C\right)}_t\\ =r_C{\alpha }_2\end{array}$

Here, $r_C$ is the distance of point C.

Substitute 200 mm for $r_C$ and $4{\text{rad/s}}^{\text{2}}$ for ${\alpha }_2$.

$\begin{array}{c}{\left(a_2\right)}_t=200\times 4\\ =800{\text{mm/s}}^{\text{2}}(\downarrow )\end{array}$

Determine the angular acceleration at point B.

${\alpha }_B=\frac{{\left(a_2\right)}_t}{\left(x_2-x_1\right)}$

Substitute $800{\text{mm/s}}^{\text{2}}$ for ${\left(a_2\right)}_t$, 200 mm for $x_2$, and 80 mm for $x_1$.

$\begin{array}{c}{\alpha }_B=\frac{800}{120}\\ =6.67{\text{rad/s}}^{\text{2}}\end{array}$

Determine the tangential component of acceleration at point B.

${\left(a_B\right)}_t=\left(x_B-x_1\right){\alpha }_B$

Substitute 120 mm for $x_B$, and 80 mm for $x_1$, and $6.67{\text{rad/s}}^{\text{2}}$ for ${\alpha }_B$.

$\begin{array}{c}{\left(a_B\right)}_t=\left(120-80\right)\times 6.67\\ =266.8{\text{mm/s}}^{\text{2}}\end{array}$

Determine the angular acceleration at point AB.

${\alpha }_{AB}=\frac{{\left(a_B\right)}_t}{r_B}$

Here, $r_B$ is the distance of point B.

Substitute $266.8{\text{mm/s}}^{\text{2}}$ for ${\left(a_B\right)}_t$ and 120 mm for $r_B$.

$\begin{array}{c}{\alpha }_{AB}=\frac{266.8}{120}\\ =2.223{\text{rad/s}}^{\text{2}}\end{array}$

Determine the acceleration of point B using the relation.

${\text{a}}_B=\left[{\alpha }_{AB}\times x_B\downarrow \right]+\left[{\omega }_{AB}^2{x}_B\leftarrow \right]$

Substitute $2.223{\text{rad/s}}^{\text{2}}$ for ${\alpha }_{AB}$, 120 mm for $x_B$, and 11.11 rad/s for ${\omega }_{AB}$.

$\begin{array}{c}{\text{a}}_B=\left[2.223\times 120\downarrow \right]+\left[{11.11}^2\times 120\leftarrow \right]\\ =\left[266.8{\text{mm/s}}^{\text{2}}\downarrow \right]+\left[14,812{\text{mm/s}}^{\text{2}}\leftarrow \right]\end{array}$

Determine the acceleration of point 1 on gear B using the relation.

${\text{a}}_1={\text{a}}_B+\left[{\alpha }_B\left(x_B-x_1\right)\uparrow \right]+\left[{\omega }_B^2\left(x_B-x_1\right)\to \right]$

Substitute $\left[266.8{\text{mm/s}}^{\text{2}}\downarrow \right]+\left[14.812{\text{mm/s}}^{\text{2}}\leftarrow \right]$ for ${\text{a}}_B$, $6.67{\text{rad/s}}^{\text{2}}$ for ${\alpha }_B$, 120 mm for $x_B$, and 80 mm for $x_1$, and $33.33\text{rad/s}$ for ${\text{ω}}_B$.

$\begin{array}{c}{\text{a}}_1=\left\{\begin{array}{l}\left[266.8\downarrow \right]+\left[14.812\leftarrow \right]+\\ \left[6.67\left(120-80\right)\uparrow \right]+\left[{33.33}^2\left(120-80\right)\to \right]\end{array}\right\}\\ =\left[266.8\downarrow \right]+\left[14,812\leftarrow \right]+\left[266.8\uparrow \right]+\left[44,435\to \right]\\ =29,623{\text{mm/s}}^{\text{2}}\times \frac{1\text{m}}{1,000\text{mm}}\\ =29.6{\text{m/s}}^2(\to )\end{array}$

Therefore, the acceleration of the point on gear B that is in contact with gear A $\left({\text{a}}_1\right)$ is $\underset{\_}{29.6{\text{m/s}}^{\text{2}}(\to )}$.