Chapter 15.3, Problem 15.98P

Two rods AB and DE are connected as shown. Knowing that point D moves to the left with a velocity of 40 in./s, determine (a) the angular velocity of each rod, (b) the velocity of point A.

To determine

(a)

The angular velocity of the rod AB at instant.

The angular velocity of the rod DE at instant.

Answer to Problem 15.98P

The angular velocity of the rod AB is $1.17647\text{rad}/\text{s}$.

The angular velocity of the rod DE is $2.5\text{rad}/\text{s}$.

Explanation of Solution

Given information:

The velocity of the point D is $40\text{in}/\text{s}$.

The following figure represents two instantaneous centers of the given system.

Figure-(1)

Write the expression for the distance between the point $C_1$ and point B.

$B{C}_1=\sqrt{{\left(EB\right)}^2+{\left(H{C}_1\right)}^2}$ ....... (I)

Here, the distance between the point $C_1$ and point B is $B{C}_1$, the distance between the point E and point B is $EB$ and the distance between the point $C_1$ and point H is $H{C}_1$.

Write the equation for the distance between the point $C_1$ and point D.

$D{C}_1=EB+DB$ ...... (II)

Here, the horizontal distance between the point $C_1$ and point D is $D{C}_1$ and the distance between the point D and point B is $DB$.

The following figure represents the instantaneous centre of the rod AB.

Figure-(2)

Write the expression for the horizontal distance between the point $C_2$ and point B.

$B{C}_2=\sqrt{{\left(C_2{C}_3\right)}^2+{\left(B{C}_3\right)}^2}$ ....... (III)

Here, the point $C_2$ and point D is $B{C}_2$, the point $C_3$ and point B is $B{C}_3$ and the distance between the point $C_3$ and the point $C_2$ is $C_2{C}_3$.

Write the expression for the velocity of the point D.

${\omega }_{DE}=\frac{v_D}{D{C}_1}$ ....... (IV)

Here, the velocity of the point D is $v_D$, the angular velocity of the rod DE is ${\omega }_{DE}$.

Write the expression for the velocity of the point B.

${\omega }_{AB}=\frac{v_B}{B{C}_1}$ ....... (V)

Here, the velocity of the point B is $v_B$, the angular velocity of the rod DE is ${\omega }_{AB}$.

Write the expression for the velocity of the point B.

$v_B={\omega }_{AB}\times B{C}_2$ ....... (VI)

Here, the angular velocity of the rod DE is ${\omega }_{AB}$.

Calculation:

Substitute $8\text{in}$ for $EB$ and $8\text{in}$ for $DB$ in Equation (II).

$\begin{array}{c}D{C}_1=8\text{in}+8\text{in}\\ \text{=16}\text{in}\end{array}$

Substitute $\text{16}\text{in}$ for $D{C}_1$ and $40\text{in}/\text{s}$ for $v_D$ in Equation (IV).

$\begin{array}{c}{\omega }_{DE}=\frac{40\text{in}/\text{s}}{16\text{in}}\\ =2.5\text{rad}/\text{s}\end{array}$

Substitute $8\text{in}$ for $EB$ and $8\text{in}$ for $H{C}_1$ in Equation (I).

$\begin{array}{c}B{C}_1=\sqrt{{\left(8\text{in}\right)}^2+{\left(8\text{in}\right)}^2}\\ =\sqrt{128{\text{in}}^2}\\ =11.313\text{in}\end{array}$

Substitute $11.313\text{in}$ for $B{C}_1$ and $2.5\text{rad}/\text{s}$ for ${\omega }_{DE}$ in Equation (V).

$\begin{array}{l}2.5\text{rad}/\text{s}=\frac{v_B}{11.313\text{in}}\\ v_B=2.5\text{rad}/\text{s}\times 11.313\text{in}\\ v_B=\text{28}\text{.2842}\text{in}/\text{s}\end{array}$

Substitute $17\text{in}$ for $B{C}_3$ and $17\text{in}$ for $C_2{C}_3$ in Equation (III).

$\begin{array}{c}B{C}_2=\sqrt{{\left(17\text{in}\right)}^2+{\left(\text{17}\text{in}\right)}^2}\\ =\sqrt{578{\text{in}}^2}\\ =24.0416{\text{in}}^2\end{array}$

Substitute $24.0416{\text{in}}^2$ for $B{C}_2$ and $\text{28}\text{.2842}\text{in}/\text{s}$ for $v_B$ in Equation (VI).

$\begin{array}{l}\text{28}\text{.2842}\text{in}/\text{s}={\omega }_{AB}\times 24.0416\text{in}\\ {\omega }_{AB}=\frac{\text{28}\text{.2842}\text{in}/\text{s}}{24.0416\text{in}}\\ {\omega }_{AB}=1.17647\text{rad}/\text{s}\end{array}$

Conclusion:

The angular velocity of the rod DE is $2.5\text{rad}/\text{s}$.

The angular velocity of the rod AB is $1.17647\text{rad}/\text{s}$.

To determine

(b)

The velocity at the point A.

Answer to Problem 15.98P

The velocity at the point A is $29.41175\text{in}/\text{s}$.

Explanation of Solution

Write the expression for the horizontal distance between the point $C_2$ and point A.

$A{C}_2=D{C}_1+C_1{C}_2$ ........ (VII)

Here, the vertical distance between the point $C_1$ and point $C_2$ is $C_1{C}_2$ and the horizontal distance between the point $C_2$ and point A is $A{C}_2$.

Write the expression for the velocity of the point A.

$v_A=\left(A{C}_2\right){\omega }_{AB}$ ....... (VIII)

Here, the velocity at the point A is $v_A$.

Calculation:

Substitute $\text{16}\text{in}$ for $D{C}_1$ and $9\text{in}$ for $C_1{C}_2$ in Equation (VII).

$\begin{array}{c}A{C}_2=16\text{in}+9\text{in}\\ =\text{25}\text{in}\end{array}$

Substitute $\text{25}\text{in}$ for $A{C}_2$ and $1.17647\text{rad}/\text{s}$ for ${\omega }_{AB}$ in Equation (VIII).

$\begin{array}{c}{v}_A=\left(25\text{in}\right)\left(1.17647\text{rad}/\text{s}\right)\\ =29.41175\text{in}/\text{s}\end{array}$

Conclusion:

The velocity at the point A is $29.41175\text{in}/\text{s}$.