Chapter 15.3, Problem 15.95P

Two 25-in. rods are pin-connected at D as shown. Knowing that B moves to the left with a constant velocity of 24 in./s, determine at the instant shown (a) the angular velocity of each rod, (b) the velocity of E.

    

To determine

(a)

The angular velocity of the rod $AB$ at instant.

The angular velocity of the rod $DE$ at instant.

Answer to Problem 15.95P

The angular velocity of the rod $AB$ at instant is $1.2\text{rad}/\text{s}$.

The angular velocity of the rod $DE$ at instant is $0.4499\text{rad}/\text{s}$.

Explanation of Solution

Given information:

The angular velocity of the rod $B$ is $24\text{in}/\text{s}$.

The following figure represents the rod $AB$.

Figure-(1)

Write the expression for the angular velocity of the rod $AB$.

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

Here, the angular velocity of the rod $AB$ is ${\omega }_{AB}$, the distance between $B$ of the rod and instantaneous center $C_{AB}$ of the rod is $B{C}_{AB}$ and the velocity of $B$ rod is $v_B$.

The following figure represents the rod $DE$.

Figure-(2)

Write the expression for the angular velocity of the rod $DE$.

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

Here, the angular velocity of the rod $DE$ is ${\omega }_{DE}$, the distance between $B$ of the rod and instantaneous center $C_{DE}$ of the rod is $D{C}_{DE}$ and the velocity at the point $D$ is $v_D$.

Write the expression for $D{C}_{AB}$.

${\left(D{C}_{AB}\right)}^2={\left(F{C}_{AB}\right)}^2+{\left(BD\right)}^2$ ....... (III)

Here, the distance between the point $D$ of the rod and instantaneous center $C_{AB}$ of the rod is $D{C}_{AB}$, the vertical distance between point $F$ and the instantaneous center $C_{AB}$ of the rod $F{C}_{AB}$ and the horizontal distance between the point $F$ and the point $D$ is $BD$.

Write the expression for the velocity at the point $D$.

$v_D=\left(D{C}_{AB}\right)\left({\omega }_{AB}\right)$ ....... (IV)

Write the expression for the $D{C}_{DE}$.

$D{C}_{DE}=\sqrt{{\left(DK\right)}^2+{\left(K{C}_{DE}\right)}^2}$ ....... (V)

Here, the distance between the point $D$ and the point $K$ is $DK$, the vertical distance between the point $K$ and the instantaneous center of the rod $C_{DE}$ is $K{C}_{DE}$.

Calculation:

Substitute $20\text{in}$ for $B{C}_{AB}$ and $24\text{in}/\text{s}$ for $B{C}_{AB}$ in Equation (I).

$\begin{array}{c}{\omega }_{AB}=\frac{24\text{in}/\text{s}}{20\text{in}}\\ =1.2\text{rad}/\text{s}\end{array}$

The angular velocity of the rod $AB$ at instant is $1.2\text{rad}/\text{s}$.

Substitute $10\text{in}$ for $F{C}_{AB}$ and $7.5\text{in}$ for $BD$ in Equation (III).

$\begin{array}{l}{\left(D{C}_{AB}\right)}^2={\left(10\text{in}\right)}^2+{\left(7.5\text{in}\right)}^2\\ {\left(D{C}_{AB}\right)}^2=100{\text{in}}^2+56.25{\text{in}}^2\\ D{C}_{AB}=\sqrt{156.25{\text{in}}^2}\\ D{C}_{AB}=12.5\text{in}\end{array}$

Substitute $12.5\text{in}$ for $D{C}_{AB}$ and $1.2\text{rad}/\text{s}$ for ${\omega }_{AB}$ in Equation (IV).

$\begin{array}{c}{v}_D=\left(12.5\text{in}\right)\left(1.2\text{rad}/\text{s}\right)\\ =15\text{in}/\text{s}\end{array}$

Substitute $20\text{in}$ for $DK$ and $26.667\text{in}$ for $K{C}_{DE}$ in Equation (V).

$\begin{array}{c}D{C}_{DE}=\sqrt{{\left(20\text{in}\right)}^2+{\left(26.667\text{in}\right)}^2}\\ =\sqrt{400{\text{in}}^2+711.1128{\text{in}}^2}\\ =\sqrt{1111.1128{\text{in}}^2}\\ =33.3334\text{in}\end{array}$

Substitute $33.3334\text{in}$ for $D{C}_{DE}$ and $15\text{in}/\text{s}$ for $v_D$ in Equation (II).

$\begin{array}{c}{\omega }_{DE}=\frac{15\text{in}/\text{s}}{33.3334\text{in}}\\ =0.4499\text{rad}/\text{s}\end{array}$

The angular velocity of the rod $DE$ at instant is $0.4499\text{rad}/\text{s}$.

Conclusion:

The angular velocity of the rod $AB$ at instant is $1.2\text{rad}/\text{s}$.

The angular velocity of the rod $DE$ at instant is $0.4499\text{rad}/\text{s}$.

To determine

(b)

The velocity of the point $E$ at instant.

Answer to Problem 15.95P

The velocity of the point $E$ at instant is $5.250\text{in}/\text{s}$.

Explanation of Solution

Given information:

The angular velocity of the rod $B$ is $24\text{in}/\text{s}$.

Write the expression for the point $E$.

$v_E=\left(E{C}_{DE}\right)\left({\omega }_{DE}\right)$ ....... (VI)

Here, the angular velocity of the point $DE$ is ${\omega }_{DE}$, the velocity of the point $E$ is $v_E$ the distance between the point $E$ and instantaneous center of the rod $DE$ is $E{C}_{DE}$.

Calculation:

Substitute $11.667\text{in}$ for $D{C}_{DE}$ and $0.4499\text{rad}/\text{s}$ for ${\omega }_{DE}$ in Equation (VI).

$\begin{array}{c}{v}_E=\left(0.4499\text{rad}/\text{s}\right)\left(11.667\text{in}\right)\\ =5.250\text{in}/\text{s}\end{array}$

The velocity of the point $E$ at instant is $5.250\text{in}/\text{s}$.

Conclusion:

The velocity of the point $E$ at instant is $5.250\text{in}/\text{s}$.