Chapter 15.3, Problem 15.97P

At the instant shown, the velocity of collar A is 0.4 m/s to the right and the velocity of collar B is 1 ms to the left, Determine (a) the angular velocity of bar AD. (b) the angular velocity of bar BD, (c) the velocity of point D.

Fig. P15.91

To determine

(a)

The angular velocity of bar AD.

Answer to Problem 15.97P

The angular velocity of bar AD is $2.488rad/s$ counter clockwise.

Explanation of Solution

Given information:

Velocity of collar A is $0.4m/s$ to the right.

Velocity of collar B is $1m/s$ to the left.

Draw a diagram locating the instantaneous center C of bar AD and I of bar BD.

The velocity directions of point A, B and D are known where AC perpendicular to $v_A$, BI perpendicular to $v_B$ and CDI perpendicular to $v_D$.

Calculation:

According to the above diagram,

The length AD

$AD=\sqrt{{0.27}^2+{0.36}^2}=0.45m$

The length BD

$BD=\sqrt{{0.18}^2+{0.135}^2}=0.225m$

The angle $\theta$

$\sin \theta =\frac{0.27}{0.45}=0.6$

The angle $\alpha$

$\sin \alpha =\frac{0.18}{0.225}=0.8$

Apply sine rule for triangle $ACD$.

$\frac{CD}{\sin \theta }=\frac{0.45}{\sin \left(90+\beta \right)}=\frac{AC}{\sin \left(90-\beta \right)}$

Rearrange

$CD=\frac{0.45\sin \theta }{\cos \beta }$

Apply sine rule for triangle $BDI$.

$\frac{DI}{\sin \alpha }=\frac{0.225}{\sin \left(90-\beta \right)}$

Rearrange

$DI=\frac{0.225\sin \alpha }{\cos \beta }$

The length AC

$AC=0.36-0.27\tan \beta$

The length BI

$BI=0.135+0.18\tan \beta$

For bar AD,

The angular velocity of bar AD

${\omega }_{AD}=\frac{v_A}{AC}=\frac{v_D}{CD}$

For bar BD,

The angular velocity of bar BD

${\omega }_{BD}=\frac{v_B}{BI}=\frac{v_D}{DI}$

According to the above equations,

$v_D=\frac{v_A\left(CD\right)}{AC}=\frac{v_B\left(DI\right)}{BI}$

Rearrange,

$AC=\frac{v_A\left(CD\right)}{DI}BI$

Substitute,

$AC=\left(\frac{0.45\sin \theta }{\cos \beta }\right)\left(\frac{\cos \beta }{0.225\sin \alpha }\right)\left(4m/s\right)BI$

Solve,

$\begin{array}{l}AC=\left(\frac{0.45\left(0.6\right)}{0.225\left(0.8\right)}\right)\left(0.4\right)BI\\ AC=\left(0.6\right)BI\end{array}$

Substitute for $AC$ and $BI$

$0.36-0.27\tan \beta =0.6\left(0.135+0.18\tan \beta \right)$

Therefore,

$\begin{array}{l}\tan \beta =0.738\\ \beta =36.43°\end{array}$

Then find $AC$

$AC=0.36-0.27\left(0.738\right)=0.16074m$

The angular velocity of bar AD

${\omega }_{AD}=\frac{v_A}{AC}=\frac{0.4m/s}{0.16074m}=2.488rad/s$

Conclusion:

According to the above explanation, the angular velocity of bar AD is $2.488rad/s$ counter clockwise.

To determine

(b)

The angular velocity of bar BD.

Answer to Problem 15.97P

The angular velocity of bar BD is $3.733rad/s$ clockwise.

Explanation of Solution

Given information:

Velocity of collar A is $0.4m/s$ to the right.

Velocity of collar B is $1m/s$ to the left.

Draw a diagram locating the instantaneous center C of bar AD and I of bar BD.

The velocity directions of point A, B, and D are known where AC perpendicular to $v_A$, BI perpendicular to $v_B$ and CDI perpendicular to $v_D$.

Calculation:

According to sub part a,

$BI=0.135+0.18\tan \beta$

And,

$\tan \beta =0.738$

Therefore,

$BI=0.135+0.18\left(0.738\right)=0.26784m$

Angular velocity of bar BD

${\omega }_{BD}=\frac{v_B}{BI}=\frac{1m/s}{0.26784m}=3.733rad/s$

Conclusion:

According to the above explanation, the angular velocity of bar BD is $3.733rad/s$ clockwise.

To determine

(c)

The velocity of point D.

Answer to Problem 15.97P

The velocity of point D is $0.835m/s$ at an angle $53.57°$ with horizontal.

Explanation of Solution

Given information:

Velocity of collar A is $0.4m/s$ to the right.

Velocity of collar B is $1m/s$ to the left.

Draw a diagram locating the instantaneous center C of bar AD and I of bar BD.

The velocity directions of point A, B, and D are known where AC perpendicular to $v_A$, BI perpendicular to $v_B$ and CDI perpendicular to $v_D$.

Calculation:

According to sub part a,

${\omega }_{AD}=2.488rad/s$

And,

$CD=\frac{0.45\sin \theta }{\cos \beta }$

And,

$\beta =36.43°$

Therefore, the length CD

$CD=\frac{0.45\sin \theta }{\cos \beta }=\frac{0.45\left(0.6\right)}{0.80458}=0.3356m$

The velocity of point D

$v_D={\omega }_{AD}\left(CD\right)$

Substitute,

$v_D=\left(2.488rad/s\right)\left(0.3356m\right)=0.835m/s$

The angle

$90°-\beta =53.57°$

Conclusion:

According to the above explanation, the velocity of point D is $0.835m/s$ at an angle $53.57°$ with horizontal.