Chapter 15.3, Problem 15.84P

Knowing that at the instant shown the angular velocity of rod BE is 4 rad/s counterclockwise, determine (a) the angular velocity of rod AD. (b) the velocity of collar D. (c) the velocity of point A.

Fig. 15.84

To determine

(a)

The angular velocity of the rod $AD$.

Answer to Problem 15.84P

The angular velocity of the rod $AD$ ,

${\omega }_{AD}=4.27 rad s^{-1}$

Explanation of Solution

Given information:

At the instant shown, the angular velocity of rob $BE$ is $4 rad s^{-1}$, counterclockwise.

Concept used:

The angular velocity ($\omega$ ),

$\omega =\frac{v}{r}$

$r-$ Radius

$v-$ Velocity

Calculation:

At the instant shown, the angular velocity of rob $BE$ is $4 rad s^{-1}$, counterclockwise.

Therefore the velocity at point $B$ should be right directed and the velocity at point $D$ should

be downward directed. (The point $D$ can only move vertically.)

Thus, the instantaneous center can be found by drawing the lines perpendicular to $v_A$ and $v_B$.

Point $C$ is the instantaneous center.

$\beta ={30}^0$

$AB=240 mm$

$BD=360 mm$

$BE=192 mm$

Consider the motion of the rod $EB$.

${\omega }_{EB}=4 rad s^{-1}$

The velocity at point $B$

$v=r\omega$

$v_B=BE\times {\omega }_{BE}$

$v_B=0.192 m\times 4 rad s^{-1}$

$v_B=0.768 m s^{-1}\to$

Then, consider the motion of the point $B$ around the instantaneous center $C$.

$BC=0.36\sin {30}^{0}m=0.18 m$

The angular velocity the rod $AD$ can be calculated as follows.

$\omega =\frac{v}{r}$

${\omega }_{AD}=\frac{v_B}{BC}$

${\omega }_{AD}=\frac{0.768 m s^{-1}}{0.18 m}$

${\omega }_{AD}=4.27 rad s^{-1}$

Thus, the angular velocity of the rod $AD$ ,

${\omega }_{AD}=4.27 rad s^{-1}$

Conclusion:

The angular velocity of the rod $AD$ ,

${\omega }_{AD}=4.27 rad s^{-1}$

To determine

(b)

The velocity of collar $D$.

Answer to Problem 15.84P

The velocity of collar $D$, $v_D=1.332 m s^{-1}$

Explanation of Solution

Given information:

At the instant shown, the angular velocity of rob $BE$ is $4 rad s^{-1}$, counterclockwise.

Concept used:

The angular velocity ($\omega$ ),

$\omega =\frac{v}{r}$

$r-$ Radius

$v-$ Velocity

Calculation:

Point $C$ is the instantaneous center.

$\beta ={30}^0$

$AB=240 mm$

$BD=360 mm$

$BE=192 mm$

Consider the motion of point $D$ around the instantaneous center $C$.

$CD=BC \cos {30}^0=0.36 m\times \cos {30}^0=0.312 m$

From part (a), ${\omega }_{AD}=4.27 rad s^{-1}$

The velocity of point $D$ can be calculated as follows.

$v=r\omega$

$v_D=CD\times {\omega }_{AD}$

$v_D=0.312 m\times 4.27 rad s^{-1}$

$v_D=1.332 m s^{-1}$

Thus, the velocity of collar $D$, $v_D=1.332 m s^{-1}$

Conclusion:

The velocity of collar $D$ ,

$v_D=1.332 m s^{-1}$

To determine

(c)

The velocity of point $A$.

Answer to Problem 15.84P

The velocity of point $A$, $v_A=1.56 m s^{-1}$

Explanation of Solution

Given information:

At the instant shown, the angular velocity of rob $BE$ is $4 rad s^{-1}$, counterclockwise.

Concept used:

The angular velocity ($\omega$ ),

$\omega =\frac{v}{r}$

$r-$ Radius

$v-$ Velocity

Calculation:

Point $C$ is the instantaneous center.

$\beta ={30}^0$

$AB=240 mm$

$BD=360 mm$

$BE=192 mm$

Consider the motion of point $A$ around the instantaneous center $C$.

$\tan \alpha =\frac{AP}{PC}$

$AP=AD\sin {30}^0=0.6\sin {30}^0 m$

$PC=AB\cos {30}^0=0.24\cos {30}^0 m$

$\tan \alpha =\frac{0.6\sin {30}^0 m}{0.24\cos {30}^0 m}=1.4434$

$\alpha ={55.3}^0$

From part (a), ${\omega }_{AD}=4.27 rad s^{-1}$

$AC=\frac{AB \cos {30}^0}{\cos {55.3}^0}=\frac{0.24\cos {30}^0}{\cos {55.3}^0} m =0.365 m$

The velocity of point $A$ can be calculated as follows.

$v=r\omega$

$v_A=AC\times {\omega }_{AD}$

$v_A=0.365 m\times 4.27 rad s^{-1}$

$v_A=1.56 m s^{-1}$

Thus, the velocity of point $A$, $v_A=1.56 m s^{-1}$

Conclusion:

The velocity of point $A$, $v_A=1.56 m s^{-1}$