Chapter 15.3, Problem 15.85P

To determine

(a)

The angular velocity of rod $AD$.

Answer to Problem 15.85P

The angular velocity of rod $AD$ , ${\omega }_{AD}=5.128 rad s^{-1}$

Explanation of Solution

Given information:

At the instant shown, the velocity of collar $D$ is $1.6 m s^{-1}$, upward.

Concept used:

The angular velocity ($\omega$ ), $\omega =\frac{v}{r}$

$r-$ Radius

$v-$ Velocity

Calculation:

At the instant shown, the velocity of collar $D$ is $1.6 m s^{-1}$, upward.

The rod $EB$ rotates around point $E$. So, at the instant shown, the velocity at point $B$ should be left directed.

Thus, the instantaneous center can be obtained 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 point $D$ around the instantaneous center $C$.

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

$v_D=1.6 m s^{-1}$

The angular velocity of $AD$ can be calculated as follows.

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

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

${\omega }_{AD}=\frac{1.6 m s^{-1}}{0.312 m}$

${\omega }_{AD}=5.128 rad s^{-1}$

Thus, the angular velocity of rod $AD$ , ${\omega }_{AD}=5.128 rad s^{-1}$

Conclusion:

The angular velocity of rod $AD$ , ${\omega }_{AD}=5.128 rad s^{-1}$

To determine

(b)

The velocity of point $B$.

Answer to Problem 15.85P

The velocity of point $B$, $v_B=0.923 m s^{-1}$

Explanation of Solution

Given information:

At the instant shown, the velocity of collar $D$ is $1.6 m s^{-1}$, upward.

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 $B$ around the instantaneous center $C$.

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

From part (a), ${\omega }_{AD}=5.128 rad s^{-1}$

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

$v=r\omega$

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

$v_B=0.18 m\times 5.128 rad s^{-1}$

$v_B=0.923 m s^{-1}$

Thus, the velocity of point $B$, $v_B=0.923 m s^{-1}$

Conclusion:

The velocity of point $B$, $v_B=0.923 m s^{-1}$

To determine

(c)

The velocity of point $A$.

Answer to Problem 15.85P

The velocity of point $A$, $v_A=1.871 m s^{-1}$

Explanation of Solution

Given information:

At the instant shown, the velocity of collar $D$ is $1.6 m s^{-1}$, upward.

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), ${\omega }_{AD}=5.128 rad s^{-1}$

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

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

$v=r\omega$

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

$v_A=0.365 m\times 5.128 rad s^{-1}$

$v_A=1.871 m s^{-1}$

Thus, the velocity of point $A$, $v_A=1.871 m s^{-1}$

Conclusion:

The velocity of point $A$, $v_A=1.871 m s^{-1}$