Chapter 15, Problem 15.257RP

Two rods AE and BD pass through holes drilled into a hexagonal block. (The holes are drilled in different planes so that the rods will not touch each other.) Knowing that rod AE has an angular velocity of 20 rad/s clockwise and an angular acceleration of 4 rad/s² counterclockwise when $\theta =90°$ , determine (a) the relative velocity of the block with respect to each rod, (b) the relative acceleration of the block with respect to each rod.

Fig. P15.257

To determine

(a)

Relative velocity of block with respect to each rod.

Answer to Problem 15.257RP

Relative velocity of block with respect to each rod

$\begin{array}{ll}AE: u_1\to =2m/s\hfill & \hfill \\ BD: u_2\nearrow 60°=0\hfill & \hfill \end{array}$

Explanation of Solution

Given information:

The angular velocity and angular acceleration of rod AE is $20rad/s,4rad/s^2$

$\theta =90°$

The Coriolis acceleration is a combination of

$a_P=a_{P^1}+a_{P/F}+a_C$

The Coriolis acceleration id defined as

$a_C=2\Omega \times v_{P/F}$

The velocity is defined as

$v=r\omega$

The normal acceleration is defined as

$a_n=r{\omega }^2$

The tangential acceleration is defined as

$a_t=r\alpha$

Calculation:

When $\theta =90°$, the above assembly will look like

Apply sine rule,

$\frac{100mm}{\sin 60°}=\frac{r_1}{\sin 30°}=\frac{r_2}{\sin 90°}$

Therefore,

$\begin{array}{l}{r}_1=57.735mm\\ r_2=115.47mm\end{array}$

The angle $A\hat{H}B$ remains at $60°$, therefore road AE and BD have common angular velocities and angular accelerations.

Assume, the relative velocity of slider H on rod AH is $u_1\to$ and acceleration is ${\dot{u}}_1\to$

$H^1$ is a point on rod AE which coincides with H

The velocity of point $H^1$

$v_H{}_{\prime }=[r_1\omega ]=[\left(57.735mm\right)(-20rad/s) ]=1154.7mm/s$

The acceleration of point $H^1$

$\begin{array}{l}{a}_H{}_{\prime }=[r_1\alpha \uparrow ]+[r_1{\omega }^2\leftarrow ]\\ =[\left(57.735mm\right)(4rad/s^2)\uparrow ]+[\left(57.735mm\right){\left(20rad/s\right)}^2\leftarrow ]\\ =[230.9mm/s^2\uparrow ]+[23094mm/s^2\leftarrow ]\\ \end{array}$

The relevant Coriolis acceleration is

$a_1=[2\omega u_1\uparrow ]=[\left(2\right)(-20)u_1\uparrow ]=40u_1\downarrow$

The velocity of point H,

$v_H =v_H{}_{\prime } +u_1\to =[1154.7mm ]+[u_1\to ]\to \left(1\right)$

The acceleration of point H

$\begin{array}{l}{a}_H =a_{H^1} +[{\dot{u}}_1\to ]+[40u_1\to ]\\ =[230.9mm/s^2\uparrow ]+[23094mm/s^2\leftarrow ]+[{\dot{u}}_1\to ]+[40u_1\to ]\to \left(2\right)\end{array}$

Assume, the relative velocity of slider H on rod BD is $\nearrow u_2$ and acceleration is $\nearrow {\dot{u}}_2$ at $60°$ with horizontal.

$H^{11}$ is a point on rod BD which coincides with H

The velocity of point $H^{11}$

$v_H{}_{\prime \prime }=r_2\omega \nwarrow 30°=\left(115.47mm\right)(-20rad/s) \nwarrow 30°=2309.4mm/s \searrow 30°$

The acceleration of point $H^{11}$

$\begin{array}{l}{a}_H{}_{\prime \prime }=r_2\alpha \nwarrow 30°+r_2{\omega }^2\swarrow 60°\\ =\left(115.47mm\right)(4rad/s^2) \nwarrow 30°+\left(115.47mm\right){\left(20rad/s\right)}^2 \swarrow 60°\\ =461.9mm/s^2 \nwarrow 30°+46188mm/s^2\searrow 60°\end{array}$

The relevant Coriolis acceleration

$a_2=2\omega u_2\nwarrow 30°=\left(2\right)(-20rad/s)u_2\nwarrow 60°=40u_2\searrow 30°$

The velocity of point H,

$v_H =v_{H^{11}} +u_2\nearrow 60°=2309.4mm/s \searrow 30°+u_2\nearrow 60°\to \left(3\right)$

The acceleration of point H

$\begin{array}{l}{a}_H =a_H{}_{\prime \prime }+{\dot{u}}_2\nearrow 60°+40u_2\searrow 30°\\ =461.9mm/s^2 \nwarrow 30°+46188mm/s^2\searrow 60°+{\dot{u}}_2\nearrow 60°+40u_2\searrow 30°\to \left(4\right)\end{array}$

Equate $v_H$ in equations 1 and 3,

$\begin{array}{l}\uparrow u_2=\frac{-1154.7+\left(2309.4\right)sin30°}{sin60°}\\ \uparrow u_2=0\end{array}$

$\begin{array}{l}\to u_1=2309.4cos30°+0\\ \to u_1=2000mm/s\end{array}$

Therefore, the relative velocities

$\begin{array}{ll}AE: u_1\to =2m/s\hfill & \hfill \\ BD: u_2\nearrow 60°=0\hfill & \hfill \end{array}$

Conclusion:

The relative velocities of each rod

$\begin{array}{ll}AE: u_1\to =2m/s\hfill & \hfill \\ BD: u_2\nearrow 60°=0\hfill & \hfill \end{array}$

To determine

(b)

Relative acceleration of block with respect to each rod.

Answer to Problem 15.257RP

Relative acceleration of block with respect to each rod $\begin{array}{ll}AE:{\dot{u}}_1\leftarrow =23.5m/s^2\hfill & \hfill \\ BD: {\dot{u}}_2\nearrow 60°=46.2m/s^2\swarrow 60°\hfill & \hfill \end{array}$

Explanation of Solution

Given information:

The angular velocity and angular acceleration of rod AE is $20rad/s,4rad/s^2$

$\theta =90°$

The Coriolis acceleration is a combination of

$a_P=a_{P^1}+a_{P/F}+a_C$

The Coriolis acceleration id defined as

$a_C=2\Omega \times v_{P/F}$

The velocity is defined as

$v=r\omega$

The normal acceleration is defined as

$a_n=r{\omega }^2$

The tangential acceleration is defined as

$a_t=r\alpha$

Calculation:

According to sub part a

$\begin{array}{l}{a}_H =a_{H^1} +[{\dot{u}}_1\to ]+[40u_1\to ]\\ =[230.9mm/s^2\uparrow ]+[23094mm/s^2\leftarrow ]+[{\dot{u}}_1\to ]+[40u_1\to ]\to \left(2\right)\end{array}$

$\begin{array}{l}{a}_H =a_H{}_{\prime \prime }+{\dot{u}}_2\nearrow 60°+40u_2\searrow 30°\\ =461.9mm/s^2 \nwarrow 30°+46188mm/s^2\searrow 60°+{\dot{u}}_2\nearrow 60°+40u_2\searrow 30°\to \left(4\right)\end{array}$

$\begin{array}{ll}AE: u_1\to =2m/s\hfill & \hfill \\ BD: u_2\nearrow 60°=0\hfill & \hfill \end{array}$

Substitute for $u_1,u_2$ in Coriolis accelerations and equate components for $a_H$

$\begin{array}{l}\uparrow 230.9mm/s^2-\left(40rad/s\right)\left(2000mm/s\right)\\ =(461.9mm/s^2)sin30°+(46188mm/s^2)sin60°+ {\dot{u}}_{2}sin60°-\left(40\right)\left(0\right)sin30°\end{array}$

Therefore

$\begin{array}{l}{\dot{u}}_2=\frac{230.9-80000-230.9+40000}{sin60°}\\ {\dot{u}}_2=-46188mm/s^2\end{array}$

$\begin{array}{l}{\dot{u}}_1-23094mm/s^2=-(461.4mm/s^2)cos30°-(46188mm/s^2)cos60°\\ -(46188mm/s^2)cos60°+0\end{array}$

Therefore

${\dot{u}}_1=-23494mm/s^2$

Conclusion:

The relative acceleration of each rod

$\begin{array}{ll}AE:{\dot{u}}_1\leftarrow =23.5m/s^2\hfill & \hfill \\ BD: {\dot{u}}_2\nearrow 60°=46.2m/s^2\swarrow 60°\hfill & \hfill \end{array}$