Chapter 15.5, Problem 15.161P

To determine

(a)

Angular acceleration of bar BD.

Answer to Problem 15.161P

The angular acceleration ${\alpha }_{BD}$ of rod BD is $54rad/s^2$ clockwise.

Explanation of Solution

Given information:

Constant angular velocity of rod AE is $6rad/s$ clockwise.

The constant relative velocity of pin P with respect to rod AE is $8rad/s$.

Plane motion of a particle relative to a rotating frame is defined as

$v_P=v_{P^1}+v_{P/F}$

In above equation

$v_P$ - Absolute velocity of particle P.

$v_{P^1}$ - Velocity of point $P^1$ of moving frame F Coinciding with P.

$v_{P/F}$ - Velocity of P relative to moving frame F.

The Coriolis acceleration is a combination of:

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

Where

$a_P$ - Absolute acceleration of particle P.

$a_{P^1}$ - Acceleration of point P¹ of moving frame F coinciding with P

$a_{P/F}$ - Acceleration of P relative to moving frame F

$a_C$ - The Coriolis acceleration

The Coriolis acceleration is defined as:

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

Calculation:

Length of AP

$AP=16in=1.333ft$

Length of BP

$BP=16\sqrt{2}in=1.333\sqrt{2}ft$

Velocity $v_{P^1}$ of coinciding point $P^1$ on rod AE

$v_P=v_{P^1}+v_{P/AE}=\left(8ft/s\right)i+\left(8ft/s\right)j\to \left(1\right)$

Velocity $v_{P/AE}$ of pin P relative to rod AE

$v_{P/AE}=\left(8ft/s\right)j$

Velocity $v_P$ of point P

$v_P=v_{P^1}+v_{P/AE}=\left(8ft/s\right)i+\left(8ft/s\right)j$

Velocity $v_{P^{11}}$ of coinciding point $P^{11}$ on rod BD:

$\begin{array}{l}{v}_{P^{11}}=\left(BP\right){\omega }_{BD}\nwarrow 45°=\left(1.333\sqrt{2}ft\right){\omega }_{BD}\nwarrow 45°\\ v_{P^{11}}=-1.333{\omega }_{BD}i+1.333{\omega }_{BD}j\end{array}$

Velocity $v_{P/BD}$ of point P relative to rod BD:

$v_{P/BD}=\left(\cos 45°\right)ui+\left(\sin 45°\right)uj$

Velocity $v_P$ of point P:

$\begin{array}{l}{v}_P=v_{P^{11}}+v_{P/BD}\\ v_P=-1.333{\omega }_{BD}i+1.333{\omega }_{BD}j+\left(\cos 45°\right)ui+\left(\sin 45°\right)uj\to \left(2\right)\end{array}$

Equate components in equations 1 and 2.

$\begin{array}{l}i:8=-1.333{\omega }_{BD}+\left(\cos 45°\right)u\\ j:8=1.333{\omega }_{BD}+\left(\sin 45°\right)u\end{array}$

Solve above equations:

$u=8\sqrt{2}ft/s$

${\omega }_{BD}=0$

Therefore:

$v_{P/BD}=\left(8ft/s\right)i+\left(8ft/s\right)j$

Acceleration $a_{P^1}$ of coinciding point $P^1$ on rod AE

$\begin{array}{l}{a}_{P^1}=\left(AP\right){\alpha }_{AE}i-\left(AP\right){\omega }_{AE}{}^{2}j\\ =0-\left(1.333\right){\left(6\right)}^{2}j\\ =-\left(48ft/s^2\right)j\end{array}$

The Coriolis acceleration $a_C$ :

$a_C=2{\omega }_{AE}\times v_{P/AE}=2\left(-6k\right)\times \left(8j\right)=\left(96ft/s^2\right)i$

The acceleration $a_P$ of point P:

$a_P=a_{P^1}+a_{P/AE}+a_C=\left(96ft/s^2\right)i-\left(48ft/s^2\right)j\to \left(1\right)$

Acceleration $a_{P^{11}}$ of coinciding point $P^1$ on rod BD

$a_{P^{11}}={\alpha }_{BD}k\times r_{B/P}-{\omega }_{BD}{}^2{r}_{B/P}=-1.333{\alpha }_{BD}i+1.333{\alpha }_{BD}j+0$

Acceleration $a_{P/BD}$ of point P relative to rod BD

$a_{P/BD}=\left(\cos 45°\right)a_{r}i+\left(\sin 45°\right)a_{r}j$

The Coriolis acceleration $a_C$ :

$a_C=2{\omega }_{BD}\times v_{P/BD}=0$

The acceleration $a_P$ of point P:

$\begin{array}{l}{a}_P=a_{P^{11}}+a_{P/BD}+a_C\\ a_P=-1.333{\alpha }_{BD}i+1.333{\alpha }_{BD}j+\left(\cos 45°\right)a_{r}i+\left(\sin 45°\right)a_{r}j\to \left(2\right)\end{array}$

Equate components in equations 1 and 2.

$\begin{array}{l}i:96=-1.333{\alpha }_{BD}+\left(\cos 45°\right)a_r\\ j:-48=1.333{\alpha }_{BD}+\left(\sin 45°\right)a_r\end{array}$

Solve above equations.

$\begin{array}{l}{\alpha }_{BD}=-54rad/s^2\\ a_r=24\sqrt{2}ft/s^2\end{array}$

Conclusion:

The angular acceleration ${\alpha }_{BD}$ of rod BD is $54rad/s^2$ clockwise.

To determine

(b)

The relative acceleration of pin P with respect to bar BD

Answer to Problem 15.161P

The relative acceleration $a_{P/BD}$ of point P with respect to rod BD is $33.94ft/s^2$ at an angle $45°$ with horizontal.

Explanation of Solution

Given information:

Constant angular velocity of rod AE is $6rad/s$ clockwise.

The constant relative velocity of pin P with respect to rod AE is $8rad/s$.

Plane motion of a particle relative to a rotating frame is defined as

$v_P=v_{P^1}+v_{P/F}$

In above equation

$v_P$ - Absolute velocity of particle P.

$v_{P^1}$ - Velocity of point $P^1$ of moving frame F Coinciding with P.

$v_{P/F}$ - Velocity of P relative to moving frame F.

The Coriolis acceleration is a combination of

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

Where

$a_P$ - Absolute acceleration of particle P.

$a_{P^1}$ - Acceleration of point P¹ of moving frame F coinciding with P

$a_{P/F}$ - Acceleration of P relative to moving frame F

$a_C$ - The Coriolis acceleration

The Coriolis acceleration is defined as

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

Calculation:

According to sub part a

We have found

$\begin{array}{l}96=-1.333{\alpha }_{BD}+\left(\cos 45°\right)a_r\to \left(1\right)\\ -48=1.333{\alpha }_{BD}+\left(\sin 45°\right)a_r\to \left(2\right)\end{array}$

Acceleration $a_{P/BD}$ of point P relative to rod BD

$a_{P/BD}=\left(\cos 45°\right)a_{r}i+\left(\sin 45°\right)a_{r}j$

Solve above equations

$\begin{array}{l}{\alpha }_{BD}=-54rad/s^2\\ a_r=24\sqrt{2}ft/s^2\end{array}$

Therefore

$a_{P/BD}=24i+24j$

The magnitude and angle $\beta$ of relative acceleration $a_{P/BD}$

$a_{P/BD}=\sqrt{{24}^2+{24}^2}=33.94ft/s^2$

$\beta ={\tan }^{-1}\left(1\right)=45°$

Conclusion:

The relative acceleration $a_{P/BD}$ of point P with respect to rod BD is $33.94ft/s^2$ at an angle $45°$ with horizontal.