Chapter 15.5, Problem 15.183P

To determine

The acceleration of pin P.

Answer to Problem 15.183P

The acceleration of pin P is $\underset{\_}{{\text{a}}_P=51.5\text{m}/{\text{s}}^{\text{2}}}$.

Explanation of Solution

Given information:

The constant angular velocity of the bar AD is ${\omega }_{AD}=4\text{rad}/\text{s}$.

The angular velocity of the bar BE is ${\omega }_{BE}=5\text{rad}/\text{s}$.

The angular acceleration of the bar BE is ${\alpha }_{BE}=-2\text{rad}/{\text{s}}^2$.

Calculation:

Calculate the slope of the bar BE $\left(\theta \right)$ as shown below.

$\begin{array}{l}\tan \theta =\frac{0.15}{0.3}\\ \theta ={\tan }^{-1}\left(0.5\right)\\ =26.565°\end{array}$

Calculate the position vector $\left(r\right)$ as shown below.

The position of P with respect to A.

$r_{P/A}=0.1\text{i}+0.15\text{j}$

The position of P with respect to B.

$r_{P/B}=-0.3\text{i}+0.15\text{j}$

Provide the angular velocities of the bar AD $\left({\omega }_{AD}\right)$ and BE $\left({\omega }_{BE}\right)$ in vector form as shown below.

$\begin{array}{l}{\omega }_{AD}=-4\text{k}\\ {\omega }_{BE}=5\text{k}\end{array}$

Provide the angular acceleration of the bar BE in vector form as shown below.

${\alpha }_{BE}=-2\text{k}$

Calculate the velocity of a point $\left(v_P\right)$ having motion relative to a rotating frame as shown below.

$v_P=v_{P^{\prime }}+v_{P/F}$ (1)

Here, $v_{P^{\prime }}$ is the velocity of the point $P^{\prime }$ in the rotating frame corresponding to point P at the instant and $v_{P/F}$ is the velocity of the point P relative to the rotating frame.

Consider that the point P in the frame AD.

Calculate the velocity component $\left(v_{P^{\prime }}\right)$ in the frame AD as shown below.

$v_{P^{\prime }}={\omega }_{AD}\times r_{P/A}$

Substitute $-4\text{k}$ for ${\omega }_{AD}$ and $0.1\text{i}+0.15\text{j}$ for $r_{P/A}$.

$\begin{array}{c}{v}_{P^{\prime }}=\left(-4\text{k}\right)\times \left(0.1\text{i}+0.15j\right)\\ =-0.4\text{j}+0.6\text{i}\\ =0.6\text{i}-0.4\text{j}\end{array}$

Calculate the velocity component $\left(v_{P/AD}\right)$ when the point P is considered in the frame AD as shown below.

$v_{P/F}=v_{P/AD}=u_1\text{j}$ (2)

Here, $u_1$ is the relative velocity of the point P relative to AD.

Calculate the velocity of a point $\left(v_P\right)$ as shown below.

Substitute $0.6\text{i}-0.4\text{j}$ for $v_{P^{\prime }}$ and $u_1\text{j}$ for $v_{P/F}$ in Equation (1).

$\begin{array}{c}{v}_P=\left(0.6\text{i}-0.4\text{j}\right)+\left(u_1\text{j}\right)\\ =0.6\text{i}+\left(u_1-0.4\right)\text{j}\end{array}$ (3)

Consider that the point P in the frame BE.

Calculate the velocity component $\left(v_{P^{\prime }}\right)$ when the point P is considered in the frame BE as shown below.

$v_{P^{\prime }}={\omega }_{BE}\times r_{P/C}$

Substitute $5\text{k}$ for ${\omega }_{BE}$ and $-0.3\text{i}+0.15\text{j}$ for $r_{P/C}$.

$\begin{array}{c}{v}_{P^{\prime }}=\left(5\text{k}\right)\times \left(-0.3\text{i}+0.15\text{j}\right)\\ =-1.5\text{j}-0.75\text{i}\\ =-0.75\text{i}-1.5\text{j}\end{array}$

Calculate the velocity component $\left(v_{P/BE}\right)$ when the point P is considered in the frame BE as shown below.

$v_{P/F}=v_{P/BE}=u_2\left(\nwarrow \theta \right)$

Here, $u_2$ is the relative velocity of the point P relative to BE.

Resolving along x and y direction.

$\begin{array}{c}{v}_{P/BE}=\left(u_2\cos \theta \right)\leftarrow +\left(u_2\sin \theta \uparrow \right)\\ =-u_2\cos \theta \text{i}+u_2\sin \theta \text{j}\end{array}$ (4)

Substitute $-0.75\text{i}-1.5\text{j}$ for $v_{P^{\prime }}$, and $-u_2\cos \theta \text{i}+u_2\sin \theta \text{j}$ for $v_{P/F}$ in Equation (1).

$\begin{array}{c}{v}_P=\left(-0.75\text{i}-1.5\text{j}\right)+\left(-u_2\cos \theta \text{i}+u_2\sin \theta \text{j}\right)\\ =\left(-0.75-u_2\cos \theta \right)\text{i}+\left(-1.5+u_2\sin \theta \right)\text{j}\end{array}$ (5)

Equating Equations (3) and (5) as shown below.

$0.6\text{i}+\left(u_1-0.4\right)\text{j}=\left(-0.75-u_2\cos \theta \right)\text{i}+\left(-1.5+u_2\sin \theta \right)\text{j}$

Resolving i and j components as shown below.

For i component.

$\begin{array}{l}0.6=-0.75-u_2\cos \theta \\ u_2=-\frac{1.35}{\cos \theta }\end{array}$

Substitute $26.565°$ for $\theta$.

$\begin{array}{c}{u}_2=-\frac{1.35}{\cos \left(26.565°\right)}\\ =-1.509\text{m}/\text{s}\end{array}$

Calculate the velocity component $\left(v_{P/BE}\right)$ BE as shown below.

Substitute $-1.509\text{m}/\text{s}$ for $u_2$ and $26.565°$ for $\theta$ in Equation (4).

$\begin{array}{c}{v}_{P/BE}=-\left(-1.509\right)\cos \left(26.565°\right)\text{i}+\left(-1.509\right)\sin \left(26.565°\right)\text{j}\\ =1.35\text{i}-0.675\text{j}\end{array}$

For j component.

$\begin{array}{l}\left(u_1-0.4\right)\text{j}=\left(-1.5+u_2\sin \theta \right)\\ u_1-0.4=-1.5+u_2\sin \theta \\ u_1=-1.1+u_2\sin \theta \\ u_1=-1.1+u_2\sin \theta \end{array}$

Substitute $-1.509\text{m}/\text{s}$ for $u_2$ and $26.565°$ for $\theta$.

$\begin{array}{c}{u}_1=-1.1+\left(-1.509\right)\sin \left(26.565°\right)\\ =-1.775\text{m}/\text{s}\end{array}$

Calculate the velocity component $\left(v_{P/AD}\right)$ as shown below.

Substitute $-1.775\text{m}/\text{s}$ for $u_1$ in Equation (2).

$v_{P/AD}=-1.775\text{j}$

Calculate the acceleration of a point $\left(a_P\right)$ relative to the rotating frame as shown below.

$a_P=a_{P^{\prime }}+a_{P/F}+a_C$ (6)

Here, $a_{P^{\prime }}$ acceleration of the point $P^{\prime }$ in the rotating frame corresponding to point P at the instant, $a_{P/F}$ is the acceleration of the point P relative to the rotating frame, and $a_C$ is the Coriolis component of acceleration.

Consider the point P in the frame AD.

Calculate the acceleration component $\left(a_{P^{\prime }}\right)$ of point P in the frame AD as shown below.

$a_{P^{\prime }}={\alpha }_{AD}\times r_{P/A}-{\omega }_{AD}^2{r}_{P/A}$

Here, ${\alpha }_{AD}$ is the angular acceleration of the rod AD.

Substitute $0$ for ${\alpha }_{AD}$, $0.1\text{i}+0.15\text{j}$ for $r_{P/A}$, and $4\text{rad}/\text{s}$ for ${\omega }_{AD}$.

$\begin{array}{c}{a}_{P^{\prime }}={\alpha }_{AD}\times r_{P/A}-{\omega }_{AD}^2{r}_{P/A}\\ =0\times \left(0.1\text{i}+0.15\text{j}\right)-{\left(4\text{rad}/\text{s}\right)}^2\left(0.1\text{i}+0.15\text{j}\right)\\ =-1.6\text{i}-2.4\text{j}\end{array}$

Calculate the acceleration component $\left(a_{P/AD}\right)$ when the point P is considered in the frame AD as shown below.

$a_{P/F}=a_{P/AD}={\dot{u}}_1\text{j}-\frac{u_1^2}{R}\text{i}$

Here, ${\dot{u}}_1$ is the acceleration of the point P relative to AD and R is the radius of rotation of point P with respect to frame AD.

Substitute $-1.775\text{m}/\text{s}$ for $u_1$ and $0.15\text{m}$ for R.

$\begin{array}{c}{a}_{P/AD}={\dot{u}}_1\text{j}-\frac{{\left(-1.775\right)}^2}{0.15}\text{i}\\ ={\dot{u}}_1\text{j}-21.0042\text{i}\end{array}$

Calculate the Coriolis component of acceleration $\left(a_C\right)$ when the point P is considered in the frame AD as shown below.

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

Substitute $-4\text{k}$ for ${\omega }_{AD}$ and $-1.775\text{j}$ for $v_{P/AD}$.

$\begin{array}{c}{a}_C=2\left(-4\text{k}\right)\times \left(-1.775\text{j}\right)\\ =-14.2\text{i}\end{array}$

Substitute $-1.6\text{i}-2.4\text{j}$ for $a_{P^{\prime }}$, ${\dot{u}}_1\text{j}-21.0042\text{i}$ for $a_{P/F}$, and $-14.2\text{i}$ for $a_C$ in Equation (6).

$\begin{array}{c}{a}_P=a_{P^{\prime }}+a_{P/F}+a_C\\ =\left(-1.6\text{i}-2.4\text{j}\right)+\left({\dot{u}}_1\text{j}-21.0042\text{i}\right)+\left(-14.2\text{i}\right)\\ =-36.8042\text{i}+\left(-2.4+{\dot{u}}_1\right)\text{j}\end{array}$ (7)

Consider the point P in the frame BE.

Calculate the acceleration component $\left(a_{P^{\prime }}\right)$ when the point P is considered in the frame BE as shown below.

$a_{P^{\prime }}={\alpha }_{BE}\times r_{P/C}-{\omega }_{BE}^2{r}_{P/C}$

Substitute $-\left(2\text{rad}/{\text{s}}^{\text{2}}\right)\text{k}$ for ${\alpha }_{BE}$, $-0.3\text{i}+0.15\text{j}$ for $r_{P/C}$, and $5\text{rad}/\text{s}$ for ${\omega }_{BE}$.

$\begin{array}{c}{a}_{P^{\prime }}=\left(-\left(2\text{rad}/{\text{s}}^{\text{2}}\right)\text{k}\right)\times \left(-0.3\text{i}+0.15\text{j}\right)-{\left(5\text{rad}/\text{s}\right)}^2\left(-0.3\text{i}+0.15\text{j}\right)\\ =0.6\text{j}+0.3\text{i}+7.5\text{i}-3.75\text{j}\\ =7.8\text{i}-3.15\text{j}\end{array}$

Calculate the acceleration component $\left(a_{P/BE}\right)$ of the point P in the frame BE ${\alpha }_{BE}$ as shown below.

$a_{P/F}=a_{P/BE}=\left({\dot{u}}_2\right)\left(\nwarrow \theta \right)$

Here, ${\dot{u}}_2$ is the acceleration of the point P relative to BE.

Resolving along x and y direction.

$\begin{array}{c}{a}_{P/BE}={\dot{u}}_2\cos \theta \leftarrow +{\dot{u}}_2\sin \theta \uparrow \\ =-{\dot{u}}_2\cos \theta \text{i}+{\dot{u}}_2\sin \theta \text{j}\end{array}$

Calculate the Coriolis component of acceleration $\left(a_C\right)$ of the point P in the frame BE as shown below.

$a_C=2{\omega }_{BE}\times v_{BE}$

Substitute $5\text{k}$ for ${\omega }_{BE}$ and $1.35\text{i}-0.675\text{j}$ for $v_{P/BE}$.

$\begin{array}{c}{a}_C=2\left(5\text{k}\right)\times \left(1.35\text{i}-0.675\text{j}\right)\\ =13.5\text{j}+6.75\text{i}\\ =6.75\text{i}+13.5\text{j}\end{array}$

Calculate the acceleration of a point $\left(a_P\right)$ as shown below.

Substitute $7.8\text{i}-3.15\text{j}$ for $a_{P^{\prime }}$, $-{\dot{u}}_2\cos \theta \text{i}+{\dot{u}}_2\sin \theta \text{j}$ for $a_{P/F}$, and $6.75\text{i}+13.5\text{j}$ for $a_C$ in Equation (6).

$\begin{array}{c}{a}_P=\left(7.8\text{i}-3.15\text{j}\right)+\left(-{\dot{u}}_2\cos \theta \text{i}+{\dot{u}}_2\sin \theta \text{j}\right)+\left(6.75\text{i}+13.5\text{j}\right)\\ =\left(14.55-{\dot{u}}_2\cos \theta \right)\text{i}+\left(10.35+{\dot{u}}_2\sin \theta \right)\text{j}\end{array}$ (8)

Equating Equations (7) and (8) as shown below.

$-36.8042\text{i}+\left(-2.4+{\dot{u}}_1\right)\text{j}=\left(14.55-{\dot{u}}_2\cos \theta \right)\text{i}+\left(10.35+{\dot{u}}_2\sin \theta \right)\text{j}$

Resolving i and j components as shown below.

$\begin{array}{l}-36.8042=14.55-{\dot{u}}_2\cos \theta \\ {\dot{u}}_2=\frac{51.3542}{\cos \theta }\end{array}$

Substitute $26.565°$ for $\theta$.

$\begin{array}{c}{\dot{u}}_2=\frac{51.3542}{\cos \left(26.565°\right)}\\ =57.416\text{m}/{\text{s}}^{\text{2}}\end{array}$

Calculate the acceleration of a point $\left(a_P\right)$ as shown below.

Substitute $57.416\text{m}/{\text{s}}^{\text{2}}$ for ${\dot{u}}_2$ and $26.565°$ for $\theta$ in Equation (8).

$\begin{array}{c}{a}_P=\left(14.55-57.416\cos \left(26.565°\right)\right)\text{i}+\left(10.35+57.416\sin \left(26.565°\right)\right)\text{j}\\ =-\left(36.804\text{m}/{\text{s}}^{\text{2}}\right)\text{i}+\left(36.027\text{m}/{\text{s}}^{\text{2}}\right)\text{j}\end{array}$

Calculate the magnitude of the acceleration $\left({\text{a}}_P\right)$ as shown below.

$\begin{array}{c}{\text{a}}_P=\sqrt{{\left(-36.804\right)}^2+{36.027}^2}\\ =\sqrt{2,652.479145}\\ =51.5\text{m}/{\text{s}}^2\end{array}$

Therefore, the acceleration of pin P is $\underset{\_}{{\text{a}}_P=51.5\text{m}/{\text{s}}^{\text{2}}}$.