Chapter 15.5, Problem 15.176P

Knowing that at the instant shown the rod attached at A has an angular velocity of 5 rad/s counterclockwise and an angular acceleration of 2 rad/s² clockwise, determine the angular velocity and the angular acceleration of the rod attached at B.

Fig. P15.176

To determine

The angular velocity and the angular acceleration of the rod at B.

Answer to Problem 15.176P

The angular velocity of the rod at B is $\underset{\_}{{\omega }_{BP}=7.86\text{rad}/\text{s}\left(\text{Counter}\text{clockwise}\right)}$.

The angular acceleration of the rod at B is $\underset{\_}{{\alpha }_{BP}=81.1\text{rad}/{\text{s}}^2\left(\text{Counter}\text{clockwise}\right)}$.

Explanation of Solution

Given information:

The angular velocity of the rod at A is ${\omega }_{AP}=5\text{rad}/\text{s}$.

The angular acceleration of the rod at A is ${\alpha }_{AP}=2\text{rad}/{\text{s}}^2$.

Calculation:

Consider the triangle ABP to get the values of AP and BP.

Calculate the angle ABP $\left(\overline{)PBA}\right)$ as shown below.

$\begin{array}{l}\overline{)PBA}+70°=180°\\ \overline{)PBA}=110°\end{array}$

Consider the sum of the angles is $180°$.

Calculate the angle APB $\left(\overline{)APB}\right)$ as shown below.

$\overline{)APB}+\overline{)PBA}+\overline{)PAB}=180°$

Substitute $110°$ for $\overline{)PBA}$, and $25°$ for $\overline{)PAB}$.

$\begin{array}{l}\overline{)APB}+110°+25°=180°\\ \overline{)APB}=45°\end{array}$

Apply law of sine for the triangle ABP as shown below.

$\frac{AB}{\sin \overline{)APB}}=\frac{AP}{\sin \overline{)PBA}}=\frac{BP}{\sin \overline{)PAB}}$

Substitute $200\text{mm}$ for AB, $45°$ for $\overline{)APB}$, $25°$ for $\overline{)PAB}$, and $110°$ for $\overline{)PBA}$.

$\frac{200}{\sin 45°}=\frac{AP}{\sin 110°}=\frac{BP}{\sin 25°}$

$\begin{array}{c}AP=265.785\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\\ =0.265785\text{m}\\ BP=119.534\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\\ =0.119534\text{m}\end{array}$

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

Position of P with respect to A.

$\begin{array}{c}{r}_{P/A}=0.265785\left(\sin 25°\text{i}-\cos 25°\text{j}\right)\\ =0.11233\text{i}-0.24088\text{j}\end{array}$

Position of P with respect to B.

$\begin{array}{c}{r}_{P/B}=0.119534\left(\sin 70°\text{i}+\cos 70°\text{j}\right)\\ =0.11233\text{i}+0.04088\text{j}\end{array}$

Provide the angular velocity of each link in vector form as shown below.

$\begin{array}{l}{\omega }_{AP}=\left(5\text{rad}/\text{s}\right)\text{k}\\ {\text{ω}}_{BP}={\omega }_{BP}\text{k}\end{array}$

Provide the angular acceleration of each link in vector form as shown below.

$\begin{array}{l}{\alpha }_{AP}=-\left(2\text{rad}/{\text{s}}^{\text{2}}\right)\text{k}\\ {\text{α}}_{BP}={\alpha }_{BP}\text{k}\end{array}$.

Calculate the velocity of point P $\left(v_P\right)$ 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 frame BP corresponding to point P and $v_{P/F}$ is the velocity of the point P relative to the frame BP.

Calculate the velocity the point P on the rod AP $\left(v_P\right)$ as shown below.

$v_P={\omega }_{AP}\times r_{P/A}$

Substitute $\left(5\text{rad}/\text{s}\right)\text{k}$ for ${\omega }_{AP}$ and $0.11233\text{i}+0.24088\text{j}$ for $r_{P/A}$.

$\begin{array}{c}{v}_P=\left(\left(5\text{rad}/\text{s}\right)\text{k}\right)\times \left(0.11233\text{i}+0.24088\text{j}\right)\\ =0.56165\text{j}-1.2044\text{i}\\ =-\left(1.2044\text{m}/\text{s}\right)\text{i}+\left(0.56165\text{m}/\text{s}\right)\text{j}\end{array}$

Calculate the acceleration at the point P $\left(v_P\right)$ on the rod AP as shown below.

$a_P={\alpha }_{AP}\times r_{P/A}-{\omega }_{AP}^2{r}_{P/A}$

Substitute $-\left(2\text{rad}/{\text{s}}^{\text{2}}\right)\text{k}$ for ${\alpha }_{AP}$, $0.11233\text{i}+0.24088\text{j}$ for $r_{P/A}$, and $\left(5\text{rad}/\text{s}\right)$ for ${\omega }_{AP}$.

$\begin{array}{c}{a}_P=\left(\begin{array}{c}\left(-\left(2\text{rad}/{\text{s}}^{\text{2}}\right)\text{k}\right)\times \left(0.11233\text{i}+0.24088\text{j}\right)-\\ {\left(5\text{rad}/\text{s}\right)}^2\left(0.11233\text{i}+0.24088\text{j}\right)\end{array}\right)\\ =-0.22466\text{j}+0.48176\text{i}-2.80825\text{i}-6.022\text{j}\\ =-\left(2.3265\text{m}/{\text{s}}^2\right)\text{i}-\left(6.2467\text{m}/{\text{s}}^2\right)\text{j}\end{array}$

Calculate the relative velocity component $\left(v_{\text{rel}}\right)$ as shown below.

$v_{\text{rel}}=u$

Here, u is the velocity of the point P relative to BP.

The angle of relative velocity with respect to the rotating rod BP is $\begin{array}{l}=90°-70°\\ =20°\end{array}$

Resolve the relative velocity along x and y directions.

$v_{\text{rel}}=u\cos 20°\text{i}+u\sin 20°\text{j}$ (2)

Calculate the relative acceleration component $\left(v_{\text{rel}}\right)$ as shown below.

$a_{\text{rel}}=\dot{u}$

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

Resolve the relative velocity along x and y directions.

$a_{\text{rel}}=\dot{u}\cos 20°\text{i}+\dot{u}\sin 20°\text{j}$

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

$v_{P^{\prime }}={\omega }_{BP}\times r_{P/B}$

Substitute ${\omega }_{BP}\text{k}$ for ${\omega }_{BP}$ and $0.11233\text{i}+0.04088\text{j}$ for $r_{P/B}$.

$\begin{array}{c}{v}_{P^{\prime }}=\left({\omega }_{BP}\text{k}\right)\times \left(0.11233\text{i}+0.0408\text{j}\right)\\ =0.11233{\omega }_{BP}\text{j}-0.04088{\omega }_{BP}\text{i}\\ =-0.04088{\omega }_{BP}\text{i}+0.11233{\omega }_{BP}\text{j}\end{array}$

Calculate the angular velocity $\left({\omega }_{BP}\right)$ as shown below.

Substitute $-\left(1.2044\text{m}/\text{s}\right)\text{i}+\left(0.56165\text{m}/\text{s}\right)\text{j}$ for $v_P$, $-0.04088{\omega }_{BP}\text{i}+0.11233{\omega }_{BP}\text{j}$  for $v_{P^{\prime }}$, and $u\cos 20°\text{i}+u\sin 20°\text{j}$ for $v_{P/F}$ in Equation (1).

$\begin{array}{l}-\left(1.2044\text{m}/\text{s}\right)\text{i}+\left(0.56165\text{m}/\text{s}\right)\text{j}=\left(\begin{array}{l}-0.04088{\omega }_{BP}\text{i}+0.11233{\omega }_{BP}\text{j}+\\ u\cos 20°\text{i}+u\sin 20°\text{j}\end{array}\right)\\ -1.2044\text{i}+0.56165\text{j}=\left(\begin{array}{l}\left(-0.04088{\omega }_{BP}+u\cos 20°\right)\text{i}+\\ \left(0.11233{\omega }_{BP}+u\sin 20°\right)\text{j}\end{array}\right)\end{array}$

Resolving i and j components as shown below.

For i component.

$-1.2044=-0.04088{\omega }_{BP}+u\cos 20°$ (3)

For i component.

$0.56165=0.11233{\omega }_{BP}+u\sin 20°$ (4)

Solving Equations (3) and (4) to get the values as shown below.

$\begin{array}{l}{\omega }_{BP}=7.8612\text{rad}/\text{s}\\ u=-0.9397\text{m}/\text{s}\end{array}$

Hence, the angular velocity of the rod at B is $\underset{\_}{{\omega }_{BP}=7.86\text{rad}/\text{s}\left(\text{Counter}\text{clockwise}\right)}$.

Calculate the relative velocity $\left(v_{\text{rel}}\right)$ as shown below.

Substitute $-0.9397\text{m}/\text{s}$ for u in Equation (2).

$\begin{array}{c}{v}_{\text{rel}}=-0.9397\cos 20°\text{i}+\left(-0.9397\right)\sin 20°\text{j}\\ =-\left(0.8830\text{m}/\text{s}\right)\text{i}-\left(0.3214\text{m}/\text{s}\right)\text{j}\end{array}$

Calculate the acceleration of the point P.

$a_P=a_{P^{\prime }}+a_{P/F}+a_C$ (5)

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

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

Substitute ${\alpha }_{BP}\text{k}$ for ${\alpha }_{BP}$, $0.11233\text{i}+0.04088\text{j}$ for $r_{P/B}$, and $7.8612\text{rad}/\text{s}$ for ${\omega }_{BP}$.

$\begin{array}{c}{a}_{P^{\prime }}={\alpha }_{BP}\text{k}\times \left(0.11233\text{i}+0.04088\text{j}\right)-{\left(7.8612\right)}^2\left(0.11233\text{i}+0.04088\text{j}\right)\\ =0.11233{\alpha }_{BP}\text{j}-0.04088{\alpha }_{BP}\text{i}-6.9418\text{i}-2.5263\text{j}\\ =-\left(0.04088{\alpha }_{BP}+6.9418\right)\text{i}+\left(0.11233{\alpha }_{BP}-2.5263\right)\text{j}\end{array}$

Calculate the Coriolis component of acceleration $\left(a_C\right)$ as shown below.

$a_C=2{\omega }_{BP}\times v_{\text{rel}}$

Substitute $\left(7.8612\text{rad}/\text{s}\right)\text{k}$ for ${\omega }_{BP}$ and $-\left(0.8830\text{m}/\text{s}\right)\text{i}-\left(0.3214\text{m}/\text{s}\right)\text{j}$ for $v_{\text{rel}}$.

$\begin{array}{c}{a}_C=2\left(\left(7.8612\text{rad}/\text{s}\right)\text{k}\right)\times \left(-\left(0.8830\text{m}/\text{s}\right)\text{i}-\left(0.3214\text{m}/\text{s}\right)\text{j}\right)\\ =-13.8828\text{j}+5.0532\text{i}\\ =5.0531\text{i}-13.8828\text{j}\end{array}$

Calculate the angular acceleration $\left({\alpha }_{BP}\right)$ as shown below.

Substitute $-\left(0.04088{\alpha }_{BP}+6.9418\right)\text{i}+\left(0.11233{\alpha }_{BP}-2.5263\right)\text{j}$ for $a_{P^{\prime }}$, $-\left(2.3265\text{m}/{\text{s}}^2\right)\text{i}-\left(6.2467\text{m}/{\text{s}}^2\right)\text{j}$ for $a_P$, $\dot{u}\cos 20°\text{i}+\dot{u}\sin 20°\text{j}$ for $a_{\text{rel}}$, and $5.0531\text{i}-13.8828\text{j}$ for $a_C$ in Equation (5).

$\begin{array}{c}\left(\begin{array}{l}-\left(2.3265\text{m}/{\text{s}}^2\right)\text{i}\\ -\left(6.2467\text{m}/{\text{s}}^2\right)\text{j}\end{array}\right)=\left(\begin{array}{l}\left(-\left(0.04088{\alpha }_{BP}+6.9418\right)\text{i}+\left(0.11233{\alpha }_{BP}-2.5263\right)\text{j}\right)+\\ \left(\dot{u}\cos 20°\text{i}+\dot{u}\sin 20°\text{j}\right)+\left(5.0531\text{i}-13.8828\text{j}\right)\end{array}\right)\\ =\left(\begin{array}{l}\left(-0.04088{\alpha }_{BP}-6.9418+\dot{u}\cos 20°+5.0531\right)\text{i}+\\ \left(0.11233{\alpha }_{BP}-2.5263+\dot{u}\sin 20°-13.8828\right)\text{j}\end{array}\right)\\ =\left(\begin{array}{c}\left(-0.04088{\alpha }_{BP}-1.8887+\dot{u}\cos 20°\right)\text{i}+\\ \left(0.11233{\alpha }_{BP}-16.4091+\dot{u}\sin 20°\right)\text{j}\end{array}\right)\end{array}$

Resolving i and j components as shown below.

For i component.

$\begin{array}{l}-2.3265=-0.04088{\alpha }_{BP}+\dot{u}\cos 20°-1.8887\\ -0.04088{\alpha }_{BP}+\dot{u}\cos 20°=-0.4395\end{array}$ (6)

For j component.

$\begin{array}{l}-6.2467=0.11233{\alpha }_{BP}+\dot{u}\sin 20°-16.4091\\ 0.11233{\alpha }_{BP}+\dot{u}\sin 20°=10.1624\end{array}$ (7)

Solving Equations (6) and (7) to get the values as shown below.

$\begin{array}{l}{\alpha }_{BP}=81.14\text{rad}/{\text{s}}^2\\ \dot{u}=3.062\text{m}/{\text{s}}^2\end{array}$

Hence, the angular acceleration of the rod at B is $\underset{\_}{{\alpha }_{BP}=81.1\text{rad}/{\text{s}}^2\left(\text{Counter}\text{clockwise}\right)}$.