Week 2 Tutorial - Kinematic Analysis of Mechanisms

Answer

1. First evaluate the constants

   \[\begin{aligned} A &= 2cd - 2ac\cos\theta_2 = 11097.32 \\ B &= -2ac\sin\theta_2 = -4113.84 \\ C &= a^2 - b^2 + c^2 + d^2 - 2ad\cos\theta_2 = -2528.36 \end{aligned}\] \[\begin{aligned} D &= 2ab\cos\theta_2 - 2bd = -16645.97 \\ E &= 2ab\sin\theta_2 = 6170.76 \\ F &= a^2 + b^2 + d^2 - c^2 - 2ad\cos\theta_2 = 13471.64 \end{aligned}\]

   Because each term is a product of two lengths, the constants $A$ to $F$ have units of $\text{mm}^2$.

2. For $\theta_4$, use

   \[t_4 = \frac{-2B \pm \sqrt{(2B)^2 - 4(C-A)(A+C)}}{2(C-A)}\]

   Therefore,

   \[\theta_{4,1}=2\tan^{-1}(t_{4,1}), \qquad \theta_{4,2}=2\tan^{-1}(t_{4,2})\]

   Numerically,

   \[t_{4,1}=0.5466, \qquad t_{4,2}=-1.1505\]

   so

   \[\theta_{4,1}=57.32^\circ, \qquad \theta_{4,2}=-98.00^\circ\]

3. For $\theta_3$, use

   \[t_3 = \frac{-2E \pm \sqrt{(2E)^2 - 4(F-D)(D+F)}}{2(F-D)}\]

   Therefore,

   \[\theta_{3,1}=2\tan^{-1}(t_{3,1}), \qquad \theta_{3,2}=2\tan^{-1}(t_{3,2})\]

   Numerically,

   \[t_{3,1}=0.1790, \qquad t_{3,2}=-0.5888\]

   so

   \[\theta_{3,1}=20.30^\circ, \qquad \theta_{3,2}=-60.98^\circ\]

4. The two configurations are

   The root pairs are selected by substituting each possible combination back into the original loop-closure equations:

   \[\begin{aligned} r_x &= a\cos\theta_2+b\cos\theta_3-c\cos\theta_4-d \\ r_y &= a\sin\theta_2+b\sin\theta_3-c\sin\theta_4 \end{aligned}\]

   A valid pair must give $r_x\approx0$ and $r_y\approx0$. Testing all four combinations gives

   $\theta_3$	$\theta_4$	$r_x$ (mm)	$r_y$ (mm)	Result
   $20.30^\circ$	$57.32^\circ$	$-0.01$	$0.01$	Closes
   $20.30^\circ$	$-98.00^\circ$	$54.32$	$146.57$	Does not close
   $-60.98^\circ$	$57.32^\circ$	$-54.34$	$-146.56$	Does not close
   $-60.98^\circ$	$-98.00^\circ$	$-0.01$	$0.00$	Closes

   The small non-zero residuals in the valid rows are caused by rounding the angles to two decimal places. This check shows that the acute $\theta_3$ root must pair with the positive $\theta_4$ root, while the two negative roots must pair together.

   Configuration	$\theta_3$	$\theta_4$
   Open	$20.30^\circ$	$57.32^\circ$
   Crossed	$-60.98^\circ$	$-98.00^\circ$

   To name the configurations, sketch the two valid geometries or calculate the joint positions. For $(20.30^\circ,57.32^\circ)$, the coupler remains above the fixed link, so this is the open configuration. For $(-60.98^\circ,-98.00^\circ)$, the coupler crosses the fixed link between the ground pivots, so this is the crossed configuration.

Answer

1. Here, $c$ is the signed vertical offset of the slider axis relative to $O_2$. Because the slider axis is below $O_2$ in this example, $c=-15\text{ mm}$. Use the position equations

   \[\sin\theta_3=\frac{a\sin\theta_2-c}{b}\] \[d=a\cos\theta_2-b\cos\theta_3\]

2. Substituting the data gives

   \[\sin\theta_3=\frac{35\sin50^\circ-(-15)}{110}=0.3801\]

3. The two angular branches are

   \[\theta_3=\sin^{-1}(0.3801)=22.34^\circ\] \[\theta_3=180^\circ-22.34^\circ=157.66^\circ\]

   The sine equation alone cannot identify the open configuration because the two supplementary angles have the same sine. They give different horizontal positions, however, because their cosines have opposite signs. Evaluate $d$ for both branches:

   \[\begin{aligned} d_{22.34^\circ} &=35\cos50^\circ-110\cos22.34^\circ \\ &=-79.25\text{ mm} \\ d_{157.66^\circ} &=35\cos50^\circ-110\cos157.66^\circ \\ &=124.24\text{ mm} \end{aligned}\]

   Here, $d$ is measured from $O_2$ along the positive slider axis. Also, $\theta_3$ describes the direction of link $b$ from slider $B$ towards joint $A$. For $\theta_3=22.34^\circ$, $\cos\theta_3>0$, so the link points to the right from $B$ to $A$ and $B$ must lie to the left of $O_2$. This gives $d<0$ and corresponds to the crossed circuit shown in the diagram. For $\theta_3=157.66^\circ$, $\cos\theta_3<0$, so the link points to the left from $B$ to $A$ and $B$ lies to the right of $O_2$. This gives $d>0$ and matches the open circuit shown in red. Therefore, under the definition given in the question, the open branch is

   \[\theta_3=157.66^\circ, \qquad d=124.24\text{ mm}.\]

4. Therefore, for the open configuration

   Quantity	Value
   $\theta_3$	$157.66^\circ$
   $d$	$124.24\text{ mm}$

Answer

1. Start from the vector loop equation

   \[\mathbf{R}_A+\mathbf{R}_{BA}-\mathbf{R}_{BO_2}=0\]

2. In complex form this becomes

   \[a e^{j\theta_2} - b e^{j\theta_3} - (d + jc) = 0\]

3. Expand with Euler’s relation $e^{j\theta}=\cos\theta+j\sin\theta$:

   \[a(\cos\theta_2+j\sin\theta_2)-b(\cos\theta_3+j\sin\theta_3)-(d+jc)=0\]

4. Separate real and imaginary parts:

   \[a\cos\theta_2-b\cos\theta_3-d=0\] \[a\sin\theta_2-b\sin\theta_3-c=0\]

5. Rearrange the imaginary equation first:

   \[a\sin\theta_2-b\sin\theta_3-c=0\] \[b\sin\theta_3=a\sin\theta_2-c\] \[\sin\theta_3=\frac{a\sin\theta_2-c}{b}\]

6. Now rearrange the real equation:

   \[a\cos\theta_2-b\cos\theta_3-d=0\] \[d=a\cos\theta_2-b\cos\theta_3\]

Answer

1. Use the velocity equations

   \[\omega_3=\omega_2\frac{a\sin(\theta_4-\theta_2)}{b\sin(\theta_3-\theta_4)}\] \[\omega_4=\omega_2\frac{a\sin(\theta_2-\theta_3)}{c\sin(\theta_4-\theta_3)}\]

   Once $\omega_3$ and $\omega_4$ are known, the point-velocity equations are

   \[\mathbf{V}_A=a\omega_2(-\sin\theta_2+j\cos\theta_2)\] \[\mathbf{V}_{BA}=b\omega_3(-\sin\theta_3+j\cos\theta_3)\] \[\mathbf{V}_B=c\omega_4(-\sin\theta_4+j\cos\theta_4)\]

2. Substituting the given configuration gives

   \[\omega_3=25\frac{40\sin(57.32^\circ-40^\circ)}{120\sin(20.30^\circ-57.32^\circ)}=-4.120456\text{ rad/s}\approx-4.12\text{ rad/s}\] \[\omega_4=25\frac{40\sin(40^\circ-20.30^\circ)}{80\sin(57.32^\circ-20.30^\circ)}=6.998396\text{ rad/s}\approx7.00\text{ rad/s}\]

3. The speed magnitudes then follow:

   \[|\mathbf{V}_A|=a\omega_2=40(25)=1000.00\text{ mm/s}\] \[|\mathbf{V}_{BA}|=b|\omega_3|=120(4.120456)=494.45\text{ mm/s}\] \[|\mathbf{V}_B|=c|\omega_4|=80(6.998396)=559.87\text{ mm/s}\]

   Therefore

   Quantity	Value
   $\omega_3$	$-4.12\text{ rad/s}$
   $\omega_4$	$7.00\text{ rad/s}$
   $\lvert \mathbf{V}_A \rvert$	$1000.00\text{ mm/s}$
   $\lvert \mathbf{V}_{BA} \rvert$	$494.45\text{ mm/s}$
   $\lvert \mathbf{V}_B \rvert$	$559.87\text{ mm/s}$

   The negative sign on $\omega_3$ shows that link 3 rotates opposite to the assumed positive angular direction.

Answer

1. Use the velocity relations

   \[\omega_3=\frac{a\cos\theta_2}{b\cos\theta_3}\omega_2\] \[\dot d=-a\omega_2\sin\theta_2+b\omega_3\sin\theta_3\]

2. Substituting the given state gives

   \[\omega_3=\frac{35\cos50^\circ}{110\cos157.66^\circ}(20)=-4.4224\text{ rad/s}\approx-4.42\text{ rad/s}\]

3. Then

   \[\dot d=-35(20)\sin50^\circ+110(-4.4224)\sin157.66^\circ=-721.14\text{ mm/s}\]

   Therefore

   Quantity	Value
   $\omega_3$	$-4.42\text{ rad/s}$
   $\dot d$	$-721.14\text{ mm/s}$

Answer

1. Start from the differentiated vector loop equation

   \[\frac{d}{dt}\left(\mathbf{R}_A+\mathbf{R}_{BA}-\mathbf{R}_{BO_2}\right)=0\]

2. Write the vectors in complex form:

   \[\mathbf{R}_A=a e^{j\theta_2}, \qquad \mathbf{R}_{BA}=-b e^{j\theta_3}, \qquad \mathbf{R}_{BO_2}=d+jc\]

3. In complex form, the differentiated loop is

   \[ja\omega_2e^{j\theta_2}-jb\omega_3e^{j\theta_3}-\dot d=0\]

4. Expand with Euler’s relation $e^{j\theta}=\cos\theta+j\sin\theta$:

   \[ja\omega_2(\cos\theta_2+j\sin\theta_2)-jb\omega_3(\cos\theta_3+j\sin\theta_3)-\dot d=0\]

5. Simplify the products by using $j(\cos\theta+j\sin\theta)=-\sin\theta+j\cos\theta$:

   \[a\omega_2(-\sin\theta_2+j\cos\theta_2)-b\omega_3(-\sin\theta_3+j\cos\theta_3)-\dot d=0\]

6. Separate real and imaginary parts:

   \[-a\omega_2\sin\theta_2+b\omega_3\sin\theta_3-\dot d=0\] \[a\omega_2\cos\theta_2-b\omega_3\cos\theta_3=0\]

7. Rearranging the real equation gives

   \[\dot d=-a\omega_2\sin\theta_2+b\omega_3\sin\theta_3\]

8. Rearranging the imaginary equation gives

   \[b\omega_3\cos\theta_3=a\omega_2\cos\theta_2\] \[\omega_3=\frac{a\cos\theta_2}{b\cos\theta_3}\omega_2\]

Answer

1. Use the solved acceleration equations from the lecture:

   \[\begin{aligned} \alpha_3={}& \frac{(a\alpha_2\sin\theta_2+a\omega_2^2\cos\theta_2+b\omega_3^2\cos\theta_3-c\omega_4^2\cos\theta_4)(c\cos\theta_4)}{(c\sin\theta_4)(b\cos\theta_3)-(b\sin\theta_3)(c\cos\theta_4)} \\ &-\frac{(c\sin\theta_4)(a\alpha_2\cos\theta_2-a\omega_2^2\sin\theta_2-b\omega_3^2\sin\theta_3+c\omega_4^2\sin\theta_4)}{(c\sin\theta_4)(b\cos\theta_3)-(b\sin\theta_3)(c\cos\theta_4)} \end{aligned}\] \[\begin{aligned} \alpha_4={}& \frac{(a\alpha_2\sin\theta_2+a\omega_2^2\cos\theta_2+b\omega_3^2\cos\theta_3-c\omega_4^2\cos\theta_4)(b\cos\theta_3)}{(c\sin\theta_4)(b\cos\theta_3)-(b\sin\theta_3)(c\cos\theta_4)} \\ &-\frac{(b\sin\theta_3)(a\alpha_2\cos\theta_2-a\omega_2^2\sin\theta_2-b\omega_3^2\sin\theta_3+c\omega_4^2\sin\theta_4)}{(c\sin\theta_4)(b\cos\theta_3)-(b\sin\theta_3)(c\cos\theta_4)} \end{aligned}\]

2. Substitute the given lengths, angles, angular velocities and $\alpha_2$ directly into the solved lecture equations:

   \[\begin{aligned} \alpha_3 &=\frac{ \begin{aligned} &\big[40(15)\sin40^\circ+40(25^2)\cos40^\circ +120((-4.12)^2)\cos20.30^\circ \\ &\qquad-80(7.00^2)\cos57.32^\circ\big](80\cos57.32^\circ) \\ &-(80\sin57.32^\circ)\big[40(15)\cos40^\circ-40(25^2)\sin40^\circ \\ &\qquad-120((-4.12)^2)\sin20.30^\circ +80(7.00^2)\sin57.32^\circ\big] \end{aligned} }{ \begin{aligned} &(80\sin57.32^\circ)(120\cos20.30^\circ) \\ &\qquad-(120\sin20.30^\circ)(80\cos57.32^\circ) \end{aligned} } \\ &=296.1074\text{ rad/s}^2 \\ &\approx296.11\text{ rad/s}^2, \end{aligned}\] \[\begin{aligned} \alpha_4 &=\frac{ \begin{aligned} &\big[40(15)\sin40^\circ+40(25^2)\cos40^\circ +120((-4.12)^2)\cos20.30^\circ \\ &\qquad-80(7.00^2)\cos57.32^\circ\big](120\cos20.30^\circ) \\ &-(120\sin20.30^\circ)\big[40(15)\cos40^\circ-40(25^2)\sin40^\circ \\ &\qquad-120((-4.12)^2)\sin20.30^\circ +80(7.00^2)\sin57.32^\circ\big] \end{aligned} }{ \begin{aligned} &(80\sin57.32^\circ)(120\cos20.30^\circ) \\ &\qquad-(120\sin20.30^\circ)(80\cos57.32^\circ) \end{aligned} } \\ &=470.1535\text{ rad/s}^2 \\ &\approx470.15\text{ rad/s}^2. \end{aligned}\]

3. Use the point-acceleration equations given in the lecture:

   \[\mathbf{A}_A=a\alpha_2(-\sin\theta_2+j\cos\theta_2)-a\omega_2^2(\cos\theta_2+j\sin\theta_2)\] \[\mathbf{A}_{BA}=b\alpha_3(-\sin\theta_3+j\cos\theta_3)-b\omega_3^2(\cos\theta_3+j\sin\theta_3)\] \[\mathbf{A}_B=c\alpha_4(-\sin\theta_4+j\cos\theta_4)-c\omega_4^2(\cos\theta_4+j\sin\theta_4)\]

4. Substitute the values directly into these three equations:

   \[\begin{aligned} \mathbf{A}_A & =40(15)(-\sin40^\circ+j\cos40^\circ) \\ &\quad -40(25^2)(\cos40^\circ+j\sin40^\circ) \\ &=-19536.78-15610.06j\text{ mm/s}^2, \end{aligned}\] \[\begin{aligned} \mathbf{A}_{BA} & =120(296.1074)(-\sin20.30^\circ+j\cos20.30^\circ) \\ &\quad -120((-4.12)^2)(\cos20.30^\circ+j\sin20.30^\circ) \\ &=-14238.04+32619.22j\text{ mm/s}^2, \end{aligned}\] \[\begin{aligned} \mathbf{A}_B & =80(470.1535)(-\sin57.32^\circ+j\cos57.32^\circ) \\ &\quad -80(7.00^2)(\cos57.32^\circ+j\sin57.32^\circ) \\ &=-33774.82+17009.16j\text{ mm/s}^2. \end{aligned}\]

   The real parts are the $x$-components and the imaginary parts are the $y$-components.

5. The results are

   Quantity	Value
   $\alpha_3$	$296.11\text{ rad/s}^2$
   $\alpha_4$	$470.15\text{ rad/s}^2$
   $A_{Ax}$	$-19536.78\text{ mm/s}^2$
   $A_{Ay}$	$-15610.06\text{ mm/s}^2$
   $A_{BAx}$	$-14238.04\text{ mm/s}^2$
   $A_{BAy}$	$32619.22\text{ mm/s}^2$
   $A_{Bx}$	$-33774.82\text{ mm/s}^2$
   $A_{By}$	$17009.16\text{ mm/s}^2$

Answer

1. Use the acceleration relations:

   \[\alpha_3=\frac{a\alpha_2\cos\theta_2-a\omega_2^2\sin\theta_2+b\omega_3^2\sin\theta_3}{b\cos\theta_3}\] \[\ddot d=-a\alpha_2\sin\theta_2-a\omega_2^2\cos\theta_2+b\alpha_3\sin\theta_3+b\omega_3^2\cos\theta_3\]

2. Substituting the data gives

   \[\alpha_3=\frac{35(12)\cos50^\circ-35(20^2)\sin50^\circ+110((-4.42)^2)\sin157.66^\circ}{110\cos157.66^\circ}=94.73\text{ rad/s}^2\] \[\ddot d=-35(12)\sin50^\circ-35(20^2)\cos50^\circ+110(94.7262)\sin157.66^\circ+110((-4.42)^2)\cos157.66^\circ=-7347.86\text{ mm/s}^2\]

   Therefore

   Quantity	Value
   $\alpha_3$	$94.73\text{ rad/s}^2$
   $\ddot d$	$-7347.86\text{ mm/s}^2$

Answer

1. Start from the twice-differentiated vector loop equation

   \[\frac{d^2}{dt^2}\left(\mathbf{R}_A+\mathbf{R}_{BA}-\mathbf{R}_{BO_2}\right)=0\]

2. Write the vectors in complex form:

   \[\mathbf{R}_A=a e^{j\theta_2}, \qquad \mathbf{R}_{BA}=-b e^{j\theta_3}, \qquad \mathbf{R}_{BO_2}=d+jc\]

3. In complex form, the twice-differentiated loop is

   \[a(j\alpha_2-\omega_2^2)e^{j\theta_2}-b(j\alpha_3-\omega_3^2)e^{j\theta_3}-\ddot d=0\]

4. Expand with Euler’s relation $e^{j\theta}=\cos\theta+j\sin\theta$:

   \[a(j\alpha_2-\omega_2^2)(\cos\theta_2+j\sin\theta_2)-b(j\alpha_3-\omega_3^2)(\cos\theta_3+j\sin\theta_3)-\ddot d=0\]

5. Simplify the products by using

   \[(j\alpha-\omega^2)(\cos\theta+j\sin\theta) = (-\alpha\sin\theta-\omega^2\cos\theta) + j(\alpha\cos\theta-\omega^2\sin\theta)\] \[\begin{aligned} &a(-\alpha_2\sin\theta_2-\omega_2^2\cos\theta_2+j(\alpha_2\cos\theta_2-\omega_2^2\sin\theta_2)) \\ &\quad-b(-\alpha_3\sin\theta_3-\omega_3^2\cos\theta_3+j(\alpha_3\cos\theta_3-\omega_3^2\sin\theta_3))-\ddot d=0 \end{aligned}\]

6. Separate real and imaginary parts:

   \[-a\alpha_2\sin\theta_2-a\omega_2^2\cos\theta_2+b\alpha_3\sin\theta_3+b\omega_3^2\cos\theta_3-\ddot d=0\] \[a\alpha_2\cos\theta_2-a\omega_2^2\sin\theta_2-b\alpha_3\cos\theta_3+b\omega_3^2\sin\theta_3=0\]

7. Rearranging the real equation gives

   \[\ddot d=-a\alpha_2\sin\theta_2-a\omega_2^2\cos\theta_2+b\alpha_3\sin\theta_3+b\omega_3^2\cos\theta_3\]

8. Rearranging the imaginary equation gives

   \[b\alpha_3\cos\theta_3=a\alpha_2\cos\theta_2-a\omega_2^2\sin\theta_2+b\omega_3^2\sin\theta_3\] \[\alpha_3=\frac{a\alpha_2\cos\theta_2-a\omega_2^2\sin\theta_2+b\omega_3^2\sin\theta_3}{b\cos\theta_3}\]