The position of the stylus tip A is controlled by the robot shown. In the position shown the stylus moves at a constant speed u = 180 mm/s relative to the solenoid BC. At the same time, arm CD rotates at the constant rate ω2 = 1.6 rad/s with respect to component DEG. Knowing that the entire robot

Question

The position of the stylus tip A is controlled by the robot shown. In the position shown the stylus moves at a constant speed u=180 mm/s relative to the solenoid BC. At the same time, arm CD rotates at the constant rate [latex]{ \omega }_{ 2 }=1.6 \ rad/s[/latex] with respect to component DEG. Knowing that the entire robot rotates about the X axis at the constant rate [latex]{ \omega }_{ 1 }=1.2 \ rad/s[/latex],determine (a) the velocity of A, (b) the acceleration of A.

Accepted Answer

Geometry:

[latex]\begin{aligned}& {\bf r}_{D/G}=-(500\;\mathrm{mm}){\bf i}+(300\;\mathrm{mm}){\bf j}+(600\;\mathrm{mm}){\bf k}\& \mathbf{r}_{A/D}=-(250~\mathrm{mm})\mathbf{i}+(600~\mathrm{mm})\mathbf{k}\& \mathbf{r}_{A/G}=\mathbf{r}_{A/D}+\mathbf{r}_{D/G}=-(750\;\mathrm{mm})\mathbf{i}+(300\;\mathrm{mm})\mathbf{j}+(600\;\mathrm{mm})\mathbf{k}\end{aligned}[/latex]

Angular velocities:

[latex]\begin{aligned}& \boldsymbol{\omega}_{1}=\omega_{1}\mathbf{i}=(1.2\;\mathrm{rad/s})\mathbf{i}\qquad(\dot{\omega}_{1}=0)\& \boldsymbol{\omega}_{2}=\omega_{2}{\bf j}=(1.6\,\mathrm{rad/s}){\bf j}\quad(\dot{\omega}_{2}=0)\end{aligned}[/latex]

Stylus motion: [latex]\quad\mathbf{u}=-u\mathbf{i}=-(180\,\mathrm{mm}/\mathbf{s})\mathbf{i}\qquad({\dot{u}}=0)[/latex]

Method 1

Let the rigid body BCD be a rotating frame of reference.

Its angular velocity is

[latex]\boldsymbol{\omega}_{C D}=\boldsymbol{\omega}_{1}+\boldsymbol{\omega}_{2}=(1.2\;\mathrm{rad/s})\mathbf{i}+(1.6\;\mathrm{rad/s}){\mathbf{j}}[/latex]

Its angular acceleration is

[latex]\begin{aligned}& \boldsymbol{\alpha}_{C D}=\boldsymbol{\omega}_{1} imes\boldsymbol{\omega}_{C D}=(1.2\;\mathrm{rad/s})\mathbf{i} imes[(1.2\;\mathrm{rad/s})\mathbf{i}+(1.6\;\mathrm{rad/s})\mathbf{j}]\& \quad\quad =(1.92\operatorname{rad/s}^{2})\mathbf{k}\end{aligned}[/latex]

Motion of the coinciding Point A′ in the frame.

[latex]\begin{aligned}& \mathbf{v}_{A^{\prime}}=\mathbf{v}_{D}+\mathbf{v}_{A^{\prime}/D}\& \mathbf{v}_{A^{\prime}}=\boldsymbol{\omega}_{1} imes\mathbf{r}_{D/G}+(\boldsymbol{\omega}_{1}+\boldsymbol{\omega}_{2}) imes\mathbf{r}_{A/D}\& \quad\quad =(1.2\operatorname{rad}{\mathfrak{s}})\mathbf{i} imes\left[-(500\operatorname{mm})\mathbf{i}+(300\operatorname{mm})\mathbf{j} ight]\& \quad\quad +[(1.2\,\mathrm{{rad/s}}){\bf{i}}+(1.6\,\mathrm{{rad/s}}){\bf{j}}]{ imes}[-(2.50\,\mathrm{{mm}}){\bf{i}}+(600\,\mathrm{{mm}}){{\bf k}}]\& \quad\quad =(360\;\mathrm{mm}/{\bf{s}}){\bf{k}}-(720\;\mathrm{mm}/{\bf{s}}){\bf{i}}+(400\;\mathrm{mm}/{\bf{s}}){\bf{k}}+(960\;\mathrm{mm/s}){\bf{i}}\& \mathbf{v}_{\boldsymbol{A}^{\prime}}=(960~\mathrm{mm/s})\mathbf{i}-(720~\mathrm{mm/s})\mathbf{j}+(760~\mathrm{mm/s})\mathbf{k}\& {\bf a}_{D}=\omega_{1} imes(\boldsymbol{\omega}_{1} imes{\bf r}_{D/G})\& \quad\quad =(1.2\,\mathrm{{rad/s}})\mathbf{i} imes\{(1.2\,\mathrm{{rad/s}})\mathbf{i} imes[-(500\,\mathrm{{mm}})\mathbf{i}+(300\,\mathrm{{mm}})\mathbf{j}]\}\& \quad\quad =-(432\operatorname{mm/s}^{2})\mathbf{j}\& {\bf a}_{A^{\prime}/D}=\boldsymbol{\alpha}_{C D} imes{\bf r}_{A/D}+\boldsymbol{\omega}_{C D} imes(\boldsymbol{\omega}_{C D} imes{\bf r}_{A/D})\& \quad\quad =(1.92\ \mathrm{rad/s}^{2})\mathbf{k} imes[-(250\ \mathrm{mm})\mathbf{i}+(600\ \mathrm{mm})\mathbf{k}]\& \quad\quad +\boldsymbol{\omega}_{C D} imes\{[(1.2\mathrm{~rad/s}){\bf{i}}+(1.6\mathrm{~rad/s}){\bf{j}}\} imes\{-(2.50\mathrm{~mm}){\bf{i}}+(600\mathrm{~mm}){{\bf k}}]\}\end{aligned}[/latex]

[latex]\begin{aligned}& {\bf a_{A^{\prime}\!{/}D}}=-(480\;\mathrm{mm/s^{2}}){\bf j}+\omega_{C D} imes[-(720\mathrm{~mm/s})\mathbf{j}+(400\mathrm{~mm/s}){\bf k}+(960\mathrm{~mm/s}){\bf i}]\& \quad\quad =-480\mathbf{j}+ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\ 1.2 & 1.6 & 0 \ 960 & -720 & 400 \end{vmatrix} \& \quad\quad =-480{\bf j}+640{\bf i}-864{\bf k}-1536{\bf k}\& \mathbf{a_{A^{\prime}\!/\!D\!}}=(640\;\mathrm{mm/s}^{2})\mathbf{i}-(960\;\mathrm{mm/s}^{2})\mathbf{j}-(2400\;\mathrm{mm/s}^{2})\mathbf{k}\& \quad {\bf a}_{A^{\prime}}={\bf a}_{D}+{\bf a}_{A^{\prime}/D}\& \quad \mathbf{a}_{A^{\prime}}=(640\;\mathrm{mm}/{\mathbf{s}}^{2}){\mathbf{i}}-(1392\;\mathrm{mm}/{\mathbf{s}}^{2}){\mathbf{j}}-(2400\;\mathrm{mm}/{\mathbf{s}}^{2}){\mathbf{k}}\end{aligned}[/latex]

Motion of Point A relative to the frame.

[latex]\begin{aligned}& \mathbf{v}_{A/F}=\mathbf{u}=-(180\,\mathrm{mm}/\mathrm{s})\mathbf{i}\& {\mathbf{a}}_{A/F}=0\end{aligned}[/latex]

(a) Velocity of A.

[latex]\begin{aligned}& \mathbf{v}_{A}=\mathbf{v}_{A^{\prime}}+\mathbf{v}_{A/F}\& \mathbf{v}_{A}=(960\;\mathrm{mm}/{\mathrm{s}})\mathbf{i}-(720\;\mathrm{mm}/{\mathrm{s}})\mathbf{j}+(760\;\mathrm{mm}/\mathrm{s})\mathbf{k}-(180\;\mathrm{mm}/\mathrm{s})\mathbf{k}\&& \mathbf{v}_{A}=(0.78~\mathrm{m}/\mathrm{s}){{\mathbf{i}}}-(0.72~\mathrm{m}/\mathrm{s}){{\mathbf{j}}}+(0.76~\mathrm{m}/\mathrm{s})\mathbf{k}\blacktriangleleft\end{aligned}[/latex]

Coriolis acceleration:

[latex]\begin{aligned}& {\bf a}_{c}=2\boldsymbol{\omega}_{C D} imes{\bf v}_{A/F}\& {\bf a}_{c}=2[(1.2\;\mathrm{{rad/s}}){\bf i}+(1.6\;\mathrm{{rad/s}}){\bf j}] imes(-180\;\mathrm{mm/s}){\bf i}\& \quad\quad =+(576\,\mathrm{mm/s}^{2}\,)\mathbf{k}\end{aligned}[/latex]

(b) Acceleration of A.

[latex]\begin{aligned}& {\bf a}_{A}={\bf a}_{A^{\prime}}+{\bf a}_{A/F}+{\bf a}_{c}\& \mathbf{a}_{A}=(640\;\mathrm{mm/s}^{2})\mathbf{i}-(1392\;\mathrm{mm/s}^{2})\mathbf{j}-(2400\;\mathrm{mm/s}^{2})\mathbf{k}+(576\;\mathrm{mm/s}^{2})\mathbf{k}\&& \mathbf{a}_{A}=(0.64\;{\mathrm{m/s}}^{2}){{\mathbf{i}}}-(1.392\;{\mathrm{m/s}}^{2}){{\mathbf{j}}}-(1.824\;{\mathrm{m/s}}^{2})\mathbf{k}\blacktriangleleft\end{aligned}[/latex]

Method 2

Use a frame of reference rotating about the x axis with angular velocity.

[latex]\boldsymbol{\omega}_{1}=\omega_{1}\mathbf{i}=(1.2\;\mathrm{rad/s})\mathbf{i}\qquad(\dot{\omega}_{1}=0)[/latex]

Motion of coinciding Point A′ in the frame.

[latex]\begin{aligned}& \mathbf{v}_{A^{\prime}}=\boldsymbol{\omega}_{1} imes\mathbf{r}_{A/G}=(1.2\;\mathrm{rad}/{\bf s})\mathbf{i} imes[-(500\;\mathrm{mm})\mathbf{i}+(300\;\mathrm{mm})\mathbf{j}+(600\;\mathrm{mm})\mathbf{k}]\& \quad\quad =(360{\mathrm{\ mm}}/{\mathrm{s}}){\mathbf{k}}-(720{\mathrm{\ mm}}/{\mathrm{s}}){\mathbf{j}}\& {\bf a}_{A^{\prime}}=\omega_{1} imes(\boldsymbol{\omega}_{1} imes{\bf r}_{A/G})=\boldsymbol{\omega}_{1} imes{\bf v}_{A^{\prime}}\& \quad\quad =(1.2\,\mathrm{rad/s})\mathbf{i} imes[(360\,\mathrm{mm/s})\mathbf{k}-(720\,\mathrm{mm/s})\mathbf{j}]\& \quad\quad =-(432\,\mathrm{mm/s^{2}}\,)\mathbf{j}-(864\,\mathrm{mm/s^{2}})\mathbf{k}\end{aligned}[/latex]

Motion of Point A relative to the frame.

[latex]\begin{aligned}& \mathbf{v}_{A/F}=\boldsymbol{\omega}_{2} imes\mathbf{r}_{A/D}+\mathbf{u}\& \quad\quad =(1.6\;\mathrm{{rad/s}})\mathbf{j} imes\lbrack-(250\;\mathrm{{mm}})\mathbf{i}+(600\;\mathrm{{mm}})\mathbf{k} brack-(180\;\mathrm{{mm}}/\mathbf{s})\mathbf{i}\& \quad\quad =(400\mathrm{\,mm}/{\mathrm{s}})\mathbf{k}+(960\mathrm{\,mm}/{\mathrm{s}})\mathbf{i}-(180\mathrm{\,mm}/{\mathrm{s}})\mathbf{i}\end{aligned}[/latex]

[latex]\mathbf{a}_{A/F}:[/latex] (Since A moves on CD, which rotates at rate [latex]\boldsymbol{\omega}_{2}[/latex] , we have a Coriolis term here).

[latex]\begin{aligned}& {\bf a}_{A/F}=\boldsymbol{\omega}_{2} imes(\boldsymbol{\omega}_{2} imes{\bf r}_{A/D})+2\boldsymbol{\omega}_{2} imes{\bf u}\& \quad\quad =\boldsymbol{\omega}_{2} imes\{(1.6\;\mathrm{rad/s})\mathbf{j} imes[-(250\;\mathrm{mm})\mathbf{i}+(600\;\mathrm{mm})\mathbf{k}\;]\}+2\boldsymbol{\omega}_{2} imes\mathbf{u}\& \quad\quad =(1.6\mathrm{~rad/s})\mathbf{j} imes[(400\mathrm{~mm/s})\mathbf{k}+(960\mathrm{~mm/s})\mathbf{i}+2(1.6\mathrm{\ rad/s})\mathbf{j} imes(-180\mathrm{~mm})\mathbf{i}\& \quad\quad =(640\;\mathrm{mm/s}^{2}){\bf i}-(1536\;\mathrm{mm/s}^{2}){\bf k}+(576\;\mathrm{mm/s}^{2}){\bf k}\& \quad\quad =(640\;\mathrm{mm/s}^{2}){\bf i}-(960\;\mathrm{mm/s}^{2}){\bf k}\end{aligned}[/latex]

(a) Velocity of A.

[latex]\begin{aligned}& \mathbf{v}_{A}=\mathbf{v}_{A^{\prime}}+\mathbf{v}_{A/F}\& \mathbf{v}_{A}=360\mathbf{k}-720\mathbf{j}+400\mathbf{k}+960\mathbf{i}-180\mathbf{i}\&& \mathbf{v}_{A}=(0.78\;\mathrm{m}/\mathrm{s})\mathbf{i}-(0.72\;\mathrm{m}/\mathrm{s})\mathbf{j}+(0.76\;\mathrm{m}/\mathrm{s})\mathbf{k}\blacktriangleleft\end{aligned}[/latex]

Coriolis acceleration:

[latex]\begin{aligned}& \mathbf{a}_{c}=2\boldsymbol{\omega}_{1} imes\mathbf{v}_{A/F}\& \mathbf{a}_{c}=2(1.2\mathrm{~rad/s})\mathbf{i} imes[(400\mathrm{~mm})\mathbf{k}+(780\mathrm{~mm}/\mathbf{s})\mathbf{i}]\& \quad\quad =-(960{\mathrm{~mm/s}}^{2})\mathbf{j}\end{aligned}[/latex]

(b) Acceleration of A.

[latex]\begin{aligned}& {\bf a}_{A}={\bf a}_{A^{\prime}}+{\bf a}_{A/F}+{\bf a}_{c}\& \mathbf{a}_{A}=-432\mathbf{j}-864\mathbf{k}+640\mathbf{i}-960\mathbf{k}-960\mathbf{j}\&& \mathbf{a}_{A}=(0.64\;{\mathrm{m/s}}^{2}){{\mathbf{i}}}-(1.392\;{\mathrm{m/s}}^{2}){{\mathbf{j}}}-(1.824\;{\mathrm{m/s}}^{2}){\mathbf{k}}\blacktriangleleft\end{aligned}[/latex]

Question 15.247: The position of the stylus tip A is controlled by the robot ...

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