Chapter 15.6, Problem 15.202P

In Prob. 15.201, the speed of point B is known to be constant. For the position shown, determine (a) the angular acceleration of the guide arm, (b) the acceleration of point C.

15.201    Several rods are brazed together to form the robotic guide arm shown that is attached to a ball-and-socket joint at O. Rod OA slides in a straight inclined slot, while rod OB slides in a slot parallel to the z axis. Knowing that at the instant shown v_B = (9 in./s)k, determine (a) the angular velocity of the guide arm, (b) the velocity of point A, (c) the velocity of point C.

Fig. P15.201

To determine

The angular acceleration of the guide arm.

Answer to Problem 15.202P

The angular acceleration of the guide arm is $\underset{\_}{\alpha =\left(1.125\text{rad}/{\text{s}}^2\right)\text{k}}$.

Explanation of Solution

Given information:

The velocity ${\text{v}}_B=\left(9\text{in}\text{.}/\text{s}\right)\text{k}$

Calculation:

Sketch the Free Body Diagram of the robotic guide arm as shown in Figure 1.

Refer to Figure 1.

The velocity at D along the x axis is ${\left(v_D\right)}_x=-2{\left(v_D\right)}_y$                                            (1)

The velocity at A along the x axis is ${\left(v_A\right)}_x=-2{\left(v_A\right)}_y$                                             (2)

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

The position of A.

$r_A=\left(10\text{in}\text{.}\right)\text{k}$

The position of B.

$r_B=\left(12\text{in}\text{.}\right)\text{j}$

The position of C.

$r_C=\left(5\text{in}\text{.}\right)\text{i}+\left(4\text{in}\text{.}\right)\text{j}+\left(2\text{in}\text{.}\right)\text{k}$

Consider the angular velocity $\omega ={\omega }_x\text{i}+{\omega }_y\text{j}+{\omega }_z\text{k}$ (3)

Consider the velocity at A ${\text{v}}_A={\left(v_A\right)}_x\text{i}+{\left(v_A\right)}_y\text{j}+{\left(v_A\right)}_z\text{k}$                                       (4)

Calculate the angular velocity of each plane using the relation of ${\text{v}}_B$ as shown below.

${\text{v}}_B=\omega \times r_B$

Substitute ${\omega }_x\text{i}+{\omega }_y\text{j}+{\omega }_z\text{k}$ for $\omega$, $\left(9\text{in}\text{.}/\text{s}\right)\text{k}$ for $v_B$, and $\left(12\text{in}\text{.}\right)\text{j}$ for $r_B$.

$\begin{array}{c}9\text{k}=\left({\omega }_x\text{i}+{\omega }_y\text{j}+{\omega }_z\text{k}\right)\times \left(\text{12j}\right)\\ =\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ {\omega }_x& {\omega }_y& {\omega }_z\\ 0& 12& 0\end{array}\right|\\ =\text{i}\left(0-12{\omega }_z\right)-\text{j}\left(0\right)+\text{k}\left(12{\omega }_x-0\right)\\ =-12{\omega }_z\text{i}+12{\omega }_x\text{k}\end{array}$

Resolving i, j and k components.

For i component.

$\begin{array}{l}-12{\omega }_z=0\\ {\omega }_z=0\end{array}$

For k component.

$\begin{array}{l}12{\omega }_x=9\\ {\omega }_x=0.75\text{rad}/\text{s}\end{array}$

Calculate the angular velocity of each plane using the relation of $v_A$ as shown below.

$v_A=\omega \times r_A$

Substitute ${\omega }_x\text{i}+{\omega }_y\text{j}+{\omega }_z\text{k}$ for $\omega$, ${\left(v_A\right)}_x\text{i}+{\left(v_A\right)}_y\text{j}+{\left(v_A\right)}_z\text{k}$ for ${\text{v}}_A$, and $\left(10\text{in}\text{.}\right)\text{k}$ for $r_A$.

$\begin{array}{c}{\left(v_A\right)}_x\text{i}+{\left(v_A\right)}_y\text{j}+{\left(v_A\right)}_z\text{k}=\left({\omega }_x\text{i}+{\omega }_y\text{j}+{\omega }_z\text{k}\right)\times \left(10\text{k}\right)\\ =\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ {\omega }_x& {\omega }_y& {\omega }_z\\ 0& 0& 10\end{array}\right|\\ =\text{i}\left(10{\omega }_y-0\right)-\text{j}\left(10{\omega }_x-0\right)+\text{k}\left(0\right)\\ =10{\omega }_y\text{i}-10{\omega }_x\text{j}\end{array}$

Substitute $0.75\text{rad}/\text{s}$ for ${\omega }_x$.

$\begin{array}{c}{\left(v_A\right)}_x\text{i}+{\left(v_A\right)}_y\text{j}+{\left(v_A\right)}_z\text{k}=10{\omega }_y\text{i}-\left(10\times 0.75\right)\text{j}\\ =10{\omega }_y\text{i}-7.5\text{j}\end{array}$

Resolving i, j and k components.

For i component.

${\left(v_A\right)}_x=10{\omega }_y$ (5)

For j component.

${\left(v_A\right)}_y=-7.5\text{in}\text{.}/\text{s}$

For k component.

${\left(v_A\right)}_z=0$

Calculate the velocity at A ${\left(v_A\right)}_x$ as shown below.

Substitute $-7.5\text{in}\text{.}/\text{s}$ for ${\left(v_A\right)}_y$ in Equation (2).

$\begin{array}{c}{\left(v_A\right)}_x=-2\times -7.5\text{in}\text{.}/\text{s}\\ =15\text{in}\text{.}/\text{s}\end{array}$

Calculate the angular velocity along y direction $\left({\omega }_y\right)$ as shown below.

Substitute $15\text{in}\text{.}/\text{s}$ for ${\left(v_A\right)}_x$ in Equation (5).

$\begin{array}{l}15=10{\omega }_y\\ {\omega }_y=1.5\text{rad}/\text{s}\end{array}$

Calculate the angular velocity $\left(\omega \right)$ as shown below.

Substitute $0.75\text{rad}/\text{s}$ for ${\omega }_x$, $1.5\text{rad}/\text{s}$ for ${\omega }_y$, and $0$ for ${\omega }_z$ in Equation (3).

$\begin{array}{c}\omega =\left(0.75\text{rad}/\text{s}\right)\text{i}+\left(1.5\text{rad}/\text{s}\right)\text{j}+\left(0\right)\text{k}\\ =\left(0.75\text{rad}/\text{s}\right)\text{i}+\left(1.5\text{rad}/\text{s}\right)\text{j}\end{array}$

Calculate the velocity of point A $\left({\text{v}}_A\right)$ as shown below.

Substitute $15\text{in}\text{.}/\text{s}$ for ${\left(v_A\right)}_x$, $-7.5\text{in}\text{.}/\text{s}$ for ${\left(v_A\right)}_y$, and $0$ for ${\left(v_A\right)}_z$ in Equation (4).

$\begin{array}{c}{\text{v}}_A=\left(15\text{in}\text{.}/\text{s}\right)\text{i}+\left(-7.5\text{in}\text{.}/\text{s}\right)\text{j}+\left(0\right)\text{k}\\ =\left(15\text{in}\text{.}/\text{s}\right)\text{i}-\left(7.5\text{in}\text{.}/\text{s}\right)\text{j}\end{array}$

Calculate the velocity of point C $\left({\text{v}}_C\right)$ as shown below.

${\text{v}}_C=\text{ω}\times r_C$

Substitute $\left(0.75\text{rad}/\text{s}\right)\text{i}+\left(1.5\text{rad}/\text{s}\right)\text{j}$ for $\omega$ and $\left(5\text{in}\text{.}\right)\text{i}+\left(4\text{in}\text{.}\right)\text{j}+\left(2\text{in}\text{.}\right)\text{k}$ for $r_C$.

$\begin{array}{c}{\text{v}}_C=\left(0.75\text{i}+1.5\text{j}\right)\times \left(\left(5\text{in}\text{.}\right)\text{i}+\left(4\text{in}\text{.}\right)\text{j}+\left(2\text{in}\text{.}\right)\text{k}\right)\\ =\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 0.75& 1.5& 0\\ 5& 4& 2\end{array}\right|\\ =\text{i}\left(3-0\right)-\text{j}\left(1.5-0\right)+\text{k}\left(3-7.5\right)\\ =3\text{i}-1.5\text{j}-4.5\text{k}\end{array}$

$=\left(3\text{in}\text{.}/\text{s}\right)\text{i}-\left(1.5\text{in}\text{.}/\text{s}\right)\text{j}-\left(4.5\text{in}\text{.}/\text{s}\right)\text{k}$

Consider the angular acceleration $\alpha ={\alpha }_z\text{i}+{\alpha }_y\text{j}+{\alpha }_z\text{k}$ (6)

Consider the acceleration at A ${\text{a}}_A={\left(a_A\right)}_x\text{i}+{\left(a_A\right)}_y\text{j}+{\left(a_A\right)}_z\text{k}$.

Consider the acceleration at B ${\text{a}}_B={\left(a_B\right)}_y\text{j}$.

Calculate the angular acceleration at B of each plane using the relation of ${\text{a}}_B$ as shown below.

${\text{a}}_B=\alpha \times r_B+\omega \times {\text{v}}_B$

Substitute ${\left(a_B\right)}_y\text{j}$ for ${\text{a}}_B$, $\left(12\text{in}\text{.}\right)\text{j}$ for $r_B$, ${\alpha }_z\text{i}+{\alpha }_y\text{j}+{\alpha }_z\text{k}$ for $\alpha$, $\left(0.75\text{rad}/\text{s}\right)\text{i}+\left(1.5\text{rad}/\text{s}\right)\text{j}$ for $\omega$, and $\left(9\text{in}\text{.}/\text{s}\right)\text{k}$ for ${\text{v}}_B$.

$\begin{array}{c}{\left(a_B\right)}_y\text{j}=\left[\left({\alpha }_z\text{i}+{\alpha }_y\text{j}+{\alpha }_z\text{k}\right)\times \left(12\text{j}\right)+\left(0.75\text{i}+1.5\text{j}\right)\times \left(9\text{k}\right)\right]\\ =\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ {\alpha }_x& {\alpha }_y& {\alpha }_z\\ 0& 12& 0\end{array}\right|+\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 0.75& 1.5& 0\\ 0& 0& 9\end{array}\right|\\ =\left[\begin{array}{l}\left(\text{i}\left(0-12{\alpha }_z\right)-\text{j}\left(0\right)+\text{k}\left(12{\alpha }_x-0\right)\right)+\\ \left(\text{i}\left(13.5-0\right)-\text{j}\left(6.75-0\right)+\text{k}\left(0\right)\right)\end{array}\right]\\ =\left(13.5-12{\alpha }_z\right)\text{i}-6.75\text{j}+12{\alpha }_x\text{k}\end{array}$

Resolving i, j and k components.

For i component.

$\begin{array}{l}13.5-12{\alpha }_z=0\\ 12{\alpha }_z=13.5\\ {\alpha }_z=1.125\text{rad}/{\text{s}}^{\text{2}}\end{array}$

For j component.

${\left(a_B\right)}_y=-6.75\text{in}\text{.}/{\text{s}}^2$

For k component.

$\begin{array}{l}12{\alpha }_x=0\\ {\alpha }_x=0\end{array}$

Calculate the angular acceleration at A of each plane using the relation of ${\text{a}}_A$ as shown below.

${\text{a}}_A=\alpha \times r_A+\omega \times {\text{v}}_A$

Substitute ${\left(a_A\right)}_x\text{i}+{\left(a_A\right)}_y\text{j}+{\left(a_A\right)}_z\text{k}$ for ${\text{a}}_A$, $\left(10\text{in}\text{.}\right)\text{k}$ for $r_A$, ${\alpha }_z\text{i}+{\alpha }_y\text{j}+{\alpha }_z\text{k}$ for $\alpha$, $\left(0.75\text{rad}/\text{s}\right)\text{i}+\left(1.5\text{rad}/\text{s}\right)\text{j}$ for $\omega$, and $\left(1.5\text{in}\text{.}/\text{s}\right)\text{i}-\left(7.5\text{in}\text{.}/\text{s}\right)\text{j}$ for ${\text{v}}_A$.

$\begin{array}{c}{\left(a_A\right)}_x\text{i}+{\left(a_A\right)}_y\text{j}+{\left(a_A\right)}_z\text{k}=\left[\begin{array}{l}\left({\alpha }_z\text{i}+{\alpha }_y\text{j}+{\alpha }_z\text{k}\right)\times \left(10\text{k}\right)+\\ \left(0.75\text{i}+1.5\text{j}\right)\times \left(15\text{i}-7.5\text{j}\right)\end{array}\right]\\ =\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ {\alpha }_x& {\alpha }_y& {\alpha }_z\\ 0& 0& 10\end{array}\right|+\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 0.75& 1.5& 0\\ 1.5& -7.5& 0\end{array}\right|\\ =\left[\begin{array}{l}\left(\text{i}\left(10{\alpha }_y-0\right)-\text{j}\left(10{\alpha }_x-0\right)+\text{k}\left(0\right)\right)+\\ \left(\text{i}\left(0\right)-\text{j}\left(0\right)+\text{k}\left(-5.625-22.5\right)\right)\end{array}\right]\\ =10{\alpha }_y\text{i}-10{\alpha }_x\text{j}-28.125\text{k}\end{array}$

Substitute 0 for ${\alpha }_x$.

$\begin{array}{c}{\left(a_A\right)}_x\text{i}+{\left(a_A\right)}_y\text{j}+{\left(a_A\right)}_z\text{k}=10{\alpha }_y\text{i}-10\left(0\right)\text{j}-28.125\text{k}\\ =10{\alpha }_y\text{i}-28.125\text{k}\end{array}$

Resolving i, j and k components.

For i component.

${\left(a_A\right)}_x=10{\alpha }_y$ (7)

For j component.

${\left(a_A\right)}_y=0$

For k component.

${\left(a_A\right)}_z=-28.125\text{in}\text{.}/{\text{s}}^2$

Refer to Figure 1.

The acceleration at A along the x axis ${\left(a_A\right)}_x=-2{\left(a_A\right)}_y$

Substitute $0$ for ${\left(a_A\right)}_y$.

${\left(a_A\right)}_x=0$

Calculate the angular acceleration along y axis $\left({\alpha }_y\right)$ as shown below.

Substitute $0$ for ${\left(a_A\right)}_x$ in Equation (7)

$\begin{array}{l}10{\alpha }_y=0\\ {\alpha }_y=0\end{array}$

Calculate the angular acceleration $\left(\alpha \right)$ as shown below.

Substitute $0$ for ${\alpha }_x$, $0$ for ${\alpha }_y$, and $1.125\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_z$ in Equation (6).

$\begin{array}{c}\alpha =0\text{i}+0\text{j}+\left(1.125\text{rad}/{\text{s}}^{\text{2}}\right)\text{k}\\ =\left(1.125\text{rad}/{\text{s}}^{\text{2}}\right)\text{k}\end{array}$

Therefore, the angular acceleration of the guide arm is $\underset{\_}{\alpha =\left(1.125\text{rad}/{\text{s}}^2\right)\text{k}}$.

To determine

Find the acceleration of point C.

Answer to Problem 15.202P

The acceleration of point C is $\underset{\_}{{\text{a}}_C=-\left(11.33\text{in}\text{.}/{\text{s}}^2\right)\text{i}+\left(9\text{in}\text{.}/{\text{s}}^2\right)\text{j}-\left(5.63\text{in}\text{.}/{\text{s}}^2\right)\text{k}}$.

Explanation of Solution

Given information:

The velocity at B is ${\text{v}}_B=\left(9\text{in}\text{.}/\text{s}\right)\text{k}$

Calculation:

Refer to part (a).

The angular velocity of the guide arm is $\omega =\left(0.75\text{rad}/\text{s}\right)\text{i}+\left(1.5\text{rad}/\text{s}\right)\text{j}$.

The angular acceleration of the guide arm is $\alpha =\left(1.125\text{rad}/{\text{s}}^2\right)\text{k}$.

The velocity at C is ${\text{v}}_C=\left(3\text{in}\text{.}/\text{s}\right)\text{i}-\left(1.5\text{in}\text{.}/\text{s}\right)\text{j}-\left(4.5\text{in}\text{.}/\text{s}\right)\text{k}$.

Calculate the acceleration at C $\left({\text{a}}_C\right)$ as shown below.

${\text{a}}_C=\alpha \times r_C+\omega \times {\text{v}}_C$

Substitute $\left(5\text{in}\text{.}\right)\text{i}+\left(4\text{in}\text{.}\right)\text{j}+\left(2\text{in}\text{.}\right)\text{k}$ for $r_C$, $\left(1.125\text{rad}/{\text{s}}^2\right)\text{k}$ for $\alpha$, $\left(0.75\text{rad}/\text{s}\right)\text{i}+\left(1.5\text{rad}/\text{s}\right)\text{j}$ for $\omega$, and $\left(3\text{in}\text{.}/\text{s}\right)\text{i}-\left(1.5\text{in}\text{.}/\text{s}\right)\text{j}-\left(4.5\text{in}\text{.}/\text{s}\right)\text{k}$ for ${\text{v}}_C$.

$\begin{array}{c}{\text{a}}_C=\left[\left(1.125\text{k}\right)\times \left(5\text{i}+4\text{j}+2\text{k}\right)+\left(0.75\text{i}+1.5\text{j}\right)\times \left(3\text{i}-1.5\text{j}-4.5\text{k}\right)\right]\\ =\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 0& 0& 1.125\\ 5& 4& 2\end{array}\right|+\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 0.75& 1.5& 0\\ 3& -1.5& -4.5\end{array}\right|\\ =\left[\begin{array}{l}\left(\text{i}\left(0-4.5\right)-\text{j}\left(0-5.625\right)+\text{k}\left(0\right)\right)+\\ \left(\text{i}\left(-6.75-0\right)-\text{j}\left(-3.375-0\right)+\text{k}\left(-1.125-4.5\right)\right)\end{array}\right]\\ =-11.25\text{i}+9\text{j}-5.625\text{k}\end{array}$

$=-\left(11.3\text{in}\text{.}/{\text{s}}^2\right)\text{i}+\left(9\text{in}\text{.}/{\text{s}}^2\right)\text{j}-\left(5.63\text{in}\text{.}/{\text{s}}^2\right)\text{k}$

Therefore, the acceleration of point C is $\underset{\_}{{\text{a}}_C=-\left(11.33\text{in}\text{.}/{\text{s}}^2\right)\text{i}+\left(9\text{in}\text{.}/{\text{s}}^2\right)\text{j}-\left(5.63\text{in}\text{.}/{\text{s}}^2\right)\text{k}}$.