Problem

Solve Example 20.5 such that the x, y, z axes move with curvilinear translation, Ω = 0 in...

Solve Example 20.5 such that the x, y, z axes move with curvilinear translation, Ω = 0 in which case the collar appears to have both an angular velocity Ωxyz = ω1 + ω2 and radial motion.

Step-by-Step Solution

Solution 1

Write the coordinate transformation equation relative to $$\mathrm{XYZ}$$ as follows:

\begin{aligned} &r_{B}=0.5 \hat{k} \mathrm{~m} \\ &v_{B}=2 \hat{j} \mathrm{~m} / \mathrm{s} \\ &a_{s}=(0.75 \hat{j}+8 \hat{k}) \mathrm{m} / \mathrm{s}^{2} \end{aligned}

Again for this transformation,

$$\Omega=0$$ and $$S=0$$

Now, for transformation of coordinates relative to $$x y z$$ (where $$x^{\prime} y^{\prime} z^{\prime}$$ remains coincident with $$\mathrm{xyz}$$ coordinate axis) the value for $$\left(r_{C B}\right)_{x=}$$ is $$0.2 \hat{j} \mathrm{~m}$$.

Calculate the value for velocity:

$$\left(\dot{r}_{C B}\right)_{x y z}=\left(\dot{r}_{C B}\right)_{x y z}+\left(\omega_{1}+\omega_{2}\right) \times\left(r_{C B}\right)_{n g z}$$

Here, $$\omega_{1}$$ and $$\omega_{2}$$ are the angular velocities and $$r_{C B}$$ is the rotation of $$C$$ with respect to $$B$$ Substitute, $$4 \hat{i}$$ for $$\omega_{1}, 5 \hat{k}$$ for $$\omega_{2}$$ and $$0.2 \hat{j}$$ for $$r_{C B}$$

\begin{aligned} &\left(v_{C B}\right)_{x=}=3 \hat{j}+(4 \hat{i}+5 \hat{k}) \times 0.2 \hat{j} \\ &\left(v_{C B}\right)_{x z}=-1 \hat{i}++3 \hat{j}+0.8 \hat{k} \end{aligned}

And:

\begin{aligned} &\left(\dot{r}_{C B}\right)_{x y}=\left(\dot{r}_{C s}\right)_{x y z}+\left(\omega_{1}+\omega_{2}\right) \times\left(\dot{r}_{C B}\right)_{x y z^{\prime}}+\left(\dot{\omega}_{1}+\dot{\omega}_{2}\right) \times\left(r_{C B}\right)_{x z}+\left(\omega_{1}+\omega_{2}\right) \times\left(\dot{r}_{C B}\right)_{x y} \\ &\left(a_{C B}\right)_{x z z}=\{2 \hat{j}+(4 \hat{i}+5 \hat{k}) \times 3 \hat{j}\}+\{(1.5 \hat{i}-6 \hat{k}) \times 0.2 \hat{j}\}+\{(4 \hat{i}+5 \hat{k}) \times(-1 \hat{i}+3 \hat{j}+0.8 \hat{k})\} \\ &\left(a_{C B}\right)_{x y y}=-28.8 \hat{i}-6.2 \hat{j}+24.3 \hat{k} \end{aligned}

So finally:

\begin{aligned} &v_{C}=v_{s}+\Omega \times\left(r_{C B}\right)_{x y z}+\left(v_{C B}\right)_{x z} \\ &v_{C}=2 \hat{j}+0+(-1 \hat{i}+3 \hat{j}+0.8 \hat{k}) \end{aligned}

Calculate the value for acceleration:

\begin{aligned} a_{C} &=a_{B}+\Omega \times\left(r_{C B}\right)_{x=7}+\Omega \times\left\{\Omega \times\left(r_{C D}\right)_{x n=}\right\}+2 \Omega \times\left(v_{C B}\right)_{x y z}+\left(a_{C B}\right)_{x=z} \\ &=(0.75 \hat{j}+8 \hat{k})+(-28.8 \hat{i}-6.2 \hat{j}+24.3 \hat{k}) \\ &=(-28.8 \hat{i}-5.45 \hat{j}+32.3 \hat{k}) \mathrm{m} / \mathrm{s}^{2} \end{aligned}