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 \(r_{C B}\)

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


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

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:

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

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