Chapter 18, Problem 18.153RP

To determine

The dynamic reaction at D (D) and E (E).

Answer to Problem 18.153RP

The dynamic reaction at D (D) and E (E) are $\underset{\_}{-\left(7.12\text{lb}\right)j+\left(4.47\text{lb}\right)k}$ and $\underset{\_}{-\left(1.822\text{lb}\right)j+\left(4.47\text{lb}\right)k}$ respectively.

Explanation of Solution

Given Information:

The weight of the disk (W) is 6 lb.

The constant angular velocity $\left({\omega }_1\right)$ with respect to arm ABC is $16\text{rad/s}$.

The constant angular velocity $\left({\omega }_1\right)$ with respect to shaft DCE is $8\text{rad/s}$.

The radius (r) of the disk is $8\text{in}\text{.}$.

The length (l) of the rod is $12\text{in}\text{.}$.

The length of the rod from disk to point B (b) and B to C (c) is $9\text{in}\text{.}$.

Assume the acceleration due to gravity (g) as $32.2{\text{ft/s}}^{\text{2}}$.

Calculation:

Write the expression for the angular velocity $\left(\Omega \right)$ of the shaft DE and arm CBA:

$\Omega ={\omega }_{2}i$

Write the expression for the angular velocity $\left(\omega \right)$:

$\omega ={\omega }_{2}i+{\omega }_{1}j$

Write the expression the centroidal moment of inertia $\left({\overline{I}}_x\right)$ of disk about x axis:

${\overline{I}}_x=\frac{1}{4}m{r}^2$

Write the expression the centroidal moment of inertia $\left({\overline{I}}_y\right)$ of disk about y axis:

${\overline{I}}_y=\frac{1}{2}m{r}^2$

Write the express the angular moment$\left(H_A\right)$ about its mass centre A:

$\begin{array}{c}{H}_A={\overline{I}}_x{\omega }_{x}i+{\overline{I}}_y{\omega }_{y}j+{\overline{I}}_z{\omega }_{z}k\\ ={\overline{I}}_x{\omega }_{2}i+{\overline{I}}_y{\omega }_{1}j\end{array}$

Substitute $\frac{1}{4}m{r}^2$ for ${\overline{I}}_x$ and $\frac{1}{2}m{r}^2$ for ${\overline{I}}_y$.

$H_A=\frac{1}{4}m{r}^2{\omega }_{2}i+\frac{1}{2}m{r}^2{\omega }_{1}j$

Let the reference frame Oxyz be rotating with angular velocity $\left(\Omega \right)$ as below:

$\Omega ={\omega }_{2}i$

Write the express the angular momentum ${\left({\dot{H}}_A\right)}_{Oxyz}$of the reference frame Oxyz.

${\left({\dot{H}}_A\right)}_{Oxyz}={\overline{I}}_x{\dot{\omega }}_{2}i+{\overline{I}}_y{\dot{\omega }}_{1}j$

Substitute $\frac{1}{4}m{r}^2$ for $I_x$ and $\frac{1}{2}m{r}^2$ for $I_y$.

${\left({\dot{H}}_A\right)}_{Oxyz}=\frac{1}{4}m{r}^2{\dot{\omega }}_{2}i+\frac{1}{2}m{r}^2{\dot{\omega }}_{1}j$

Write the express the rate of change of angular $\left({\dot{H}}_A\right)$ momentum:

${\dot{H}}_A={\left({\dot{H}}_A\right)}_{Oxyz}+\Omega \times H_A$

Substitute $\frac{1}{4}m{r}^2{\dot{\omega }}_{2}i+\frac{1}{2}m{r}^2{\dot{\omega }}_{1}j$ for ${\left({\dot{H}}_A\right)}_{Oxyz}$, ${\omega }_{2}i$ for $\Omega$ and $\frac{1}{4}m{r}^2{\omega }_{2}i+\frac{1}{2}m{r}^2{\omega }_{1}j$ for $H_A$.

$\begin{array}{c}{\dot{H}}_A=\frac{1}{4}m{r}^2{\dot{\omega }}_{2}i+\frac{1}{2}m{r}^2{\dot{\omega }}_{1}j+{\omega }_{2}i\times \frac{1}{4}m{r}^2{\omega }_{2}i+\frac{1}{2}m{r}^2{\omega }_{1}j\\ =\frac{1}{4}m{r}^2{\dot{\omega }}_{2}i+\frac{1}{2}m{r}^2{\dot{\omega }}_{1}j+\frac{1}{2}m{r}^2{\omega }_1{\omega }_{2}k\end{array}$

Write the expression for the position vector $\left(r_{A/O}\right)$:

$r_{A/O}=\left(-bk+cj\right)$

Write the expression for the velocity $\left(v_A\right)$ of the disk A:

$v_A={\omega }_{2}i+r_{A/O}$

Substitute $\left(-bk+cj\right)$ and $r_{A/O}$.

$\begin{array}{c}{v}_A={\omega }_{2}i+\left(-bk+cj\right)\\ =b{\omega }_{2}j+c{\omega }_{2}k\end{array}$

Write the expression for the acceleration $\left(a_A\right)$ at point A:

$a_A={\dot{\omega }}_{2}j\times r_{A/O}+{\omega }_{2}j+v_A$

Substitute $b{\omega }_{2}j+c{\omega }_{2}k$ for $v_A$ and $\left(-bk+cj\right)$ for $r_{A/O}$.

$\begin{array}{c}{a}_A={\dot{\omega }}_{2}j\times \left(-bk+cj\right)+{\omega }_{2}j+b{\omega }_{2}j+c{\omega }_{2}k\\ =\left(b{\dot{\omega }}_2-c{\omega }_2^2\right)j+\left(c{\dot{\omega }}_2+b{\omega }_2\right)k\end{array}$

Show the impulse momentum diagram as in Figure (1).

Write the expression for the sum of the forces:

$\begin{array}{l}{\displaystyle \sum F}=m{a}_A\\ D_{y}j+D_{z}k+E_{y}j+E_{z}k=m{a}_A\end{array}$

Substitute $\left(b{\dot{\omega }}_2-c{\omega }_2^2\right)j+\left(c{\dot{\omega }}_2+b{\omega }_2\right)k$ for $a_A$.

$D_{y}j+D_{z}k+E_{y}j+E_{z}k=m\left(b{\dot{\omega }}_2-c{\omega }_2^2\right)j+\left(c{\dot{\omega }}_2+b{\omega }_2\right)k$

Resolve the i and k component,

$D_y+E_y=m\left(b{\dot{\omega }}_2-c{\omega }_2^2\right)$ (1)

$D_z+E_z=m\left(c{\dot{\omega }}_2+b{\omega }_2^2\right)$ (2)

Express the moment about the point D.

$\begin{array}{c}{\displaystyle \sum M_D}=(M_0)i+2li\times \left(E_{y}j+E_{z}k\right)\\ =M_0)i-2l{E}_{z}j+2l{E}_{y}k\end{array}$

Write the expression for the position vector $\left(r_{A/D}\right)$ for AD:

$r_{A/D}=li+cj-bk$

Write the expression for the sum of the moment about D:

${\displaystyle \sum M_D}={\dot{H}}_A+r_{A/D}\times m{a}_A$

Substitute $\frac{1}{4}m{r}^2{\dot{\omega }}_{2}i+\frac{1}{2}m{r}^2{\dot{\omega }}_{1}j+\frac{1}{2}m{r}^2{\omega }_1{\omega }_{2}k$ for ${\dot{H}}_A$, $M_0)i-2l{E}_{z}j+2l{E}_{y}k$ for ${\displaystyle \sum M_D}$, $li+cj-bk$ for $r_{A/D}$ and $\left(b{\dot{\omega }}_2-c{\omega }_2^2\right)j+\left(c{\dot{\omega }}_2+b{\omega }_2\right)k$ for $a_A$.

$\begin{array}{c}{M}_0)i-2l{E}_{z}j+2l{E}_{y}k=\left\{\begin{array}{c}\frac{1}{4}m{r}^2{\dot{\omega }}_{2}i+\frac{1}{2}m{r}^2{\dot{\omega }}_{1}j+\frac{1}{2}m{r}^2{\omega }_1{\omega }_{2}k+\\ \left(li+cj-bk\right)\times m\left(\begin{array}{l}\left(b{\dot{\omega }}_2-c{\omega }_2^2\right)j\\ +\left(c{\dot{\omega }}_2+b{\omega }_2\right)k\end{array}\right)\end{array}\right\}\\ (M_0)i-2l{E}_{z}j+2l{E}_{y}k=\left\{\begin{array}{c}m\left(\frac{1}{4}{r}^2+b^2+c^2\right){\dot{\omega }}_{2}i+m\left(\frac{1}{2}{r}^2{\dot{\omega }}_1-lc{\dot{\omega }}_2-lb{\omega }_2^2\right)j\\ +m\left(\frac{1}{2}{r}^2{\omega }_1{\omega }_2+lb{\dot{\omega }}_2-lc{\omega }_2^2\right)k\end{array}\right\}\end{array}$

Resolve the component i, j and k component.

For i component,

$M_0=m\left(\frac{1}{4}{r}^2+b^2+c^2\right){\dot{\omega }}_2$ (3)

For j component,

$\begin{array}{c}{E}_z=\frac{m}{2l}\left(-\frac{1}{2}{r}^2{\dot{\omega }}_1+lc{\dot{\omega }}_2+lb{\omega }_2^2\right)\\ =\frac{m}{2l}\left(lc{\dot{\omega }}_2+lb{\omega }_2^2-\frac{1}{2}{r}^2{\dot{\omega }}_1\right)\end{array}$ (4)

For k component,

$E_y=\frac{m}{2l}\left(\frac{1}{2}{r}^2{\omega }_1{\omega }_2+lb{\dot{\omega }}_2-lc{\omega }_2^2\right)$ (5)

Calculate the mass of the disk (m) using the relation:

$m=\frac{W}{g}$

Substitute 6lb for W and $32.2{\text{ft/s}}^{\text{2}}$ for g.

$\begin{array}{c}m=\frac{6}{32.2}\\ =0.186335\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}\end{array}$

Differentiate the angular velocity of ${\omega }_1$.

${\dot{\omega }}_1=0$

Differentiate the angular velocity of ${\omega }_2$.

${\dot{\omega }}_2=0$

Calculate the z component dynamic reaction $\left(E_z\right)$ at point E:

Substitute $0.186335\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}$ for m, 0 for ${\dot{\omega }}_2$, 0 for ${\dot{\omega }}_1$, 8 rad/s for , $8\text{in}\text{.}$ for r, 12 in. for l, $9\text{in}\text{.}$ for b and $9\text{in}\text{.}$ for c in Equation (4).

$\begin{array}{c}{E}_y=\frac{0.186335}{2\left(12\text{in}\text{.}\times \frac{\text{1}\text{ft}}{\text{12}\text{in}\text{.}}\right)}\left(0+\left(12\text{in}\text{.}\times \frac{\text{1}\text{ft}}{\text{12}\text{in}\text{.}}\right)\left(9\text{in}\text{.}\times \frac{\text{1}\text{ft}}{\text{12}\text{in}\text{.}}\right){\left(8\right)}^2-0\right)\\ =0.0931675\left(48\right)\\ =4.47\text{lb}\end{array}$

Calculate the y component dynamic reaction $\left(E_y\right)$ at point E:

Substitute $0.186335\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}$ for m, 0 for ${\dot{\omega }}_2$, 0 for ${\dot{\omega }}_1$, $16\text{rad/s}$ for ${\omega }_1$, 8 rad/s for ${\omega }_2$, $8\text{in}\text{.}$ for r, 12 in. for l, $9\text{in}\text{.}$ for b, and $9\text{in}\text{.}$ for c in Equation (5).

$\begin{array}{c}{E}_y=\frac{0.186335}{2\left(12\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}\left(\frac{1}{2}{\left(8\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}^2\left(16\times 8\right)+0-\left(12\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)\left(9\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right){\left(8\right)}^2\right)\\ =0.0931675\times \left(28.44-48\right)\\ =-1.822\text{lb}\end{array}$

Calculate the y component dynamic reaction $\left(D_y\right)$ at point D:

Substitute $9\text{in}\text{.}$ for b and c, $0.186335\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}$ for m, 0 for ${\dot{\omega }}_2$, 8 rad/s for ${\omega }_2$, and $-1.822\text{lb}$ for $E_y$ in Equation (1).

$\begin{array}{l}{D}_y+\left(-1.822\right)=0.186335\left( \left(9\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)\left(0\right)-\left(9\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right){\left(8\right)}^2\right)\\ D_y+\left(-1.822\right)=-8.94408\\ D_y=-8.94408+1.822\\ D_y=-7.12\text{lb}\end{array}$

Calculate the z component dynamic reaction $\left(D_z\right)$ at point D:

Substitute $9\text{in}\text{.}$ for b and c , $0.186335\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}$ for m, 0 for ${\dot{\omega }}_2$, 8 rad/s for ${\omega }_2$, and 4.47 lb for $E_z$ in Equation (2).

$\begin{array}{l}{D}_z+4.47=0.186335\left(\left(9\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)\left(0\right)+\left(9\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)8^2\right)\\ D_z+4.47=8.944\\ D_z=8.944-4.47\\ D_z=4.47\text{lb}\end{array}$

Calculate the dynamic reaction (D) at D:

$D=D_{y}j+D_{z}k$

Substitute $-7.12\text{lb}$ for $D_y$ and $4.47\text{lb}$ for $D_z$.

$D=-\left(7.12\text{lb}\right)j+\left(4.47\text{lb}\right)k$

Calculate the dynamic reaction (E) at E:

$E=E_{y}j+E_{z}k$

Substitute $-1.822\text{lb}$ for $E_y$ and $4.47\text{lb}$ for $E_z$.

$E=-\left(1.822\text{lb}\right)j+\left(4.47\text{lb}\right)k$

Thus, the dynamic reaction at D (D) and E (E) are $\underset{\_}{-\left(7.12\text{lb}\right)j+\left(4.47\text{lb}\right)k}$ and $\underset{\_}{-\left(1.822\text{lb}\right)j+\left(4.47\text{lb}\right)k}$.