Chapter 18.2, Problem 18.103P

A 2.5-kg homogeneous disk of radius 80 mm rotates with an angular velocity ω₁ with respect to arm ABC, which is welded to a shaft DCE rotating as shown at the constant rate ω₂ = 12 rad/s. Friction in the bearing at A causes ω₁ to decrease at the rate of 15 rad/s². Determine the dynamic reactions at D and E at a time when ω₁ has decreased to 50 rad/s.

Fig. P18.103 and P18.104

To determine

The dynamic reactions at D and E at a time when ${\omega }_1$ has decreased to 50 rad/s.

Answer to Problem 18.103P

The dynamic reactions at Dat a time when ${\omega }_1$ has decreased to 50 rad/s is $\underset{\_}{-\left(22.0\text{N}\right)i+\left(26.8\text{N}\right)j}$.

The dynamic reactions at Eat a time when ${\omega }_1$ has decreased to 50 rad/s is $\underset{\_}{-\left(21.2\text{N}\right)i-\left(5.20\text{N}\right)j}$.

Explanation of Solution

Given information:

The mass (m) of the disk is 2.5kg.

The radius (r) of the disk Ais 80 mm.

The angular velocity $\left({\omega }_2\right)$ of the shaft DCE is12 rad/s.

The decreasing acceleration $\left({\dot{\omega }}_1\right)$ of the shaft DCE is $15{\text{rad/s}}^2$.

Calculation:

The angular velocity $\left({\omega }_x\right)$ of disk A along the x-axis is zero.

Write the equation of angular velocity of disk A$\left({\omega }_y\right)$ along the y-axis:

${\omega }_y={\omega }_1$

Write the equation of angular velocity $\left({\omega }_z\right)$ of disk A along the z-axis:

${\omega }_z={\omega }_2$

Find the equation of angular velocity $\left(\omega \right)$ of disk.

$\omega ={\omega }_{x}i+{\omega }_{y}j+{\omega }_{z}k$

Substitute 0 for ${\omega }_x$, ${\omega }_1$ for ${\omega }_y$, and ${\omega }_2$ for ${\omega }_z$.

$\begin{array}{c}\omega =\left(0\right)i+\left({\omega }_1\right)j+\left({\omega }_2\right)k\\ ={\omega }_{1}j+{\omega }_{2}k\end{array}$

Find the equation of angular momentum about A $\left(H_A\right)$ about A.

$H_A={\overline{I}}_x{\omega }_{x}i+{\overline{I}}_y{\omega }_{y}j+{\overline{I}}_z{\omega }_{z}k$

Here, ${\overline{I}}_x$ is the moment of inertia in the x direction, ${\overline{I}}_y$ is the moment of inertia in the y direction, and ${\overline{I}}_z$ is the moment of inertia in the z direction.

Substitute 0 for ${\omega }_x$, ${\omega }_1$ for ${\omega }_y$, and ${\omega }_2$ for ${\omega }_z$.

$\begin{array}{c}{H}_A={\overline{I}}_x\left(0\right)i+{\overline{I}}_y{\omega }_{1}j+{\overline{I}}_z{\omega }_{2}k\\ ={\overline{I}}_y{\omega }_{1}j+{\overline{I}}_z{\omega }_{2}k\end{array}$ (1)

Find the rate of change of angular momentum ${\left({\dot{H}}_A\right)}_{Axyz}$ about the reference frame.

${\left({\dot{H}}_A\right)}_{Axyz}={\overline{I}}_y{\dot{\omega }}_{1}j+{\overline{I}}_z{\dot{\omega }}_{2}k$

Here, ${\dot{\omega }}_1$ is the angular acceleration disk, and ${\dot{\omega }}_2$ is the acceleration of shaft CBD and arm.

Write the equation of vector form of angular velocity $\left(\Omega \right)$ of the reference frame Axyz.

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

Write the equation of the rate of change of angular momentum about A$\left({\dot{H}}_A\right)$.

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

Substitute ${\overline{I}}_y{\dot{\omega }}_{1}j+{\overline{I}}_z{\dot{\omega }}_{2}k$ for ${\left({\dot{H}}_A\right)}_{Axyz}$, ${\omega }_{2}j$ for $\Omega$, and ${\overline{I}}_y{\dot{\omega }}_{1}j+{\overline{I}}_z{\dot{\omega }}_{2}k$ for ${\dot{H}}_A$.

$\begin{array}{c}{\dot{H}}_A=\left({\overline{I}}_y{\dot{\omega }}_{1}j+{\overline{I}}_z{\dot{\omega }}_{2}k\right)+{\omega }_{2}k\times \left({\overline{I}}_y{\omega }_{1}j+{\overline{I}}_z{\omega }_{2}k\right)\\ =\left({\overline{I}}_y{\dot{\omega }}_{1}j+{\overline{I}}_z{\dot{\omega }}_{2}k\right)-{\overline{I}}_y{\omega }_1{\omega }_{2}i+0\\ =-{\overline{I}}_y{\omega }_1{\omega }_{2}i+{\overline{I}}_y{\dot{\omega }}_{1}j+{\overline{I}}_z{\dot{\omega }}_{2}k\end{array}$ (2)

Write the equation mass moment of inertia $\left({\overline{I}}_y\right)$ along y-axis.

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

Write the equation mass moment of inertia $\left({\overline{I}}_z\right)$ along z-axis.

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

Substitute $\frac{1}{2}m{r}^2$ for ${\overline{I}}_y$ and $\frac{1}{4}m{r}^2$ for ${\overline{I}}_z$ in Equation (2).

${\dot{H}}_A=-\frac{1}{2}m{r}^2{\omega }_1{\omega }_{2}i+\frac{1}{2}m{r}^2{\dot{\omega }}_{1}j+\frac{1}{4}m{r}^2{\dot{\omega }}_{2}k$ (3)

Find the position vector $\left({\text{r}}_{A/C}\right)$ of A with respect to C.

$r_{A/C}=bi-cj$

Here, b is the horizontal distance and c is the vertical distance.

Write the equation of velocity $\left(v_A\right)$ of the mass center A of the disk.

$v_A={\omega }_{2}k\times r_{A/C}$

Substitute $bi-cj$ for ${\text{r}}_{A/C}$.

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

Write the equation of acceleration of the mass center A of the disk.

$a_A={\dot{\omega }}_{2}k\times r_{A/C}+{\omega }_{2}k\times {\text{v}}_A$

Substitute $c{\omega }_{2}i+b{\omega }_{2}j$ for ${\text{v}}_A$ and $bi-cj$ for ${\text{r}}_{A/C}$.

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

Sketch the free body diagram and kinetic diagram of the system as shown in Figure (1).

Refer Figure (1),

Apply Newton’s law of motion.

$\begin{array}{l}\Sigma F=m\overline{a}\\ D_{x}i+D_{y}j+E_{x}i+E_{y}j=m{a}_A\end{array}$

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

$\begin{array}{c}{D}_{x}i+D_{y}j+E_{x}i+E_{y}j=m\left[\left(c{\dot{\omega }}_2-b{\omega }^2{}_2\right)i+\left(b{\dot{\omega }}_2+c{\omega }^2{}_2\right)j\right]\\ =m\left(c{\dot{\omega }}_2-b{\omega }^2{}_2\right)i+m\left(b{\dot{\omega }}_2+c{\omega }^2{}_2\right)j\end{array}$ (4)

Equate i-vector coefficients in Equation (4).

$D_x+E_x=m\left(c{\dot{\omega }}_2-b{\omega }^2{}_2\right)$ (5)

Equate j-vector coefficients in Equation (4).

$D_y+E_y=m\left(b{\dot{\omega }}_2+c{\omega }^2{}_2\right)$ (6)

Find the rate of change of angular momentum about E $\left({\dot{H}}_E\right)$.

${\dot{H}}_E={\dot{H}}_A+{\text{r}}_{A/E}\times m{\text{a}}_A$

Here, ${\text{r}}_{A/E}$ is the position vector of E with respect to A.

Substitute $-\frac{1}{2}m{r}^2{\omega }_1{\omega }_{2}i+\frac{1}{2}m{r}^2{\dot{\omega }}_{1}j+\frac{1}{4}m{r}^2{\dot{\omega }}_{2}k$ for ${\dot{H}}_A$, $bi-cj+lk$ for ${\text{r}}_{A/E}$, and $\left(c{\dot{\omega }}_2-b{\omega }^2{}_2\right)i+\left(b{\dot{\omega }}_2+c{\omega }^2{}_2\right)j$ for ${\text{a}}_A$.

${\dot{H}}_E=\left[\begin{array}{l}\left(-\frac{1}{2}m{r}^2{\omega }_1{\omega }_{2}i+\frac{1}{2}m{r}^2{\dot{\omega }}_{1}j+\frac{1}{4}m{r}^2{\dot{\omega }}_{2}k\right)+\\ \left(bi-cj+lk\right)\times m\left(\left(c{\dot{\omega }}_2-b{\omega }^2{}_2\right)i+\left(b{\dot{\omega }}_2+c{\omega }^2{}_2\right)j\right)\end{array}\right]$

Apply matrix multiplication,

$\begin{array}{c}{\dot{H}}_E=\left[\begin{array}{l}-\frac{1}{2}m{r}^2{\omega }_1{\omega }_{2}i+\frac{1}{2}m{r}^2{\dot{\omega }}_{1}j+\frac{1}{4}m{r}^2{\dot{\omega }}_{2}k+-lm\left(b{\dot{\omega }}_2+c{\omega }^2{}_2\right)i\\ +lm\left(c{\dot{\omega }}_2-b{\omega }^2{}_2\right)j+bm\left(b{\dot{\omega }}_2+c{\omega }^2{}_2\right)k\\ +cm\left(c{\dot{\omega }}_2-b{\omega }^2{}_2\right)k\end{array}\right]\\ =\left[\begin{array}{l}m\left(-\frac{1}{2}{r}^2{\omega }_1{\omega }_2-bl{\dot{\omega }}_2-cl{\omega }^2{}_2\right)i\\ +m\left(\frac{1}{2}{r}^2{\dot{\omega }}_1+cl{\dot{\omega }}_1-bl{\omega }^2{}_2\right)j\\ +m\left(\frac{1}{4}{r}^2+b^2+c^2\right){\dot{\omega }}_{2}k\end{array}\right]\end{array}$ (7)

Take moment about E.

$\begin{array}{c}{M}_E=M_{0}k+2lk\times \left(D_{x}i+D_{y}j\right)\\ =-2l{D}_{y}i+2l{D}_{x}j+M_{0}k=\end{array}$ (8)

Here, $C_z$ is the dynamic reaction at C along z-axis, $M_0$ is the couple at O, and $C_x$ is the dynamic reaction at C along x-axis.

The moment at E is equal to the rate of change of angular momentum at E.

Equate Equation (7) and (8).

$2l{D}_{x}j-2l{D}_{y}i+M_{0}k=\left[\begin{array}{l}m\left(-\frac{1}{2}{r}^2{\omega }_1{\omega }_2-bl{\dot{\omega }}_2-cl{\omega }^2{}_2\right)i+\\ m\left(\frac{1}{2}{r}^2{\dot{\omega }}_1+cl{\dot{\omega }}_2-bl{\omega }^2{}_2\right)j+\\ m\left(\frac{1}{4}{r}^2+b^2+c^2\right){\dot{\omega }}_{2}k\end{array}\right]$ (9)

Convert the unit of radius from mm to m.

$\begin{array}{c}r=\left(80\text{mm}\right)\left(\frac{\text{1}\text{m}}{1,000\text{mm}}\right)\\ =0.08\text{m}\end{array}$

Convert the unit of b from mm to m.

$\begin{array}{c}b=\left(120\text{mm}\right)\left(\frac{\text{1}\text{m}}{1,000\text{mm}}\right)\\ =0.120\text{m}\end{array}$

Convert the unit of c from mm to m.

$\begin{array}{c}c=\left(60\text{mm}\right)\left(\frac{\text{1}\text{m}}{1,000\text{mm}}\right)\\ =0.06\text{m}\end{array}$

Convert the unit of l from mm to m.

$\begin{array}{c}l=\left(150\text{mm}\right)\left(\frac{\text{1}\text{m}}{1,000\text{mm}}\right)\\ =0.15\text{m}\end{array}$

Find the component of dynamic reaction $\left(D_x\right)$ at D along x-axis.

Equate $j$ vector in Equation (9).

$\begin{array}{l}2l{D}_x=m\left(\frac{1}{2}{r}^2{\dot{\omega }}_1+cl{\dot{\omega }}_2-bl{\omega }^2{}_2\right)\\ D_x=\frac{m}{2l}\left(\frac{1}{2}{r}^2{\dot{\omega }}_1+cl{\dot{\omega }}_2-bl{\omega }^2{}_2\right)\end{array}$ (10)

Substitute 2.5 kg for $m$, 0.08 m for r, 0.120 m for b, 0.06 m for c, 0 for ${\dot{\omega }}_2$, 0.15 m for l, $12\text{rad}/\text{s}$ for ${\omega }_2$, $-15{\text{rad/s}}^2$ for ${\dot{\omega }}_1$, and $12\text{rad}/\text{s}$ for ${\omega }_2$.

$\begin{array}{c}{D}_x=\frac{2.5}{2\left(0.15\right)}\left(\begin{array}{l}\frac{1}{2}{\left(0.08\right)}^2\left(-15\right)+\left(0.06\right)\left(0.15\right)\left(0\right)\\ -\left(0.12\right)\left(0.15\right){\left(12\right)}^2\end{array}\right)\\ =8.3333\left(-0.048+0-2.592\right)\\ =-22\text{N}\end{array}$

Find the component of dynamic reaction $\left(D_y\right)$ at D along y-axis.

Equate $i$ vector in Equation (9).

$\begin{array}{l}-2l{D}_y=m\left(-\frac{1}{2}{r}^2{\omega }_1{\omega }_2-bl{\dot{\omega }}_2-cl{\omega }^2{}_2\right)\\ D_y=\frac{m}{2l}\left(\frac{1}{2}{r}^2{\omega }_1{\omega }_2+bl{\dot{\omega }}_2+cl{\omega }^2{}_2\right)\end{array}$ (11)

Substitute 2.5 kg for $m$, 0.08 m for r, 0.120 m for b, 0.06 m for c, $0$ for ${\dot{\omega }}_2$, 0.15 m for l, $50\text{rad}/\text{s}$ for ${\omega }_1$, and $12\text{rad}/\text{s}$ for ${\omega }_2$.

$\begin{array}{c}{D}_y=\frac{2.5}{2\left(0.15\right)}\left(\begin{array}{l}\frac{1}{2}{\left(0.08\right)}^2\left(12\right)\left(50\right)\\ +\left(0.12\right)\left(0.15\right)\left(0\right)\\ +\left(0.06\right)\left(0.15\right){\left(12\right)}^2\end{array}\right)\\ =8.3333\left(1.92+0+1.296\right)\\ =26.8\text{N}\end{array}$

Find the dynamic reaction at D using the equation:

$D=D_{x}i+D_{y}j$

Substitute $-22\text{N}$ for $D_x$ and $26.8\text{N}$ for $D_y$.

$D=-\left(22\text{N}\right)i+\left(26.8\text{N}\right)j$

Thus, the dynamic reaction at D is $\underset{\_}{-\left(22\text{N}\right)i+\left(26.8\text{N}\right)j}$.

Find the component of dynamic reaction $\left(E_x\right)$ at E along x-axis.

Substitute Equation (11) in (5).

$\begin{array}{l}\frac{m}{2l}\left(\frac{1}{2}{r}^2{\dot{\omega }}_1+cl{\dot{\omega }}_2-bl{\omega }^2{}_2\right)+E_x=m\left(c{\dot{\omega }}_2-b{\omega }^2{}_2\right)\\ E_x=mc{\dot{\omega }}_2-mb{\omega }^2{}_2-\frac{mc{\dot{\omega }}_2}{2}+\frac{mb{\omega }^2{}_2}{2}-\left(\frac{1}{2}{r}^2{\dot{\omega }}_1\right)\left(\frac{m}{2l}\right)\\ E_x=\frac{mc{\dot{\omega }}_2}{2}-\frac{mb{\omega }^2{}_2}{2}-\left(\frac{m}{2l}\right)\frac{1}{2}{r}^2{\dot{\omega }}_1\\ E_x=\left(\frac{m}{2l}\right)\left(-\frac{1}{2}{r}^2{\dot{\omega }}_1+cl{\dot{\omega }}_2-bl{\omega }^2{}_2\right)\end{array}$

Substitute 2.5 kg for $m$, 0.08 m for r, 0.120 m for b, 0.06 m for c,$0$ for ${\dot{\omega }}_2$, 0.15 m for l, $12\text{rad}/\text{s}$ for ${\omega }_2$, $-15{\text{rad/s}}^2$ for ${\dot{\omega }}_1$, and $12\text{rad}/\text{s}$ for ${\omega }_2$.

$\begin{array}{c}{E}_x=\frac{2.5}{2\left(0.15\right)}\left(\begin{array}{l}-\frac{1}{2}{\left(0.08\right)}^2\left(-15\right)+\left(0.06\right)\left(0.15\right)\left(0\right)\\ -\left(0.12\right)\left(0.15\right){\left(12\right)}^2\end{array}\right)\\ =8.3333\left(0.048+0-2.592\right)\\ =-21.2\text{N}\end{array}$

Find the component of dynamic reaction $\left(E_y\right)$ at E along y-axis.

Substitute Equation (12) in (6).

$\begin{array}{l}\frac{m}{2l}\left(\frac{1}{2}{r}^2{\omega }_1{\omega }_2+bl{\dot{\omega }}_2+cl{\omega }^2{}_2\right)+E_y=m\left(b{\dot{\omega }}_2+c{\omega }^2{}_2\right)\\ E_y=mb{\dot{\omega }}_2+mc{\omega }^2{}_2-\frac{mb{\dot{\omega }}_2}{2}-\frac{mc{\omega }^2{}_2}{2}-\left(\frac{m}{2l}\right)\left(\frac{1}{2}{r}^2{\omega }_1{\omega }_2\right)\\ =\frac{mb{\dot{\omega }}_2}{2}+\frac{mc{\omega }^2{}_2}{2}-\left(\frac{m}{2l}\right)\left(\frac{1}{2}{r}^2{\omega }_1{\omega }_2\right)\\ =\left(\frac{m}{2l}\right)\left(-\frac{1}{2}{r}^2{\omega }_1{\omega }_2+bl{\dot{\omega }}_2+cl{\omega }^2{}_2\right)\end{array}$

Substitute 2.5 kg for $m$, 0.08 m for r, 0.120 m for b, 0.06 m for c, $0$ for ${\dot{\omega }}_2$, 0.15 m for l, $50\text{rad}/\text{s}$ for ${\omega }_1$, and $12\text{rad}/\text{s}$ for ${\omega }_2$.

$\begin{array}{c}{E}_y=\frac{2.5}{2\left(0.15\right)}\left(\begin{array}{l}-\frac{1}{2}{\left(0.08\right)}^2\left(12\right)\left(50\right)+\\ \left(0.12\right)\left(0.15\right)\left(0\right)+\\ \left(0.06\right)\left(0.15\right){\left(12\right)}^2\end{array}\right)\\ =8.3333\left(-1.92+0+1.296\right)\\ =-5.20\text{N}\end{array}$

Find the dynamic reaction at E using the equation:

$E=E_{x}i+E_{y}j$

Substitute $-21.2\text{N}$ for $E_x$ and $-5.20\text{N}$ for $E_y$.

$D=-\left(21.2\text{N}\right)i-\left(5.20\text{N}\right)j$

Thus, the dynamic reaction at D is $\underset{\_}{-\left(21.2\text{N}\right)i-\left(5.20\text{N}\right)j}$.