Chapter 18.2, Problem 18.104P

A 2.5-kg homogeneous disk of radius 80 mm rotates at the constant rate ω₁ = 50 rad/s with respect to arm ABC, which is welded to a shaft DCE. Knowing that at the instant shown, shaft DCE has an angular velocity ω₂ = (12 rad/s)k and an angular acceleration α₂ = (8 rad/s²)k, determine (a) the couple that must be applied to shaft DCE to produce that acceleration, (b) the corresponding dynamic reactions at D and E.

Fig. P18.103 and P18.104

To determine

The couple which must be applied to shaft DCE to produce that acceleration.

Answer to Problem 18.104P

The couple which must be applied to shaft DCE to produce that acceleration is $\underset{\_}{\left(0.392\text{N}\cdot \text{m}\right)k}$.

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 angular acceleration $\left({\alpha }_2\right)$ of the shaft DCE is $\left(8{\text{rad/s}}^2\right)k$.

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 couple $\left(M_0\right)$ which must be applied to shaft DCE to produce that acceleration.

Equate $k$ vector in Equation (9).

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

Substitute 2.5 kg for $m$, 0.08 m for r, 0.120 m for b, 0.06 m for c, and $8{\text{rad}/\text{s}}^2$ for ${\dot{\omega }}_2$

$\begin{array}{c}{M}_0=\left(2.5\right)\left[\frac{1}{4}{\left(0.08\right)}^2+{\left(0.12\right)}^2+{\left(0.06\right)}^2\right]\left(8\right)\\ =2.5\left(0.0016+0.0144+0.0036\right)8\\ =\left(0.392\text{N}\cdot \text{m}\right)k\end{array}$

Thus, the couple which must be applied to shaft DCE to produce that acceleration is $\underset{\_}{\left(0.392\text{N}\cdot \text{m}\right)k}$.

To determine

Find the corresponding dynamic reactions at D and E.

Answer to Problem 18.104P

The dynamic reaction at D is $\underset{\_}{-\left(21.0\text{N}\right)i+\left(28.0\text{N}\right)j}$.

The dynamic reaction at E is $\underset{\_}{-\left(21.0\text{N}\right)i-\left(4.00\text{N}\right)j}$.

Explanation of Solution

Calculation:

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, $8{\text{rad}/\text{s}}^2$ for ${\dot{\omega }}_2$, 0.15 m for l, $12\text{rad}/\text{s}$ for ${\omega }_2$, 0 for ${\dot{\omega }}_1$,and $12\text{rad}/\text{s}$ for ${\omega }_2$.

$\begin{array}{c}{D}_x=-\frac{2.5}{2\left(0.15\right)}\left(\begin{array}{l}\frac{1}{2}{\left(0.12\right)}^2\left(0\right)+\left(0.06\right)\left(0.15\right)\left(8\right)\\ -\left(0.12\right)\left(0.15\right){\left(12\right)}^2\end{array}\right)\\ =-8.3333\left(0.072-2.592\right)\\ =-21\text{N}\end{array}$

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, $8{\text{rad}/\text{s}}^2$ for ${\dot{\omega }}_2$, 0.15 m for l, $50\text{rad}/\text{s}$ for ${\omega }_1$, and $12\text{rad}/\text{s}$ for ${\omega }_2$.

$\begin{array}{c}{D}_y=\frac{2.5}{2\left(0.15\right)}\left(\begin{array}{l}\frac{1}{2}{\left(0.08\right)}^2\left(12\right)\left(50\right)\\ +\left(0.12\right)\left(0.15\right)\left(8\right)\\ +\left(0.06\right)\left(0.15\right){\left(12\right)}^2\end{array}\right)\\ =8.3333\left(1.92+0.144+1.296\right)\\ =28\text{N}\end{array}$

Find the dynamic reaction at D using the equation:

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

Substitute $-21\text{N}$ for $D_x$ and $28\text{N}$ for $D_y$.

$D=-\left(21\text{N}\right)i+\left(28\text{N}\right)j$

Thus, the dynamic reaction at D is $\underset{\_}{-\left(21.0\text{N}\right)i+\left(28.0\text{N}\right)j}$.

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

Substitute Equation (11) in (5).

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

Substitute 2.5 kg for $m$, 0.08 m for r, 0.120 m for b, 0.06 m for c,$8{\text{rad}/\text{s}}^2$ for ${\dot{\omega }}_2$, 0.15 m for l, $12\text{rad}/\text{s}$ for ${\omega }_2$, 0 for ${\dot{\omega }}_1$, and $12\text{rad}/\text{s}$ for ${\omega }_2$.

$\begin{array}{c}{E}_x=-\frac{2.5}{2\left(0.15\right)}\left(\begin{array}{l}-\frac{1}{2}{\left(0.12\right)}^2\left(0\right)+\left(0.06\right)\left(0.15\right)\left(8\right)\\ -\left(0.12\right)\left(0.15\right){\left(12\right)}^2\end{array}\right)\\ =-8.3333\left(0.072-2.592\right)\\ =-21\text{N}\end{array}$

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, $8{\text{rad}/\text{s}}^2$ for ${\dot{\omega }}_2$, 0.15 m for l, $50\text{rad}/\text{s}$ for ${\omega }_1$, $12\text{rad}/\text{s}$ for ${\omega }_2$, and $12\text{rad}/\text{s}$ for ${\omega }_2$.

$\begin{array}{c}{E}_y=\frac{2.5}{2\left(0.15\right)}\left(\begin{array}{l}-\frac{1}{2}{\left(0.08\right)}^2\left(12\right)\left(50\right)+\\ \left(0.12\right)\left(0.15\right)\left(8\right)+\\ \left(0.06\right)\left(0.15\right){\left(12\right)}^2\end{array}\right)\\ =8.3333\left(-1.92+0.144+1.296\right)\\ =-4.00\text{N}\end{array}$

Find the dynamic reaction at E using the equation:

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

Substitute $-21\text{N}$ for $E_x$ and $-4.00\text{N}$ for $E_y$.

$D=-\left(21\text{N}\right)i-\left(4.00\text{N}\right)j$

Thus, the dynamic reaction at E is $\underset{\_}{-\left(21.0\text{N}\right)i-\left(4.00\text{N}\right)j}$.