Chapter 18.2, Problem 18.102P

To determine

(a)

The couple $M_0$ .

Answer to Problem 18.102P

The couple $M_0$ is $0.2115\text{ft}\cdot \text{lb}$.

Explanation of Solution

Given information:

Angular velocity of disk in z-direction is $60\text{rad}/\text{s}$, radius of the disk is $3\text{in}$ ,length of $BC$ member is $4\text{in}$.

The figure is represented below.

Figure (1)

Write the equation for the mass of the disk.

$m=\frac{W}{g}$ .........(1)

Here, weight of disk is $W$ and acceleration due to gravity is $g$.

Write the expression for the angular momentum about point $A$.

${\text{H}}_A={\overline{I}}_x{\omega }_x\cdot i+{\overline{I}}_y{\omega }_y\cdot j+{\overline{I}}_z{\omega }_z\cdot k$ .........(2)

Here, mass moment of inertia about the x-axis is ${\overline{I}}_x$, mass moment of inertia about the y-axis is ${\overline{I}}_y$, mass moment of inertia about the z-axis is ${\overline{I}}_z$, angular velocity of disk about x axis is ${\omega }_x$, angular velocity of disk about y axis is ${\omega }_y$ .and angular velocity of disk about z axis is ${\omega }_z$.

Write the expression for the angular velocity of disk in x direction.

${\omega }_x=0$ .........(3)

Substitute $0$ for ${\omega }_x$, ${\omega }_2$ for ${\omega }_y$, and ${\omega }_1$ for ${\omega }_z$ .in Equation (2).

$\begin{array}{c}{\text{H}}_A={\overline{I}}_x\times 0\times \cdot i+{\overline{I}}_y{\omega }_2\cdot j+{\overline{I}}_z{\omega }_1\cdot k\\ {\text{H}}_A={\overline{I}}_y{\omega }_2\cdot j+{\overline{I}}_z{\omega }_1\cdot k\end{array}$ .........(4)

Here, ${\omega }_2$ is the angular velocity in y-direction and ${\omega }_1$ is the angular velocity in z-direction.

Write the expression for angular velocity in vector form.

$\Omega ={\omega }_{2}j$ ..........(5)

Write the expression for rate of angular velocity of the reference frame $A_{xyz}$.

${\left({\text{<mover accent="’true'"><mo>H</mo><mo>˙</mo></mover>}}_A\right)}_{xyz}={\overline{I}}_y{\dot{\omega }}_2\cdot j+{\overline{I}}_z{\dot{\omega }}_1\cdot k$ .........(6)

Write the expression for rate of total angular velocity.

${\text{<mover accent="’true'"><mo>H</mo><mo>˙</mo></mover>}}_A={\left(\text{<mover accent="’true'"><mo>H</mo><mo>˙</mo></mover>}\right)}_{Ax\text{'}y\text{'}z\text{'}}+\Omega \times {\text{H}}_A$ .........(7)

Substitute $\left({\overline{I}}_y{\omega }_2\cdot j+{\overline{I}}_z{\omega }_1\cdot k\right)$ for ${\text{H}}_A$, ${\overline{I}}_y{\dot{\omega }}_2\cdot j+{\overline{I}}_z{\dot{\omega }}_1\cdot k$ for ${\left({\text{<mover accent="’true'"><mo>H</mo><mo>˙</mo></mover>}}_A\right)}_{xyz}$ and ${\omega }_{2}j$ for $\Omega$ in Equation (7).

${\text{<mover accent="’true'"><mo>H</mo><mo>˙</mo></mover>}}_A={\overline{I}}_y{\dot{\omega }}_2\cdot j+{\overline{I}}_z{\dot{\omega }}_1\cdot k+{\omega }_{2}j\times \left({\overline{I}}_y{\omega }_2\cdot j+{\overline{I}}_z{\omega }_1\cdot k\right)$ .........(8)

Write the expression for Matrix multiplication of the vector product for Equation (8).

$\begin{array}{c}{\text{<mover accent="’true'"><mo>H</mo><mo>˙</mo></mover>}}_A=\left({\overline{I}}_y{\dot{\omega }}_2\cdot j+{\overline{I}}_z{\dot{\omega }}_1\cdot k\right)+{\overline{I}}_z{\omega }_1{\omega }_{2}i\\ ={\overline{I}}_z{\omega }_1{\omega }_{2}i+{\overline{I}}_y{\dot{\omega }}_2\cdot j+{\overline{I}}_z{\dot{\omega }}_1\cdot k\end{array}$ .........(9)

Write the expression for the mass moment of inertia about the y-direction.

${\overline{I}}_y=\frac{1}{4}m{r}^2$ .........(10)

Here mass of the disk is $m$ and radius of the disk is $r$.

Write the expression for the mass moment of inertia about the z- direction.

${\overline{I}}_z=\frac{1}{2}m{r}^2$ .........(11)

Substitute $\frac{1}{4}m{r}^2$ for ${\overline{I}}_y$ and $\frac{1}{2}m{r}^2$ for ${\overline{I}}_z$ in Equation (9).

${\text{<mover accent="’true'"><mo>H</mo><mo>˙</mo></mover>}}_A=\frac{1}{2}m{r}^2\times {\omega }_1{\omega }_{2}i+\frac{1}{4}m{r}^2\times {\dot{\omega }}_2\cdot j+\frac{1}{2}m{r}^2\times {\dot{\omega }}_1\cdot k$ .........(12)

Write the expression for the velocity of mass centre of the disk.

$\overline{v}={\omega }_{2}j\times ci$ .........(13)

Here, velocity of mass centre is $\overline{v}$ and distance between $B$ and $A$ is $c$.

Write the expression for the matrix multiplication of the vector product for Equation (13).

$\overline{v}=-c{\omega }_{2}k$ .........(14)

Write the expression for the acceleration of the mass centre of the disk.

$\overline{a}={\dot{\omega }}_{2}j\times ci+{\dot{\omega }}_{2}j\times \overline{v}$ .........(15)

Write the expression for the matrix multiplication of the vector product for Equation (15).

$\overline{a}=-c{\dot{\omega }}_{2}k-c{\omega }_2^{2}i$ .........(16)

Write the expression for the the sum of the forces acting on the system.

$F=C_{x}i+C_{z}k{\text{+D}}_{x}i+{\text{D}}_{z}k$ .........(17)

Write the expression for the force in terms of mass and acceleration.

$F=m\overline{a}$ .........(18)

Substitute $C_{x}i+C_{z}k{\text{+D}}_{x}i+{\text{D}}_{z}k$ for $F$ in Equation (18).

$C_{x}i+C_{z}k{\text{+D}}_{x}i+{\text{D}}_{z}k\text{=m}\text{<mover accent="’true'"><mo>a</mo><mo>¯</mo></mover>}$ .........(19)

Here, force at $C$ in x-direction is $C_x$, force at $C$ in z-direction is $C_z$. force at $D$ in x-direction is $D_x$, and force at $D$ in z-direction is $D_z$.

Substitute $-c{\dot{\omega }}_{2}k-c{\omega }_2^{2}i$ for $\overline{a}$ in Equation (19)

$\begin{array}{c}{C}_{x}i+C_{z}k{\text{+D}}_{x}i+{\text{D}}_{z}k\text{=m}\times \left(-c{\dot{\omega }}_{2}k-c{\omega }_2^{2}i\right)\\ C_{x}i+C_{z}k{\text{+D}}_{x}i+{\text{D}}_{z}k=-\text{m}c{\dot{\omega }}_{2}k-\text{m}c{\omega }_2^{2}i\end{array}$ .........(20)

Compare the coefficients of the unit vector of $i$ on both side of Equation (20).

$C_x+{\text{D}}_x=-\text{m}c{\omega }_2^2$ .........(21)

Compare the coefficients of the unit vector of $i$ on both side of Equation.

$C_z{\text{+D}}_z=-\text{m}c{\dot{\omega }}_2$ .........(22)

Write the expression for the rate of angular momentum about $D$.

${\dot{H}}_D={\dot{H}}_A+r_{A/D}\times m\overline{a}$ .........(23)

Here, distance between $A$ and $D$ is $r_{A/D}$.

Write the expression for $r_{A/D}$ in vector form.

$r_{A/D}=-ci-bj$

Here, distance from the centre of disk to point $B$ is $c$ and distance of $BD$ is $b$.

Substitute $\left(\frac{1}{2}m{r}^2\times {\omega }_1{\omega }_{2}i+\frac{1}{4}m{r}^2\times {\dot{\omega }}_2\cdot j+\frac{1}{2}m{r}^2\times {\dot{\omega }}_1\cdot k\right)$ for ${\dot{H}}_A$, $\left(-ci-bj\right)$ for $r_{A/D}$ and ${\dot{\omega }}_{2}j\times ci+{\dot{\omega }}_{2}j\times \overline{v}$ for $\overline{a}$ in Equation(23).

${\dot{H}}_D=\left(\begin{array}{l}\frac{1}{2}m{r}^2\times {\omega }_1{\omega }_{2}i\\ +\frac{1}{4}m{r}^2\times {\dot{\omega }}_2\cdot j\\ +\frac{1}{2}m{r}^2\times {\dot{\omega }}_1\cdot k\end{array}\right)+\left(-ci-bj\right)\times m\left({\dot{\omega }}_{2}j\times ci+{\dot{\omega }}_{2}j\times \overline{v}\right)$ .........(24)

Write the expression for the matrix multiplication for vector product for equation (24).

$\begin{array}{c}{\dot{H}}_D=\left[\begin{array}{c}\left(\frac{1}{2}m{r}^2\times {\omega }_1{\omega }_{2}i+\frac{1}{4}m{r}^2\times {\dot{\omega }}_2\cdot j+\frac{1}{2}m{r}^2\times {\dot{\omega }}_1\cdot k\right)\\ -mb{\dot{\omega }}_{2}i+m{c}^2{\dot{\omega }}_{2}j+mbc{\omega }_2^{2}k\end{array}\right]\\ {\dot{H}}_D=\left[\begin{array}{c}m\left(\frac{1}{2}{r}^2{\omega }_1{\omega }_2-bc{\dot{\omega }}_2\right)i\\ +m\left(\frac{1}{4}{r}^2+c^2\right){\dot{\omega }}_{2}j+m\left(\frac{1}{2}{r}^2{\dot{\omega }}_1+bc{\omega }_2^2\right)k\end{array}\right]\end{array}$ .........(25)

Write the expression for the moment about $D$

$M_D=M_{0}j+2bj\times \left(C_{x}i+C_{z}k\right)$ .........(26)

Here, length of $BC$ member is $b$ and moment couple when system is at rest is $M_0$.

Write the expression for the matrix multiplication for the vector product for equation (26).

$M_D=2b{C}_{z}i+M_{0}j-2b{C}_{x}k$ .........(27)

Here $2b$ is the length of $DC$.

The sum of the moment at $D$ is equal to the rate of change of angular momentum at $D$.

$M_D={\dot{H}}_D$ .........(29)

Substitute $2b{C}_{z}i+M_{0}j-2b{C}_{x}k$ for $M_D$ in equation (25).

$\left[\begin{array}{l}2b{C}_{z}i\\ +M_{0}j\\ -2b{C}_{x}k\end{array}\right]=\left[\begin{array}{c}m\left(\frac{1}{2}{r}^2{\omega }_1{\omega }_2-bc{\dot{\omega }}_2\right)i\\ +m\left(\frac{1}{4}{r}^2+c^2\right){\dot{\omega }}_{2}j\\ +m\left(\frac{1}{2}{r}^2{\dot{\omega }}_1+bc{\omega }_2^2\right)k\end{array}\right]$ .........(30)

Compare the coefficients of the unit vector of $j$ on both side of Equation (30).

$M_0=m\left(\frac{1}{4}{r}^2+c^2\right){\dot{\omega }}_2$ .........(31)

Compare the coefficients of the unit vector of $k$ on both side of Equation (30).

$\begin{array}{l}-2b{C}_x=m\left(\frac{1}{2}{r}^2{\dot{\omega }}_1+bc{\omega }_2^2\right)\\ C_x=\frac{m}{-2b}\left(\frac{1}{2}{r}^2{\dot{\omega }}_1+bc{\omega }_2^2\right)\end{array}$ .........(32).

Compare the coefficients of the unit vector of $i$ on both side of Equation (30).

$\begin{array}{c}2b{C}_z=m\left(\frac{1}{2}{r}^2{\omega }_1{\omega }_2-bc{\dot{\omega }}_2\right)\\ C_z=\frac{m}{2b}\left(\frac{1}{2}{r}^2{\omega }_1{\omega }_2-bc{\dot{\omega }}_2\right)\end{array}$ .........(33)

Substitute $\frac{m}{-2b}\left(\frac{1}{2}{r}^2{\dot{\omega }}_1+bc{\omega }_2^2\right)$ for $C_x$ in Equation (21).

$\begin{array}{l}\left[\begin{array}{l}\frac{m}{-2b}\left(\frac{1}{2}{r}^2{\dot{\omega }}_1+bc{\omega }_2^2\right)\\ +D_x\end{array}\right]=-mc{\omega }_2^2\\ D_x=\frac{m}{2b}\left(\frac{1}{2}{r}^2{\dot{\omega }}_1+bc{\omega }_2^2\right)-mc{\omega }_2^2\\ D_x=\frac{mc{\omega }_2^2}{2}+\frac{m}{2b}\frac{1}{2}{r}^2{\dot{\omega }}_1\\ D_x=-\left(\frac{m}{2b}\right)\left(-\frac{1}{2}{r}^2{\dot{\omega }}_1+bc{\omega }_2^2\right)\end{array}$ .........(34)

Substitute $\frac{m}{2b}\left(\frac{1}{2}{r}^2{\omega }_1{\omega }_2-bc{\dot{\omega }}_2\right)$ for $C_z$ in Equation (22).

$\begin{array}{l}\left[\begin{array}{l}\frac{m}{2b}\left(\frac{1}{2}{r}^2{\omega }_1{\omega }_2-bc{\dot{\omega }}_2\right)\\ +D_z\end{array}\right]=-mc{\dot{\omega }}_2\\ D_z=-mc{\dot{\omega }}_2-\frac{m}{2b}\left(\frac{1}{2}{r}^2{\omega }_1{\omega }_2-bc{\dot{\omega }}_2\right)\\ D_z=-\frac{mc{\dot{\omega }}_2}{2}-\left(\frac{m}{2b}\right)\left(\frac{1}{2}{r}^2{\omega }_1{\omega }_2\right)\\ D_z=-\left(\frac{m}{2b}\right)\left(\frac{1}{2}{r}^2{\omega }_1{\omega }_2+bc{\dot{\omega }}_2\right)\end{array}$ .........(35)

Calculation:

Substitute $\text{6}\text{lb}$ for $W$ and $32.2\text{ft}/{\text{s}}^2$ for $g$ in Equation (1).

$\begin{array}{l}m=\frac{\text{6}\text{lb}}{32.2\text{ft}/{\text{s}}^2}\\ =0.1863\frac{\text{lb}\cdot {\text{s}}^2}{\text{ft}}\end{array}$

Substitute values of $0.1863\frac{\text{lb}\cdot {\text{s}}^2}{\text{ft}}$ for $m$, $\left(3\text{in}\right)$ for $r$, $\left(5\text{in}\right)$ for $c$ and $6\text{rad}/{\text{s}}^2$ for ${\dot{\omega }}_2$ in Equation (31).

$\begin{array}{c}{M}_0=0.1863\frac{\text{lb}\cdot {\text{s}}^2}{\text{ft}}\left(\frac{1}{4}{\left(3\text{in}\right)}^2+{\left(5\text{in}\right)}^2\right)\times 6\text{rad}/{\text{s}}^2\\ M_0=0.1863\frac{\text{lb}\cdot {\text{s}}^2}{\text{ft}}\times \left(\frac{1}{4}{\left(3\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)\right)}^2+{\left(5\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)\right)}^2\right)\times 6\text{rad}/{\text{s}}^2\\ M_0=0.1863\frac{\text{lb}\cdot {\text{s}}^2}{\text{ft}}\times \left({\left(0.25\text{ft}\right)}^2+{\left(0.4166\text{ft}\right)}^2\right)\times 6\text{rad}/{\text{s}}^2\\ M_0=0.212\text{ft}\cdot \text{lb}\end{array}$

Thus value of couple $M_0$ is $0.212\text{ft}\cdot \text{lb}$.

Conclusion:

The value of couple is $0.212\text{ft}\cdot \text{lb}$.

To determine

(b)

The dynamic reaction at $C$.

The dynamic reaction at $D$.

Answer to Problem 18.102P

The dynamic reaction at $C$ .is $-\left(12.58\text{lb}\right)i-\left(9.43\text{lb}\right)k$ .

The dynamic reaction at $D$ .is $-\left(12.58\text{lb}\right)i-\left(9.43\text{lb}\right)k$ .

Explanation of Solution

Given information:

Write the expression for the angular velocity in terms of time in y-direction.

${\omega }_2={\left({\omega }_2\right)}_0+{\dot{\omega }}_{2}t$ .........(36)

Here time is $t$.

Calculation:

Substitute $0$ for ${\left({\omega }_2\right)}_0$, $6\text{rad}/{\text{s}}^2$ for ${\dot{\omega }}_2$ and $3\text{s}$ for $t$ in Equation (36).

$\begin{array}{c}{\omega }_2=0+7.089\text{rad}/{\text{s}}^2\times 2\text{s}\\ \text{=}\text{14}\text{.179}\text{rad}/\text{s}\end{array}$

Substitute values of $0.1863\frac{\text{lb}\cdot {\text{s}}^2}{\text{ft}}$ for $m$, $\left(3\text{in}\right)$ for $r$, $\left(4\text{in}\right)$ for $b$, $\left(5\text{in}\right)$ for $c$, $0$ for ${\dot{\omega }}_1$, and $\text{18}\text{rad}/\text{s}$ for ${\omega }_2$ in Equation (32).

$\begin{array}{c}{C}_x=\left[\frac{0.1863\frac{\text{lb}\cdot {\text{s}}^2}{\text{ft}}}{-2\left(4\text{in}\right)}\times \left(\begin{array}{l}\frac{1}{2}{\left(3\text{in}\right)}^2\left(0\right)\\ +\left(4\text{in}\right)\left(5\text{in}\right)\left(18\text{rad}/{\text{s}}^2\right)\end{array}\right)\right]\\ C_x=\left[\frac{0.1863\frac{\text{lb}\cdot {\text{s}}^2}{\text{ft}}}{-2\left(4\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)\right)}\times \left(\begin{array}{l}\frac{1}{2}{\left(3\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)\right)}^2\left(0\right)\\ +\left(4\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)\right)\left(5\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)\right)\left(18\text{rad}/{\text{s}}^2\right)\end{array}\right)\right]\\ =-0.2795\left(0+44.999\right)\\ =-12.58\text{lb}\end{array}$

Substitute values of $0.1863\frac{\text{lb}\cdot {\text{s}}^2}{\text{ft}}$ for $m$, $\left(3\text{in}\right)$ for $r$, $\left(4\text{in}\right)$ for $b$, $\left(5\text{in}\right)$ for $c$, $60\text{rad}/\text{s}$ for ${\omega }_1$, $\text{18}\text{rad}/\text{s}$ for ${\omega }_2$ and $0$ for ${\dot{\omega }}_2$ in Equation (33).

$\begin{array}{c}{C}_z=\left[\frac{0.1863\frac{\text{lb}\cdot {\text{s}}^2}{\text{ft}}}{2\left(4\text{in}\right)}\times \left(\begin{array}{l}\frac{1}{2}{\left(3\text{in}\right)}^2\left(60\text{rad}/\text{s}\right)\left(18\text{rad}/\text{s}\right)\\ -\left(4\text{in}\right)\left(5\text{in}\right)\times 0\end{array}\right)\right]\\ C_z=\left[\frac{0.1863\frac{\text{lb}\cdot {\text{s}}^2}{\text{ft}}}{2\left(4\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)\right)}\times \left(\begin{array}{l}\frac{1}{2}{\left(3\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)\right)}^2\left(60\text{rad}/\text{s}\right)\left(18\text{rad}/\text{s}\right)\\ -\left(4\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)\right)\left(5\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)\right)\times 0\end{array}\right)\right]\\ =0.2795\left(33.75+0\right)\\ =-9.43\text{lb}\end{array}$

Hence, dynamic reaction at $C$ is $-\left(12.58\text{lb}\right)i-\left(9.43\text{lb}\right)k$.

Substitute values of $0.1863\frac{\text{lb}\cdot {\text{s}}^2}{\text{ft}}$ for $m$, $\left(3\text{in}\right)$ for $r$, $\left(4\text{in}\right)$ for $b$, $\left(5\text{in}\right)$ for $c$, $0$ for ${\dot{\omega }}_1$, and $18\text{rad}/\text{s}$ for ${\omega }_2$, Equation (34).

$\begin{array}{l}{D}_x=-\left(\frac{0.1863\frac{\text{lb}\cdot {\text{s}}^2}{\text{ft}}}{2\left(4\text{in}\right)}\right)\left(\begin{array}{l}-\frac{1}{2}{\left(3\text{in}\right)}^2\times 0\\ +\left(4\text{in}\right)\left(5\text{in}\right){\left(18\text{rad}/\text{s}\right)}^2\end{array}\right)\\ D_x=-\left(\frac{0.1863\frac{\text{lb}\cdot {\text{s}}^2}{\text{ft}}}{2\left(4\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)\right)}\right)\left(\begin{array}{l}-\frac{1}{2}{\left(3\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)\right)}^2\times 0\\ +\left(4\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)\right)\left(5\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)\right){\left(18\text{rad}/\text{s}\right)}^2\end{array}\right)\\ D_x=-0.2795\left(0+44.999\right)\text{lb}\\ D_x\text{=}-\text{12}\text{.58}\text{lb}\end{array}$

Substitute values of $0.1863\frac{\text{lb}\cdot {\text{s}}^2}{\text{ft}}$ for $m$, $\left(3\text{in}\right)$ for $r$, $\left(4\text{in}\right)$ for $b$, $\left(5\text{in}\right)$ for $c$, $60\text{rad}/\text{s}$ for ${\omega }_1$, $\text{18}\text{rad}/\text{s}$ for ${\omega }_2$ and $0$ for ${\dot{\omega }}_2$ in Equation (35).

$\begin{array}{l}{D}_z=-\left(\frac{0.1863\frac{\text{lb}\cdot {\text{s}}^2}{\text{ft}}}{2\left(4\text{in}\right)}\right)\left(\begin{array}{l}\frac{1}{2}{\left(3\text{in}\right)}^2\times \left(60\text{rad}/\text{s}\right)\times \left(18\text{rad}/\text{s}\right)\\ +\left(4\text{in}\right)\left(5\text{in}\right)\times 0\end{array}\right)\\ D_z=-\left(\frac{0.1863\frac{\text{lb}\cdot {\text{s}}^2}{\text{ft}}}{2\left(4\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)\right)}\right)\left(\begin{array}{l}\frac{1}{2}{\left(3\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)\right)}^2\times \left(60\text{rad}/\text{s}\right)\times \left(18\text{rad}/\text{s}\right)\\ +\left(4\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)\right)\left(5\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)\right)\times 0\end{array}\right)\\ D_z=-9.43\text{lb}\end{array}$

Hence, dynamic reaction at $D$ is $-\left(12.58\text{lb}\right)i-\left(9.43\text{lb}\right)k$

Conclusion:

The dynamic reactions at $C$ is $-\left(12.58\text{lb}\right)i-\left(9.43\text{lb}\right)k$.

The dynamic reactions at $D$ is $-\left(12.58\text{lb}\right)i-\left(9.43\text{lb}\right)k$.