Chapter 15.2, Problem 15.45P

The disk shown moves in the xy plane. Knowing that (v_A)_y = −7 m/s, (v_B)_x = −7.4 m/s, and (v_C)_x = −1.4 m/s, detemine (a) the velocity of point O, (b) the point of the disk with zero velocity.

Fig. P15.44 and P15.45

To determine

Find the angular velocity of point O.

Answer to Problem 15.45P

The angular velocity of point O is $\underset{\_}{\left(-1.4\text{i}-1\text{j}\right)\text{m/s}}$.

Explanation of Solution

Given information:

Show the velocity ${\text{v}}_A,{\text{v}}_B,{\text{v}}_C$ as follows:

$\begin{array}{c}{\text{v}}_A={\left(v_A\right)}_x\text{i+}{\left(v_A\right)}_y\text{j}\\ \text{=}{\left(v_A\right)}_x\text{i}-7\text{j}\end{array}$

$\begin{array}{c}{\text{v}}_B={\left(v_B\right)}_x\text{i+}{\left(v_B\right)}_y\text{j}\\ \text{=}-7.4\text{i+}{\left(v_B\right)}_y\text{j}\end{array}$

$\begin{array}{c}{\text{v}}_C={\left(v_C\right)}_x\text{i+}{\left(v_C\right)}_y\text{j}\\ \text{=}-1.4\text{i+}{\left(v_C\right)}_y\text{j}\end{array}$

Consider the velocity of point C and B with respect to A is denoted by ${\text{v}}_{C/A}$ and ${\text{v}}_{B/A}$.

Consider the position of point C, B, O with respect to A is denoted by ${\text{r}}_{C/A}$, ${\text{r}}_{B/A}$, ${\text{r}}_{O/A}$.

Calculation:

Show the position ${\text{r}}_{C/A}$, ${\text{r}}_{B/A}$, ${\text{r}}_{O/A}$ as follows:

$\begin{array}{c}{\text{r}}_{C/A}=\left(1200\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)\text{i}\\ \text{=1}\text{.2}\text{i}\end{array}$

$\begin{array}{c}{\text{r}}_{B/A}=\left(600\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)\text{i}+\left(600\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)\text{j}\\ =\text{0}\text{.6}\text{i}+\text{0}\text{.6j}\end{array}$

$\begin{array}{c}{\text{r}}_{O/A}=\left(600\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)\text{i}\\ =\text{0}\text{.6}\text{i}\end{array}$

Calculate the velocity ${\text{v}}_B$ using the relation.

$\begin{array}{c}{\text{v}}_B={\text{v}}_A+{\text{v}}_{B/A}\\ ={\text{v}}_A+\omega \text{k}\times {\text{r}}_{B/A}\end{array}$

Substitute $\left[{\left(v_A\right)}_x\text{i}-7\text{j}\right]$ for ${\text{v}}_A$, $\left[\text{0}\text{.6}\text{i}+\text{0}\text{.6j}\right]$ for ${\text{r}}_{B/A}$, and $\left[-7.4\text{i}+{\left(v_B\right)}_y\text{j}\right]$ for ${\text{v}}_B$.

$\begin{array}{l}\left[-7.4\text{i}+{\left(v_B\right)}_y\text{j}\right]=\left[{\left(v_A\right)}_x\text{i}-7\text{j}\right]+\omega \text{k}\times \left[\text{0}\text{.6}\text{i}+\text{0}\text{.6j}\right]\\ \left[-7.4\text{i}+{\left(v_B\right)}_y\text{j}\right]=\left[{\left(v_A\right)}_x\text{i}-7\text{j}\right]+0.6\omega \text{j}-0.6\omega \text{i}\end{array}$ (1)

Show the components of i on both sides of the Equation (1).

$-7.4={\left(v_A\right)}_x-0.6\omega$ (2)

Show the components of j on both sides of the Equation (1).

${\left(v_B\right)}_y=-7+0.6\omega$ (3)

Calculate the velocity ${\text{v}}_C$ using the relation.

$\begin{array}{c}{\text{v}}_C={\text{v}}_A+{\text{v}}_{C/A}\\ ={\text{v}}_A+\omega \text{k}\times {\text{r}}_{C/A}\end{array}$

Substitute $\left[{\left(v_A\right)}_x\text{i}-7\text{j}\right]$ for ${\text{v}}_A$, $\text{1}\text{.2}\text{i}$ for ${\text{r}}_{C/A}$, and $-1.4\text{i}+{\left(v_C\right)}_y\text{j}$ for ${\text{v}}_C$.

$\begin{array}{l}\left[-1.4\text{i+}{\left(v_C\right)}_y\text{j}\right]=\left[{\left(v_A\right)}_x\text{i}-7\text{j}\right]+\omega \text{k}\times \left[\text{1}\text{.2}\text{i}\right]\\ \left[-1.4\text{i+}{\left(v_C\right)}_y\text{j}\right]=\left[{\left(v_A\right)}_x\text{i}-7\text{j}\right]+1.2\omega \text{j}\end{array}$ (4)

Show the components of the i on both sides of the Equation (4).

$-1.4={\left(v_A\right)}_x$ (5)

Show the components of the j on both sides of the Equation (4).

${\left(v_C\right)}_y=-7+1.2\omega$ (6)

Refer Equation (2).

$\begin{array}{l}-7.4={\left(v_A\right)}_x-0.6\omega \\ \omega =\frac{7.4+{\left(v_A\right)}_x}{0.6}\end{array}$

Substitute $-1.4$ for ${\left(v_A\right)}_x$.

$\begin{array}{c}\omega =\frac{7.4-1.4}{0.6}\\ =10\text{rad/s}\left(\text{Counterclockwise}\right)\end{array}$

Calculate the velocity $\left({\text{v}}_A\right)$ using the relation:

${\text{v}}_A={\left(v_A\right)}_x\text{i}-7\text{j}$

Substitute $-1.4\text{m/s}$ for ${\left(v_A\right)}_x$.

${\text{v}}_A=-1.4\text{i}-7\text{j}$

Consider the velocity at the point O is denoted by $\left({\text{v}}_O\right)$.

Consider the velocity of the point O with respect to A is denoted by $\left({\text{v}}_{O/A}\right)$.

Calculate the velocity of the point O using the relation:

$\begin{array}{c}{\text{v}}_O={\text{v}}_A+{\text{v}}_{O/A}\\ ={\text{v}}_A+\text{ω}\times {\text{r}}_{O/A}\\ ={\text{v}}_A+\omega \text{k}\times {\text{r}}_{O/A}\end{array}$

Substitute $\text{0}\text{.6}\text{i}$ for ${\text{r}}_{O/A}$, $\left[-1.4\text{i}-7\text{j}\right]$ for ${\text{v}}_A$ and $10\text{rad/s}$ for $\omega$.

$\begin{array}{c}{\text{v}}_O=\left[-1.4\text{i}-7\text{j}\right]+10\text{k}\times \text{0}\text{.6}\text{i}\\ =-1.4\text{i}-7\text{j+}6\text{j}\\ \text{=}-1.4\text{i}-1\text{j}\end{array}$

Thus, the velocity of the point O is $\underset{\_}{\left(-1.4\text{i}-1\text{j}\right)\text{m/s}}$.

To determine

Find the point in the disk with zero velocity.

Answer to Problem 15.45P

The point on the disk with zero velocity is $\underset{\_}{x=100\text{mm,}y=-140\text{mm}}$.

Explanation of Solution

Given information:

Calculation:

Consider the point on the disk with zero velocity is $P\left(x,y\right)$.

Show the distance $\left({\text{r}}_{O/P}\right)$ of the point P with respect to point O as follows:

${\text{r}}_{O/P}=x\text{i+yj}$

The velocity $\left({\text{v}}_P\right)$ at point P is zero.

Show the velocity $\left({\text{v}}_{\text{P}}\right)$ of the point P as follows:

${\text{v}}_P={\text{v}}_O+\omega \text{k}\times {\text{r}}_{O/P}$

Substitute 0 for ${\text{v}}_P$, $10\text{rad/s}$ for $\omega$, $\left(-1.4\text{i}-1\text{j}\right)$ for ${\text{v}}_O$, and $x\text{i}+\text{yj}$ for ${\text{r}}_{O/P}$.

$\begin{array}{c}0=\left(-1.4\text{i}-1\text{j}\right)+10\text{k}\times \left(x\text{i+yj}\right)\\ =\left(-1.4\text{i}-1\text{j}\right)+10x\text{j}-10y\text{i}\end{array}$ (7)

Equate i component on both sides of the Equation (7).

$\begin{array}{c}0=-1.4-10y\\ y=-0.14\text{m}\times \frac{1000\text{mm}}{1\text{m}}\\ =-140\text{mm}\end{array}$

Equate j component on both sides of the Equation (7).

$\begin{array}{c}0=-1+10x\\ x=0.1\text{m}\times \frac{1000\text{mm}}{1\text{m}}\\ =100\text{mm}\end{array}$

Thus, the point with zero velocity is $\underset{\_}{x=100\text{mm,}y=-140\text{mm}}$.