Chapter 15.7, Problem 15.245P

(a)

To determine

Find the velocity and acceleration of point E.

Answer to Problem 15.245P

The velocity of point E is $\underset{\_}{{\text{v}}_E=\left(0.610\text{m}/\text{s}\right)\text{k}}$.

The acceleration of point E is $\underset{\_}{{\text{a}}_E=-\left(0.88\text{m}/{\text{s}}^2\right)\text{i}-\left(1.17\text{m}/{\text{s}}^2\right)\text{j}}$.

Explanation of Solution

Given information:

The radius of disk is $r=130\text{mm}$.

The length of rod CD is $l_{CD}=500\text{mm}$.

The constant rate of rotation of the rod-and-disk with respect to arm AB is ${\omega }_2=3\text{rad}/\text{s}$.

The constant rate of rotation of the rod-and-disk is ${\omega }_1=4\text{rad}/\text{s}$.

Calculation:

Calculate the position of E with respect to B $\left(r_{E/B}\right)$ as shown below.

$\begin{array}{c}{r}_{E/B}=\left(\left(250\text{mm}\right)\text{i}+\left(130\text{mm}\right)\text{j}\right)\times \frac{1\text{m}}{1,000\text{mm}}\\ =\left(0.25\text{m}\right)\text{i}+\left(0.13\text{m}\right)\text{j}\end{array}$

Calculate the position of E with respect to D $\left(r_{E/D}\right)$ as shown below.

$\begin{array}{c}{r}_{E/D}=\left(\left(130\text{mm}\right)\text{j}\right)\times \frac{1\text{m}}{1,000\text{mm}}\\ =\left(0.13\text{m}\right)\text{j}\end{array}$

Calculate the rate of rotation of the frame Bxyz $\left(\Omega \right)$ as shown below.

$\Omega ={\omega }_1\text{j}$

Substitute $4\text{rad}/\text{s}$ for ${\omega }_1$.

$\Omega =\left(4\text{rad}/\text{s}\right)\text{j}$

The angular velocity of motion relative to x axis is ${\omega }_2\text{i}=\left(3\text{rad}/\text{s}\right)\text{i}$.

Calculate the velocity at the point $E^{\prime }$ $\left({\text{v}}_{E^{\prime }}\right)$ as shown below.

${\text{v}}_{E^{\prime }}=\Omega \times r_{E/B}$

Substitute $\left(4\text{rad}/\text{s}\right)\text{j}$ for $\Omega$ and $\left(0.25\text{m}\right)\text{i}+\left(0.13\text{m}\right)\text{j}$ for $r_{E/B}$.

$\begin{array}{c}{\text{v}}_{E^{\prime }}=\left(\text{4j}\right)\times \left(\text{0}\text{.25i}+0.13\text{j}\right)\\ =-1\text{k}\\ =-\left(1\text{m}/\text{s}\right)\text{k}\end{array}$

Calculate the velocity at the point E with respect to the frame $\left({\text{v}}_{E/F}\right)$ as shown below.

${\text{v}}_{E/F}={\omega }_2\text{i}\times r_{E/D}$

Substitute $\left(3\text{rad}/\text{s}\right)$ for ${\omega }_2$ and $\left(0.13\text{m}\right)\text{j}$ for $r_{E/D}$.

$\begin{array}{c}{\text{v}}_{E/F}=\left(\text{3i}\right)\times \left(\text{0}\text{.13j}\right)\\ =0.39\text{k}\\ =\left(0.39\text{m}/\text{s}\right)\text{k}\end{array}$

Calculate the velocity at the point E $\left({\text{v}}_E\right)$ as shown below.

${\text{v}}_E={\text{v}}_{E^{\prime }}+{\text{v}}_{E/F}$

Substitute $-\left(1\text{m}/\text{s}\right)\text{k}$ for ${\text{v}}_{E^{\prime }}$ and $\left(0.39\text{m}/\text{s}\right)\text{k}$ for ${\text{v}}_{E/F}$.

$\begin{array}{c}{\text{v}}_E=-1\text{k}+0.39\text{k}\\ =-0.61\text{k}\\ =-\left(0.61\text{m}/\text{s}\right)\text{k}\end{array}$

Hence, the velocity of point E is $\underset{\_}{{\text{v}}_E=-\left(0.61\text{m}/\text{s}\right)\text{k}}$.

Calculate the acceleration at the point $E^{\prime }$ $\left({\text{a}}_{E^{\prime }}\right)$ as shown below.

${\text{a}}_{E^{\prime }}=\Omega \times {\text{v}}_{E^{\prime }}$

Substitute $\left(4\text{rad}/\text{s}\right)\text{j}$ for $\Omega$ and $-\left(1\text{m}/\text{s}\right)\text{k}$ for ${\text{v}}_E$.

$\begin{array}{c}{\text{a}}_{E^{\prime }}=\left(\text{4j}\right)\times \left(-1\text{k}\right)\\ =-4\text{i}\\ =-\left(4\text{m}/{\text{s}}^2\right)\text{i}\end{array}$

Calculate the acceleration at E with respect to the frame $\left({\text{a}}_{E/F}\right)$ as shown below.

${\text{a}}_{E/F}={\omega }_2\text{i}\times {\text{v}}_{E/F}$

Substitute $\left(3\text{rad}/\text{s}\right)$ for ${\omega }_2$ and $\left(0.39\text{m}/\text{s}\right)\text{k}$ for ${\text{v}}_{E/F}$.

$\begin{array}{c}{\text{a}}_{E/F}=3\text{i}\times \text{0}\text{.39k}\\ =-1.17\text{j}\\ =-\left(1.17\text{m}/{\text{s}}^2\right)\text{j}\end{array}$

Calculate the Coriolis acceleration $\left({\text{a}}_{\text{Coriolis}}\right)$ as shown below.

${\text{a}}_{\text{Coriolis}}=2\Omega \times {\text{v}}_{E/F}$

Substitute $\left(4\text{rad}/\text{s}\right)\text{j}$ for $\Omega$ and $\left(0.39\text{m}/\text{s}\right)\text{k}$ for ${\text{v}}_{E/F}$.

$\begin{array}{c}{\text{a}}_{\text{Coriolis}}=2\left(\text{4j}\right)\times \left(\text{0}\text{.39k}\right)\\ =3.12\text{i}\\ =\left(3.12\text{m}/{\text{s}}^2\right)\text{i}\end{array}$

Calculate the acceleration at E $\left({\text{a}}_E\right)$ as shown below.

${\text{a}}_E={\text{a}}_{E^{\prime }}+{\text{a}}_{E/F}+{\text{a}}_{\text{Coriolis}}$

Substitute $-\left(4\text{m}/{\text{s}}^2\right)\text{i}$ for ${\text{a}}_{E^{\prime }}$, $-\left(1.17\text{m}/{\text{s}}^2\right)\text{j}$ for ${\text{a}}_{E/F}$, and $\left(3.12\text{m}/{\text{s}}^2\right)\text{i}$ for ${\text{a}}_{\text{Coriolis}}$.

$\begin{array}{c}{\text{a}}_E=-\left(\text{4i}\right)+\left(-1.17\text{j}\right)+\left(3.12\text{i}\right)\\ =-0.88\text{i}-1.17\text{j}\\ =-\left(0.88\text{m}/{\text{s}}^2\right)\text{i}-\left(1.17\text{m}/{\text{s}}^2\right)\text{j}\end{array}$

Therefore, the acceleration of A is $\underset{\_}{-\left(0.88\text{m}/{\text{s}}^2\right)\text{i}-\left(1.17\text{m}/{\text{s}}^2\right)\text{j}}$.

(b)

To determine

Find the velocity and acceleration of point F.

Answer to Problem 15.245P

The velocity of point F is $\underset{\_}{{\text{v}}_F=\left(0.52\text{m}/\text{s}\right)\text{i}-\left(0.39\text{m}/\text{s}\right)\text{j}-\left(1\text{m}/\text{s}\right)\text{k}}$.

The acceleration of point F is $\underset{\_}{{\text{a}}_F=-\left(4\text{m}/{\text{s}}^2\right)\text{i}-\left(3.25\text{m}/{\text{s}}^2\right)\text{k}}$.

Explanation of Solution

Given information:

The radius of disk is $r=130\text{mm}$.

The length of rod CD is $l_{CD}=500\text{mm}$.

The constant rate of rotation of the rod-and-disk with respect to arm AB is ${\omega }_2=3\text{rad}/\text{s}$.

The constant rate of rotation of the rod-and-disk is ${\omega }_1=4\text{rad}/\text{s}$.

Calculation:

Calculate the position of F with respect to B $\left(r_{F/B}\right)$ as shown below.

$\begin{array}{c}{r}_{F/B}=\left(\left(250\text{mm}\right)\text{i}+\left(130\text{mm}\right)\text{k}\right)\times \frac{1\text{m}}{1,000\text{mm}}\\ =\left(0.25\text{m}\right)\text{i}+\left(0.13\text{m}\right)\text{k}\end{array}$

Calculate the position of F with respect to D $\left(r_{F/D}\right)$ as shown below.

$\begin{array}{c}{r}_{F/D}=\left(\left(130\text{mm}\right)\text{k}\right)\times \frac{1\text{m}}{1,000\text{mm}}\\ =\left(0.13\text{m}\right)\text{k}\end{array}$

Calculate the rate of rotation of the frame Bxyz $\left(\Omega \right)$ as shown below.

$\Omega ={\omega }_1\text{j}$

Substitute $4\text{rad}/\text{s}$ for ${\omega }_1$.

$\Omega =\left(4\text{rad}/\text{s}\right)\text{j}$

The angular velocity of motion relative to x axis is ${\omega }_2\text{i}=\left(3\text{rad}/\text{s}\right)\text{i}$.

Calculate the velocity at the point $F^{\prime }$ $\left({\text{v}}_{F^{\prime }}\right)$ as shown below.

${\text{v}}_{F^{\prime }}=\Omega \times r_{F/B}$

Substitute $\left(4\text{rad}/\text{s}\right)\text{j}$ for $\Omega$ and $\left(0.25\text{m}\right)\text{i}+\left(0.13\text{m}\right)\text{k}$ for $r_{F/B}$.

$\begin{array}{c}{\text{v}}_{F^{\prime }}=\left(\text{4j}\right)\times \left(\text{0}\text{.25i}+0.13\text{k}\right)\\ =-1\text{k}+0.52\text{i}\\ =\left(0.52\text{m}/\text{s}\right)\text{i}-\left(1\text{m}/\text{s}\right)\text{k}\end{array}$

Calculate the velocity at the point F with respect to the frame $\left({\text{v}}_{F/F}\right)$ as shown below.

${\text{v}}_{F/F}={\omega }_2\text{i}\times r_{F/D}$

Substitute $\left(3\text{rad}/\text{s}\right)$ for ${\omega }_2$ and $\left(0.13\text{m}\right)\text{k}$ for $r_{F/D}$.

$\begin{array}{c}{\text{v}}_{F/F}=\left(\text{3i}\right)\times \left(\text{0}\text{.13k}\right)\\ =-0.39\text{j}\\ =-\left(0.39\text{m}/\text{s}\right)\text{j}\end{array}$

Calculate the velocity at the point F $\left({\text{v}}_F\right)$ as shown below.

${\text{v}}_F={\text{v}}_{F^{\prime }}+{\text{v}}_{F/F}$

Substitute $\left(0.52\text{m}/\text{s}\right)\text{i}-\left(1\text{m}/\text{s}\right)\text{k}$ for ${\text{v}}_{F^{\prime }}$ and $-\left(0.39\text{m}/\text{s}\right)\text{j}$ for ${\text{v}}_{F/F}$.

$\begin{array}{c}{\text{v}}_F=\left(0.52\text{i}-1\text{k}\right)+\left(-0.39\text{j}\right)\\ =\left(0.52\text{m}/\text{s}\right)\text{i}-\left(0.39\text{m}/\text{s}\right)\text{j}-\left(1\text{m}/\text{s}\right)\text{k}\end{array}$

Hence, the velocity of point F is $\underset{\_}{{\text{v}}_F=\left(0.52\text{m}/\text{s}\right)\text{i}-\left(0.39\text{m}/\text{s}\right)\text{j}-\left(1\text{m}/\text{s}\right)\text{k}}$.

Calculate the acceleration at the point $F^{\prime }$ $\left({\text{a}}_{F^{\prime }}\right)$ as shown below.

${\text{a}}_{F^{\prime }}=\Omega \times {\text{v}}_{F^{\prime }}$

Substitute $\left(4\text{rad}/\text{s}\right)\text{j}$ for $\Omega$ and $\left(0.52\text{m}/\text{s}\right)\text{i}-\left(1\text{m}/\text{s}\right)\text{k}$ for ${\text{v}}_F$.

$\begin{array}{c}{\text{a}}_{F^{\prime }}=\left(\text{4j}\right)\times \left(0.52\text{i}-1\text{k}\right)\\ =-2.08\text{k}-4\text{i}\\ =-\left(4\text{m}/{\text{s}}^2\right)\text{i}-\left(2.08\text{m}/{\text{s}}^2\right)\text{k}\end{array}$

Calculate the acceleration at F with respect to the frame $\left({\text{a}}_{F/F}\right)$ as shown below.

${\text{a}}_{F/F}={\omega }_2\text{i}\times {\text{v}}_{F/F}$

Substitute $\left(3\text{rad}/\text{s}\right)$ for ${\omega }_2$ and $-\left(0.39\text{m}/\text{s}\right)\text{j}$ for ${\text{v}}_{F/F}$.

$\begin{array}{c}{\text{a}}_{F/F}=3\text{i}\times \left(-\text{0}\text{.39j}\right)\\ =-1.17\text{k}\\ =-\left(1.17\text{m}/{\text{s}}^2\right)\text{k}\end{array}$

Calculate the Coriolis acceleration $\left({\text{a}}_{\text{Coriolis}}\right)$ as shown below.

${\text{a}}_{\text{Coriolis}}=2\Omega \times {\text{v}}_{F/F}$

Substitute $\left(4\text{rad}/\text{s}\right)\text{j}$ for $\Omega$ and $-\left(0.39\text{m}/\text{s}\right)\text{j}$ for ${\text{v}}_{F/F}$.

$\begin{array}{c}{\text{a}}_{\text{Coriolis}}=2\left(\text{4j}\right)\times \left(-\text{0}\text{.39j}\right)\\ =0\end{array}$

Calculate the acceleration at F $\left({\text{a}}_F\right)$ as shown below.

${\text{a}}_F={\text{a}}_{F^{\prime }}+{\text{a}}_{F/F}+{\text{a}}_{\text{Coriolis}}$

Substitute $-\left(4\text{m}/{\text{s}}^2\right)\text{i}-\left(2.08\text{m}/{\text{s}}^2\right)\text{k}$ for ${\text{a}}_{F^{\prime }}$, $-\left(1.17\text{m}/{\text{s}}^2\right)\text{k}$ for ${\text{a}}_{F/F}$, and $0$ for ${\text{a}}_{\text{Coriolis}}$.

$\begin{array}{c}{\text{a}}_F=\left(-4\text{i}-2.08\text{k}\right)+\left(-1.17\text{k}\right)\\ =-4\text{i}-3.25\text{k}\\ =-\left(4\text{m}/{\text{s}}^2\right)\text{i}-\left(3.25\text{m}/{\text{s}}^2\right)\text{k}\end{array}$

Therefore, the acceleration of F is $\underset{\_}{{\text{a}}_F=-\left(4\text{m}/{\text{s}}^2\right)\text{i}-\left(3.25\text{m}/{\text{s}}^2\right)\text{k}}$.