Chapter 15.7, Problem 15.244P

(a)

To determine

The acceleration of end B of the rod if $\theta =0°$.

Answer to Problem 15.244P

The acceleration of end B is $\underset{\_}{{\text{a}}_B=r{\omega }_2^2\sin 30°\text{j}-\left(r{\omega }_2^2\cos 30°+2r{\omega }_1{\omega }_2\right)\text{k}}$.

Explanation of Solution

Given information:

The side length of the square plate is $2r$.

The shaft rotates with a constant angular velocity as ${\omega }_1$.

The constant angular velocity with respect to the plate is ${\omega }_2$.

Calculation:

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

$r_{A/O}=r\left(\text{sin30}°\text{j}-\cos 30°\text{k}\right)$

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

$r_{B/O}=r_{A/O}+r_{B/A}$ (1)

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

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

Calculate the motion relative to the frame $\left({\omega }_2\right)$ as shown below.

${\text{ω}}_2={\omega }_2\left(\cos 30°\text{j}+\sin 30°\text{k}\right)$

For $\theta =0$.

Sketch the Free Body Diagram of the plate, the rod AB is located at an angle $\theta =0$ as shown in Figure 1.

Refer to Figure 1.

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

$r_{B/A}=r\left(-\sin 30°\text{j}+\cos 30°\text{k}\right)$

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

Substitute $r\left(\sin 30°\text{j}-\cos 30°\text{k}\right)$ for $r_{A/O}$ and $r\left(-\sin 30°\text{j}+\cos 30°\text{k}\right)$ for $r_{B/A}$ in Equation (1).

$\begin{array}{c}{r}_{B/O}=r\left(\sin 30°\text{j}-\cos 30°\text{k}\right)+r\left(-\sin 30°\text{j}+\cos 30°\text{k}\right)\\ =0\end{array}$

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

${\text{v}}_{B^{\prime }}=\Omega \times r_{B/O}$ (2)

Substitute ${\omega }_1\text{j}$ for $\Omega$ and $0$ for $r_{B/O}$ in Equation (2).

$\begin{array}{c}{\text{v}}_{B^{\prime }}=\Omega \times 0\\ =0\end{array}$

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

${\text{v}}_{B/F}={\text{ω}}_2\times r_{B/A}$ (3)

Substitute ${\omega }_2\left(\cos 30°\text{j}+\sin 30°\text{k}\right)$ for ${\text{ω}}_2$ and $r\left(-\sin 30°\text{j}+\cos 30°\text{k}\right)$ for $r_{B/A}$ in Equation (3).

$\begin{array}{c}{\text{v}}_{B/F}={\omega }_2\left(\cos 30°\text{j}+\sin 30°\text{k}\right)\times r\left(-\sin 30°\text{j}+\cos 30°\text{k}\right)\\ =r{\omega }_2\left({\cos }^{2}30°\text{i}+{\sin }^{2}30°\text{i}\right)\\ =r{\omega }_2\text{i}\end{array}$

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

${\text{a}}_{B^{\prime }}=\Omega \times {\text{v}}_{B^{\prime }}$ (4)

Substitute ${\omega }_1\text{j}$ for $\Omega$ and $0$ for ${\text{v}}_{B^{\prime }}$ in Equation (4).

$\begin{array}{c}{\text{a}}_{B^{\prime }}={\omega }_1\text{j}\times \left(0\right)\\ =0\end{array}$

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

${\text{a}}_{B/F}={\omega }_2\times {\text{v}}_{B/F}$ (5)

Substitute ${\omega }_2\left(\cos 30°\text{j}+\sin 30°\text{k}\right)$ for ${\text{ω}}_2$ and $r{\omega }_2\text{i}$ for ${\text{v}}_{B/F}$ in Equation (5).

$\begin{array}{c}{\text{a}}_{B/F}={\omega }_2\left(\cos 30°\text{j}+\sin 30°\text{k}\right)\times \left(r{\omega }_2\text{i}\right)\\ =r{\omega }_2^2\left(-\cos 30°\text{k}+\sin 30°\text{j}\right)\\ =r{\omega }_2^2\left(\sin 30°\text{j}-\cos 30°\text{k}\right)\end{array}$

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

${\text{a}}_{\text{Coriolis}}=2\Omega \times {\text{v}}_{B/F}$ (6)

Substitute ${\omega }_1\text{j}$ for $\Omega$ and $r{\omega }_2\text{i}$ for ${\text{v}}_{B/F}$ in Equation (6).

$\begin{array}{c}{\text{a}}_{\text{Coriolis}}=2\left({\omega }_1\text{j}\right)\times \left(\text{r}{\omega }_2\text{i}\right)\\ =-2r{\omega }_1{\omega }_2\text{k}\end{array}$

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

${\text{a}}_B={\text{a}}_{B^{\prime }}+{\text{a}}_{B/F}+{\text{a}}_{\text{Coriolis}}$ (7)

Substitute $0$ for ${\text{a}}_{B^{\prime }}$, $r{\omega }_2^2\left(\sin 30°\text{j}-\cos 30°\text{k}\right)$ for ${\text{a}}_{B/F}$, and $-2r{\omega }_1{\omega }_2\text{k}$ for ${\text{a}}_{\text{Coriolis}}$ in Equation (7).

$\begin{array}{c}{\text{a}}_B=\text{0}+\left(r{\omega }_2^2\left(\sin 30°\text{j}-\cos 30°\text{k}\right)\right)+\left(-2r{\omega }_1{\omega }_2\text{k}\right)\\ =r{\omega }_2^2\sin 30°\text{j}-\left(r{\omega }_2^2\cos 30°+2r{\omega }_1{\omega }_2\right)\text{k}\end{array}$

Hence, the acceleration of B is $\underset{\_}{{\text{a}}_B=r{\omega }_2^2\sin 30°\text{j}-\left(r{\omega }_2^2\cos 30°+2r{\omega }_1{\omega }_2\right)\text{k}}$.

(b)

To determine

The acceleration of end B of the rod if $\theta =90°$.

Answer to Problem 15.244P

The acceleration of end B is $\underset{\_}{{\text{a}}_B=-r\left({\omega }_1^2+{\omega }_2^2+2{\omega }_1{\omega }_2\cos 30°\right)\text{i}+r{\omega }_1^2\text{cos30}°\text{k}}$.

Explanation of Solution

Given information:

The side length of the square plate is $2r$.

The shaft rotates with a constant angular velocity as ${\omega }_1$.

The constant angular velocity with respect to the plate is ${\omega }_2$.

Calculation:

For $\theta =90$.

Sketch the Free Body Diagram of the plate, the rod AB is located at an angle $\theta =90°$ as shown in Figure 2.

Refer to Figure 2.

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

$r_{B/A}=r\text{i}$

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

Substitute $r\left(\sin 30°\text{j}-\cos 30°\text{k}\right)$ for $r_{A/O}$ and $r\text{i}$ for $r_{B/A}$ in Equation (1).

$\begin{array}{c}{r}_{B/O}=r\left(\sin 30°\text{j}-\cos 30°\text{k}\right)+r\text{i}\\ =r\text{i}+r\sin 30°\text{j}-r\cos 30°\text{k}\end{array}$

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

Substitute ${\omega }_1\text{j}$ for $\Omega$ and $r\text{i}+r\sin 30°\text{j}-r\cos 30°\text{k}$ for $r_{B/O}$ in Equation (2).

$\begin{array}{c}{\text{v}}_{B^{\prime }}={\omega }_1\text{j}\times \left(r\text{i}+r\sin 30°\text{j}-r\cos 30°\text{k}\right)\\ =-r{\omega }_1\text{k}-r{\omega }_1\cos 30°\text{i}\\ =-r{\omega }_1\cos 30°\text{i}-r{\omega }_1\text{k}\end{array}$

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

Substitute ${\omega }_2\left(\cos 30°\text{j}+\sin 30°\text{k}\right)$ for ${\text{ω}}_2$ and $r\text{i}$ for $r_{B/A}$ in Equation (3).

$\begin{array}{c}{\text{v}}_{B/F}={\omega }_2\left(\cos 30°\text{j}+\sin 30°\text{k}\right)\times r\text{i}\\ =r{\omega }_2\left(-\cos 30°\text{k}+\sin 30°\text{j}\right)\\ =r{\omega }_2\sin 30°\text{j}-r{\omega }_2\cos 30°\text{k}\end{array}$

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

Substitute ${\omega }_1\text{j}$ for $\Omega$ and $-r{\omega }_1\cos 30°\text{i}-r{\omega }_1\text{k}$ for ${\text{v}}_{B^{\prime }}$ in Equation (4).

$\begin{array}{c}{\text{a}}_{B^{\prime }}={\omega }_1\text{j}\times \left(-r{\omega }_1\cos 30°\text{i}-r{\omega }_1\text{k}\right)\\ =r{\omega }_1^2\cos 30°\text{k}-r{\omega }_1^2\text{i}\\ =-r{\omega }_1^2\text{i}+r{\omega }_1^2\cos 30°\text{k}\end{array}$

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

Substitute ${\omega }_2\left(\cos 30°\text{j}+\sin 30°\text{k}\right)$ for ${\text{ω}}_2$ and $r{\omega }_2\sin 30°\text{j}-r{\omega }_2\cos 30°\text{k}$ for ${\text{v}}_{B/F}$ in Equation (5).

$\begin{array}{c}{\text{a}}_{B/F}={\omega }_2\left(\cos 30°\text{j}+\sin 30°\text{k}\right)\times \left(r{\omega }_2\sin 30°\text{j}-r{\omega }_2\cos 30°\text{k}\right)\\ =r{\omega }_2^2\left(-{\cos }^{2}30°\text{i}-{\sin }^{2}30°\text{i}\right)\\ =-r{\omega }_2^2\left({\cos }^{2}30°+{\sin }^{2}30°\right)\text{i}\\ =-r{\omega }_2^2\text{i}\end{array}$

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

Substitute ${\omega }_1\text{j}$ for $\Omega$ and $r{\omega }_2\sin 30°\text{j}-r{\omega }_2\cos 30°\text{k}$ for ${\text{v}}_{B/F}$ in Equation (6).

$\begin{array}{c}{\text{a}}_{\text{Coriolis}}=2\left({\omega }_1\text{j}\right)\times \left(r{\omega }_2\sin 30°\text{j}-r{\omega }_2\cos 30°\text{k}\right)\\ =-2r{\omega }_1{\omega }_2\cos 30°\text{i}\end{array}$

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

Substitute $-r{\omega }_1^2\text{i}+r{\omega }_1^2\cos 30°\text{k}$ for ${\text{a}}_{B^{\prime }}$, $-r{\omega }_2^2\text{i}$ for ${\text{a}}_{B/F}$, and $-2r{\omega }_1{\omega }_2\cos 30°\text{i}$ for ${\text{a}}_{\text{Coriolis}}$ in Equation (7).

$\begin{array}{c}{\text{a}}_B=\left(-r{\omega }_1^2\text{i}+r{\omega }_1^2\cos 30°\text{k}\right)+\left(-r{\omega }_2^2\text{i}\right)+\left(-2r{\omega }_1{\omega }_2\cos 30°\text{i}\right)\\ =-\left(r{\omega }_1^2+r{\omega }_2^2+2r{\omega }_1{\omega }_2\cos 30°\right)\text{i}+r{\omega }_1^2\cos 30°\text{k}\end{array}$

Hence, the acceleration of B is $\underset{\_}{{\text{a}}_B=-\left(r{\omega }_1^2+r{\omega }_2^2+2r{\omega }_1{\omega }_2\cos 30°\right)\text{i}+r{\omega }_1^2\cos 30°\text{k}}$.

(c)

To determine

The acceleration of end B of the rod if $\theta =180°$.

Answer to Problem 15.244P

The acceleration of end B is $\underset{\_}{{\text{a}}_B=-r{\omega }_2^2\sin 30°\text{j}+r\left(2{\omega }_1^2\cos 30°+{\omega }_2^2\cos 30°+2{\omega }_1{\omega }_2\right)\text{k}}$.

Explanation of Solution

Given information:

The side of the square plate of side $2r$.

The shaft rotates with a constant angular velocity ${\omega }_1$.

The constant angular velocity with respect to the plate is ${\omega }_2$.

Calculation:

For $\theta =180$.

Sketch the Free Body Diagram of the plate, the rod AB is located at an angle $\theta =180°$ as shown in Figure 3.

Refer to Figure 3.

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

$r_{B/A}=r\left(\sin 30°\text{j}-\cos 30°\text{k}\right)$

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

Substitute $r\left(\sin 30°\text{j}-\cos 30°\text{k}\right)$ for $r_{A/O}$ and $r\left(\sin 30°\text{j}-\cos 30°\text{k}\right)$ for $r_{B/A}$ in Equation (1).

$\begin{array}{c}{r}_{B/O}=r\left(\sin 30°\text{j}-\cos 30°\text{k}\right)+r\left(\sin 30°\text{j}-\cos 30°\text{k}\right)\\ =2r\left(\sin 30°\text{j}-\cos 30°\text{k}\right)\end{array}$

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

Substitute ${\omega }_1\text{j}$ for $\Omega$ and $2r\left(\sin 30°\text{j}-\cos 30°\text{k}\right)$ for $r_{B/O}$ in Equation (2).

$\begin{array}{c}{\text{v}}_{B^{\prime }}={\omega }_1\text{j}\times \left(2r\left(\sin 30°\text{j}-\cos 30°\text{k}\right)\right)\\ =-2r{\omega }_1\cos 30°\text{i}\end{array}$

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

Substitute ${\omega }_2\left(\cos 30°\text{j}+\sin 30°\text{k}\right)$ for ${\text{ω}}_2$ and $r\left(\sin 30°\text{j}-\cos 30°\text{k}\right)$ for $r_{B/A}$ in Equation (3).

$\begin{array}{c}{\text{v}}_{B/F}={\omega }_2\left(\cos 30°\text{j}+\sin 30°\text{k}\right)\times r\left(\sin 30°\text{j}-\cos 30°\text{k}\right)\\ =r{\omega }_2\left(-{\cos }^{2}30°\text{i}-{\sin }^{2}30°\text{i}\right)\\ =-r{\omega }_2\left({\cos }^{2}30°+{\sin }^{2}30°\right)\text{i}\\ =-r{\omega }_2\text{i}\end{array}$

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

Substitute ${\omega }_1\text{j}$ for $\Omega$ and $-2r{\omega }_1\cos 30°\text{i}$ for ${\text{v}}_{B^{\prime }}$ in Equation (4).

$\begin{array}{c}{\text{a}}_{B^{\prime }}={\omega }_1\text{j}\times \left(-2r{\omega }_1\cos 30°\text{i}\right)\\ =2r{\omega }_1^2\cos 30°\text{k}\end{array}$

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

Substitute ${\omega }_2\left(\cos 30°\text{j}+\sin 30°\text{k}\right)$ for ${\text{ω}}_2$ and $-r{\omega }_2\text{i}$ for ${\text{v}}_{B/F}$ in Equation (5).

$\begin{array}{c}{\text{a}}_{B/F}={\omega }_2\left(\cos 30°\text{j}+\sin 30°\text{k}\right)\times \left(-r{\omega }_2\text{i}\right)\\ =r{\omega }_2^2\left(\cos 30°\text{k}-\sin 30°\text{j}\right)\\ =r{\omega }_2^2\left(-\sin 30°\text{j}+\cos 30°\text{k}\right)\end{array}$

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

Substitute ${\omega }_1\text{j}$ for $\Omega$ and $-r{\omega }_2\text{i}$ for ${\text{v}}_{B/F}$ in Equation (6).

$\begin{array}{c}{\text{a}}_{\text{Coriolis}}=2\left({\omega }_1\text{j}\right)\times \left(-r{\omega }_2\text{i}\right)\\ =2r{\omega }_1{\omega }_2\text{k}\end{array}$

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

Substitute $2r{\omega }_1^2\cos 30°\text{k}$ for ${\text{a}}_{B^{\prime }}$, $r{\omega }_2^2\left(-\sin 30°\text{j}+\cos 30°\text{k}\right)$ for ${\text{a}}_{B/F}$, and $2r{\omega }_1{\omega }_2\text{k}$ for ${\text{a}}_{\text{Coriolis}}$ in Equation (7).

$\begin{array}{c}{\text{a}}_B=\left(2r{\omega }_1^2\cos 30°\text{k}\right)+\left(r{\omega }_2^2\left(-\sin 30°\text{j}+\cos 30°\text{k}\right)\right)+\left(2r{\omega }_1{\omega }_2\text{k}\right)\\ =-r{\omega }_2^2\sin 30°\text{j}+r\left(2{\omega }_1^2\cos 30°+{\omega }_2^2\cos 30°+2{\omega }_1{\omega }_2\right)\text{k}\end{array}$

Hence, the acceleration of B is $\underset{\_}{{\text{a}}_B=-r{\omega }_2^2\sin 30°\text{j}+r\left(2{\omega }_1^2\cos 30°+{\omega }_2^2\cos 30°+2{\omega }_1{\omega }_2\right)\text{k}}$.