Chapter 15.2, Problem 15.71P

The 80-mm-radius wheel shown rolls to the left with a velocity of 900 mm/s. Knowing that the distance AD is 50 mm, determine the velocity of the collar and the angular velocity of rod AB when (a) β = 0, (b) β = 90°.

Fig. P15.71

To determine

Find the velocity of the collar and the angular velocity of rod AB when $\beta =0$.

Answer to Problem 15.71P

The velocity of the collar and the angular velocity of rod AB when $\beta =0$.

 are $\underset{\_}{v_B\approx \text{338}\text{mm/s}(\leftarrow )}$ and $\underset{\_}{2.37\text{rad/s}\left(\text{Clockwise}\right)}$.

Explanation of Solution

Given information:

The radius of the wheel is $r=80\text{mm}$.

The velocity of the wheel is $v_D=900\text{mm/s}$.

Calculation:

Calculate the angular velocity $\left({\text{ω}}_{AD}\right)$ of AD using the relation:

${\text{ω}}_{AD}=-\frac{v_D}{r}$

Substitute $900\text{mm/s}$ for $v_D$ and $80\text{mm}$ for $r$.

$\begin{array}{c}{\text{ω}}_{AD}=-\frac{900\text{mm/s}}{80\text{mm}}\\ =-11.25\text{rad/s}\\ \text{=}11.25\text{k}\text{rad/s}\end{array}$

Consider the value of the angle $\beta =0$.

Consider the position of point A with respect to point D is denoted by ${\text{r}}_{A/D}=-50\text{j}\text{mm}$.

Show the relation between the velocity at A and D as follows:

${\text{v}}_A={\text{v}}_D+{\omega }_{AD}\times {\text{r}}_{A/D}$

Substitute $-900\text{i}\text{mm/s}$ for ${\text{v}}_D$, $11.25\text{k}\text{rad/s}$ for ${\omega }_{AD}$, and $-50\text{j}\text{mm}$ for ${\text{r}}_{A/D}$.

$\begin{array}{l}{\text{v}}_A=-900\text{i}\text{mm/s}+\left(11.25\text{k}\text{rad/s}\right)\times \left(-50\text{j}\text{mm}\right)\\ {\text{v}}_A=-900\text{i}\text{mm/s}+\text{562}\text{.5i}\text{mm/s}\\ {\text{v}}_A=-900\text{i}\text{mm/s}+\text{562}\text{.5i}\text{mm/s}\\ {\text{v}}_A=-337.5\text{i}\text{mm/s}\end{array}$

Consider the position of the point B with respect to A is denoted by ${\text{r}}_{B/A}=\left(214\text{i}+1300\text{j}\right)\text{mm}$.

Show the relation between the velocity at A and B using the relation:

$\begin{array}{l}{\text{v}}_B={\text{v}}_A+{\text{ω}}_{AB}\times {\text{r}}_{B/A}\\ {\text{v}}_B={\text{v}}_A+\left({\omega }_{AB}\text{k}\right)\times {\text{r}}_{B/A}\end{array}$

Substitute $\left(214\text{i}+130\text{j}\right)\text{mm}$ for ${\text{r}}_{B/A}$ and $-337.5\text{i}\text{mm/s}$ for ${\text{v}}_A$.

$\begin{array}{l}{\text{v}}_B=-337.5\text{i}+{\omega }_{AB}\text{k}\times \left(214\text{i}+130\text{j}\right)\\ v_B\text{i}=-337.5\text{i}+\left(214{\omega }_{AB}\text{j}-130{\omega }_{AB}\text{i}\right)\\ v_B\text{i}=\left(-337.5+130{\omega }_{AB}\right)\text{i}+214{\omega }_{AB}\text{j}\end{array}$ (1)

Equate j component of Equation (1) as follows:

$\begin{array}{l}0=214{\omega }_{AB}\\ {\omega }_{AB}=0\end{array}$

Thus, the angular velocity of the rod AB is $\underset{\_}{0}$.

Equate i component of Equation (1) as follows:

$`v_B=\left(-337.5+130{\omega }_{AB}\right)$

Substitute 0 for ${\omega }_{AB}$.

$\begin{array}{l}{v}_B=\left(-337.5+130\times 0\right)\\ v_B=-337.5\text{mm/s}\\ v_B\approx -\text{338}\text{mm/s}\\ v_B\approx \text{338}\text{mm/s}(\leftarrow )\end{array}$

Thus, the velocity of the collar B is $\underset{\_}{v_B\approx \text{338}\text{mm/s}(\leftarrow )}$.

To determine

 Find the velocity of the collar and the angular velocity of rod AB when $\beta =90°$.

Answer to Problem 15.71P

The velocity of the collar and the angular velocity of rod AB when $\beta =90°$.

 are $\underset{\_}{`v_B\approx 710\text{mm/s}(\leftarrow )}$ and $\underset{\_}{0}$.

Explanation of Solution

Given information:

The radius of the wheel is $r=80\text{mm}$.

The velocity of the wheel is $v_D=900\text{mm/s}$.

Calculation:

Calculate the angular velocity $\left({\text{ω}}_{AD}\right)$ of AD using the relation:

${\text{ω}}_{AD}=-\frac{v_D}{r}$

Substitute $900\text{mm/s}$ for $v_D$ and $80\text{mm}$ for $r$.

$\begin{array}{c}{\text{ω}}_{AD}=-\frac{900\text{mm/s}}{80\text{mm}}\\ =-11.25\text{rad/s}\\ \text{=}11.25\text{k}\text{rad/s}\end{array}$

Consider the value of the angle $\beta =90°$.

Consider the position of point A with respect to point D is denoted by ${\text{r}}_{A/D}=50\text{i}\text{mm}$.

Show the relation between the velocity at A and D as follows:

${\text{v}}_A={\text{v}}_D+{\omega }_{AD}\times {\text{r}}_{A/D}$

Substitute $-900\text{i}\text{mm/s}$ for ${\text{v}}_D$, $11.25\text{k}\text{rad/s}$ for ${\omega }_{AD}$, and $50\text{i}\text{mm}$ for ${\text{r}}_{A/D}$.

$\begin{array}{l}{\text{v}}_A=-900\text{i}\text{mm/s}+\left(11.25\text{k}\text{rad/s}\right)\times \left(50\text{i}\text{mm}\right)\\ {\text{v}}_A=-900\text{i}\text{mm/s}+\text{562}\text{.5j}\text{mm/s}\end{array}$

Consider the position of the point B with respect to A is denoted by ${\text{r}}_{B/A}=\left(237\text{i}+80\text{j}\right)\text{mm}$.

Show the relation between the velocity at A and B using the relation:

$\begin{array}{l}{\text{v}}_B={\text{v}}_A+{\text{ω}}_{AB}\times {\text{r}}_{B/A}\\ {\text{v}}_B={\text{v}}_A+\left({\omega }_{AB}\text{k}\right)\times {\text{r}}_{B/A}\end{array}$

Substitute $\left(237\text{i}+80\text{j}\right)\text{mm}$ for ${\text{r}}_{B/A}$ and $\left(-900\text{i}\text{mm/s}+\text{562}\text{.5j}\text{mm/s}\right)$ for ${\text{v}}_A$.

$\begin{array}{l}{\text{v}}_B=\left(-900\text{i}+\text{562}\text{.5j}\right)+{\omega }_{AB}\text{k}\times \left(237\text{i}+80\text{j}\right)\\ v_B\text{i}=-900\text{i}+\text{562}\text{.5j}+\left(237{\omega }_{AB}\text{j}-80{\omega }_{AB}\text{i}\right)\\ v_B\text{i}=\left(-900-80{\omega }_{AB}\right)\text{i}+\left(\text{562}\text{.5}+237{\omega }_{AB}\right)\text{j}\end{array}$ (1)

Equate j component of Equation (1) as follows:

$\begin{array}{l}0=\left(\text{562}\text{.5}+237{\omega }_{AB}\right)\\ {\omega }_{AB}=-\frac{\text{562}\text{.5}}{237}\\ {\omega }_{AB}=-2.37\text{rad/s}\\ {\omega }_{AB}=2.37\text{rad/s}\left(\text{Clockwise}\right)\end{array}$

Thus, the angular velocity of the rod AB is $\underset{\_}{2.37\text{rad/s}\left(\text{Clockwise}\right)}$.

Equate i component of Equation (1) as follows:

$v_B=\left(-900-80{\omega }_{AB}\right)$

Substitute $-2.37\text{rad/s}$ for ${\omega }_{AB}$.

$\begin{array}{l}`v_B=\left(-900-80\times \left(-2.37\text{rad/s}\right)\right)\\ `v_B=-710\text{mm/s}\\ `v_B\approx 710\text{mm/s}(\leftarrow )\end{array}$

Thus, the velocity of the collar B is $\underset{\_}{`v_B\approx 710\text{mm/s}(\leftarrow )}$.