Chapter 15.5, Problem 15.154P

Pin P is attached to the wheel shown and slides in a slot cut in bar BD. The wheel rolls to the right without slipping with a constant angular velocity of 20 rad/s. Knowing that x = 480 mm when θ = 0, determine the angular velocity of the bar and the relative velocity of pin P with respect to the rod when (a) θ = 0, (b) θ = 90°.

Fig. P15.154

To determine

Find the angular velocity of the bar when $\theta =0°$.

Find the relative velocity of the pin P with respect to the rod when $\theta =0°$.

Answer to Problem 15.154P

The angular velocity of the bar when $\theta =0°$ is $\underset{\_}{{\omega }_{BD}=3.81\text{rad/s}\left(\text{Clockwise}\right)}$.

The relative velocity of the pin P with respect to the rod when $\theta =0°$ is $\underset{\_}{6.53\text{m/s}\left(\measuredangle \text{16}\text{.26}°\right)}$.

Explanation of Solution

Given information:

The constant angular velocity of the wheel is ${\omega }_{AC}=20\text{rad/s}$.

The distance BA is ${\left(x_A\right)}_0=480\text{mm}$.

The distance AP is $e=140\text{mm}$.

The radius of the wheel is $r=200\text{mm}$.

Consider the relative velocity of the pin P with respect to the rod is denoted by u.

Calculation:

Show the wheel and the rod arrangement as shown in Figure 1.

Refer Figure 1.

Consider the coordinates of the point A, B, C, and P are $\left(x_A,y_A\right),\left(x_B,y_B\right),\left(x_C,y_C\right)$ and $\left(x_P,y_P\right)$ respectively.

Calculate the value of the distance $x_A,y_A,x_B,y_B,x_C,y_C,x_P,y_P$ as follows:

$\begin{array}{l}{x}_A={\left(x_A\right)}_0+r\theta \\ y_A=r\end{array}$

$\begin{array}{l}{x}_B=0\\ y_B=r\end{array}$

$\begin{array}{l}{x}_C=x_A={\left(x_A\right)}_0+r\theta \\ y_C=0\end{array}$

$\begin{array}{l}{x}_P=x_A+e\sin \theta \\ y_P=r+e\cos \theta \end{array}$

Consider the angular velocity of the wheel AC is ${\text{ω}}_{AC}={\omega }_{AC}\left(\text{Clockwise}\right)$.

Consider the angular velocity of the rod BD is ${\text{ω}}_{BD}={\omega }_{BD}\left(\text{Clockwise}\right)$.

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

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

$\begin{array}{c}{\text{v}}_P=\left({\text{v}}_A\right)+\left[{\text{v}}_{P/A}\right]\\ =\left[r{\omega }_{AC}(\to )\right]+\left[e{\omega }_{AC}\left(\begin{array}{l}\text{Acting at }\theta \text{clockwise }\\ \text{below the horizontal}\end{array}\right)\right]\\ =r{\omega }_{AC}(\to )+e{\omega }_{AC}\left(\begin{array}{l}\text{Acting at }\theta \text{clockwise }\\ \text{below the horizontal}\end{array}\right)\end{array}$ (1)

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

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

$\begin{array}{c}{\text{v}}_P=\left({\text{v}}_{P^{\prime }}\right)+\left[{\text{v}}_{P/F}\right]\\ =\left[x_p{\omega }_{BD}(\downarrow )+e\cos \theta {\omega }_{BD}(\to )\right]+\left[u\cos \beta (\to )+u\sin \beta (\uparrow )\right]\\ =x_p{\omega }_{BD}(\downarrow )+e\cos \theta {\omega }_{BD}(\to )+u\cos \beta (\to )+u\sin \beta (\uparrow )\end{array}$ (2)

Equate Equation (1) and (2).

$r{\omega }_{AC}(\to )+e{\omega }_{AC}\left(\begin{array}{l}\text{Acting at }\theta \text{clockwise }\\ \text{below the horizontal}\end{array}\right)=x_p{\omega }_{BD}(\downarrow )+u\cos \beta (\to )+u\sin \beta (\uparrow )$ (3)

Equate the horizontal component of the Equation (3) as follows:

Take the direction towards right as positive.

$\begin{array}{l}r{\omega }_{AC}+e{\omega }_{AC}\cos \theta =e\cos \theta {\omega }_{BD}+u\cos \beta \\ \left(r+e\cos \theta \right){\omega }_{AC}=e\cos \theta {\omega }_{BD}+u\cos \beta \end{array}$ (4)

Equate the vertical component of the Equation (3) as follows:

Take the direction towards downward as positive.

$\begin{array}{l}e{\omega }_{AC}\sin \theta =x_p{\omega }_{BD}-u\sin \beta \\ e{\omega }_{AC}\sin \theta =\left\{\left[{\left(x_A\right)}_0+r\theta \right]+e\sin \theta \right\}{\omega }_{BD}-u\sin \beta \end{array}$ (5)

Calculate the value of the angle $\theta$ using the relation:

$\begin{array}{c}\tan \beta =\frac{e\cos \theta }{x_p}\\ \tan \beta =\frac{e\cos \theta }{x_p}\\ =\frac{e\cos \theta }{x_A+e\sin \theta }\\ =\frac{e\cos \theta }{\left[{\left(x_A\right)}_0+r\theta \right]+e\sin \theta }\end{array}$

Substitute $0°$ for $\theta$, $480\text{mm}$ for ${\left(x_A\right)}_0$, $140\text{mm}$ for $e$, and $200\text{mm}$ for $r$.

$\begin{array}{c}\tan \beta =\frac{140\cos \left(0°\right)}{480+200\times 0+140\sin \left(0°\right)}\\ \tan \beta =\frac{140}{480}\\ \beta ={\tan }^{-1}\left(\frac{140}{480}\right)\\ \beta =16.26°\end{array}$

Modify Equation (4) as follows:

Substitute $0°$ for $\theta$, $16.26°$ for $\beta$, $480\text{mm}$ for ${\left(x_A\right)}_0$, $140\text{mm}$ for $e$, and $200\text{mm}$ for $r$, and $20\text{rad/s}$ for ${\omega }_{AC}$ in Equation (4).

$\begin{array}{l}\left[200+140\cos \left(0°\right)\right]\times 20=140\cos \left(0°\right){\omega }_{BD}+u\cos \left(16.26°\right)\\ 6800=140{\omega }_{BD}+0.96u\\ u=\frac{6800-140{\omega }_{BD}}{0.96}\end{array}$ (6)

Modify Equation (5) as follows:

Substitute $0°$ for $\theta$, $16.26°$ for $\beta$, $140\text{mm}$ for $e$, and $200\text{mm}$ for $r$, and $20\text{rad/s}$ for ${\omega }_{AC}$ in Equation (5).

$\begin{array}{l}140\times 20\sin \left(0°\right)=\left[480+200\times 0+140\sin \left(0°\right)\right]{\omega }_{BD}-u\sin \left(16.26°\right)\\ 0=480{\omega }_{BD}+0.28u\end{array}$

Substitute $\frac{6800-140{\omega }_{BD}}{0.96}$ for u in above Equation.

$\begin{array}{l}0=480{\omega }_{BD}-0.28\left(\frac{6800-140{\omega }_{BD}}{0.96}\right)\\ 0=480{\omega }_{BD}-1983.33+40.83{\omega }_{BD}\\ 520.83{\omega }_{BD}=1983.33\\ {\omega }_{BD}=\frac{1983.33}{520.83}\end{array}$

${\omega }_{BD}=3.81\text{rad/s}\left(\text{Clockwise}\right)$

Thus, the angular velocity of the bar when $\theta =0°$ is $\underset{\_}{{\omega }_{BD}=3.81\text{rad/s}\left(\text{Clockwise}\right)}$.

Calculate the relative velocity of the pin P with respect to the rod when $\theta =0°$ as follows:

Substitute $3.81\text{rad/s}$ for ${\omega }_{BD}$ in Equation (5).

$\begin{array}{c}u=\frac{6800-140\times 3.81}{0.96}\\ =6530\text{mm/s}\times \left(\frac{1\text{m}}{1000\text{mm}}\right)\\ =6.53\text{m/s}\left(\measuredangle \text{16}\text{.26}°\right)\end{array}$

Thus, the relative velocity of the pin P with respect to the rod when $\theta =0°$ is $\underset{\_}{6.53\text{m/s}\left(\measuredangle \text{16}\text{.26}°\right)}$.

To determine

Find the angular velocity of the bar when $\theta =90°$.

Find the relative velocity of the pin P with respect to the rod when $\theta =90°$.

Answer to Problem 15.154P

The angular velocity of the bar when $\theta =90°$ is $\underset{\_}{3\text{rad/s}\left(\text{Clockwise}\right)}$.

The relative velocity of the pin P with respect to the rod when $\theta =90°$ is $\underset{\_}{4\text{m/s}(\to )}$.

Explanation of Solution

Given information:

Calculation:

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

Refer to Part (a).

Calculate the value of the angle $\theta$ using the relation:

$\begin{array}{c}\tan \beta =\frac{e\cos \theta }{x_p}\\ =\frac{e\cos \theta }{x_A+e\sin \theta }\\ =\frac{e\cos \theta }{\left[{\left(x_A\right)}_0+r\theta \right]+e\sin \theta }\end{array}$

Substitute $90°$ for $\theta$, $480\text{mm}$ for ${\left(x_A\right)}_0$, $140\text{mm}$ for $e$, and $200\text{mm}$ for $r$.

$\begin{array}{c}\tan \beta =\frac{140\cos \left(90°\right)}{480+200\times \left(90°\times \frac{\pi }{180°}\right)+140\sin \left(90°\right)}\\ \tan \beta =0\\ \beta ={\tan }^{-1}\left(0\right)\\ \beta =0°\end{array}$

Modify Equation (4) as follows:

Substitute $90°$ for $\theta$, $0°$ for $\beta$, $480\text{mm}$ for ${\left(x_A\right)}_0$, $140\text{mm}$ for $e$, and $200\text{mm}$ for $r$, and $20\text{rad/s}$ for ${\omega }_{AC}$ in Equation (4).

$\begin{array}{l}\left[200+140\cos \left(90°\right)\right]\times 20=140\cos \left(90°\right){\omega }_{BD}+u\cos \left(0°\right)\\ 4000=u\\ u=4000\text{mm/s}\times \left(\frac{1\text{m}}{1000\text{mm}}\right)\\ u=4\text{m/s}(\to )\end{array}$

Thus, the relative velocity of the pin P with respect to the rod when $\theta =90°$ is $\underset{\_}{4\text{m/s}(\to )}$.

Modify Equation (5) as follows:

Substitute $90°$ for $\theta$, $0°$ for $\beta$, $140\text{mm}$ for $e$, and $200\text{mm}$ for $r$, and $20\text{rad/s}$ for ${\omega }_{AC}$ in Equation (5).

$\begin{array}{l}140\times 20\sin \left(90°\right)=\left[480+200\times \left(90°\times \frac{\pi }{180°}\right)+140\sin \left(90°\right)\right]{\omega }_{BD}-u\sin \left(0°\right)\\ 2800=\left(480+100\pi +140\right){\omega }_{BD}\\ {\omega }_{BD}=\frac{2800}{\left(480+100\pi +140\right)}\\ {\omega }_{BD}=3\text{rad/s}\left(\text{Clockwise}\right)\end{array}$

Thus, the angular velocity of the bar when $\theta =90°$ is $\underset{\_}{3\text{rad/s}\left(\text{Clockwise}\right)}$.