Chapter 15.2, Problem 15.57P

Knowing that the disk has a constant angular velocity of 15 rad/s clockwise, determine the angular velocity of bar BD and the velocity of collar D when (a) θ = 0, (b) θ = 90°, (c) θ = 180°.

Fig. P15.57 and P15.58

To determine

Find the angular velocity of the bar BD and velocity of the collar D when $\theta =0°$.

Answer to Problem 15.57P

The the angular velocity of the bar BD and velocity of the collar D when $\theta =0°$ are $\underset{\_}{\text{4}\text{.38}\text{rad/s}\left(\text{Clockwise}\right)}$ and $\underset{\_}{12.25\text{in}\text{./s}(\uparrow )}$.

Explanation of Solution

Given information:

The constant angular velocity of the disk is ${\text{ω}}_{\text{A}}=15\text{rad/s}\left(\text{Clockwise}\right)$.

The distance AB is $2.8\text{in}\text{.}$

Calculation:

Show the disk as shown in Figure 1.

Refer to Figure 1.

Calculate the velocity at B using the relation:

$v_B=\left(AB\right){\omega }_A$

Substitute $2.8\text{in}\text{.}$ for AB and $15\text{rad/s}$ for ${\text{ω}}_{\text{A}}$.

$\begin{array}{c}{v}_B=2.8\times 15\\ =42\text{in}\text{./s}\end{array}$

Consider that $\theta =0°$.

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

${\text{v}}_B=42\text{in}\text{./s}(\to )$

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

$\begin{array}{l}\sin \beta =\frac{2.8}{10}\\ \beta ={\sin }^{-1}\left(\frac{2.8}{10}\right)\\ \beta =16.260°\end{array}$

Consider bar BD.

Show the velocity diagram as shown in Figure 2.

Refer to Figure 2.

Show the relation between the velocity ${\text{v}}_D$ and ${\text{v}}_B$ as follows:

${\text{v}}_D={\text{v}}_B+{\text{v}}_{D/B}$

Calculate the velocity of point D with respect to B using the relation:

$\begin{array}{l}{v}_{D/B}\cos \beta =42\\ v_{D/B}=\frac{42}{\cos \beta }\end{array}$

Substitute $16.260°$ for $\beta$.

$\begin{array}{c}{v}_{D/B}\cos \beta =42\\ v_{D/B}=\frac{42}{\cos \left(16.260°\right)}\\ v_{D/B}=43.75\text{in}\text{./s}\end{array}$

Calculate the angular velocity of bar BD using the relation:

$\begin{array}{l}{v}_{D/B}={\omega }_{D/B}\left(DB\right)\\ {\omega }_{D/B}=\frac{v_{D/B}}{DB}\end{array}$

Substitute $43.75\text{in}\text{./s}$ for $v_{D/B}$ and $10\text{in}\text{.}$ for DB.

$\begin{array}{c}{\omega }_{D/B}=\frac{43.75}{10}\\ =4.375\text{rad/s}\\ \approx \text{4}\text{.38}\text{rad/s}\end{array}$

Thus, the angular velocity of bar BD is $\underset{\_}{\text{4}\text{.38}\text{rad/s}}$.

Calculate the velocity of the collar $\left(v_D\right)$ using the relation:

$v_D=v_B\tan \beta$

Substitute $16.260°$ for $\beta$ and $42\text{in}\text{./s}$ for ${\text{v}}_B$.

$\begin{array}{c}{v}_D=42\tan \left(16.260°\right)\\ =12.25\text{in}\text{./s}(\uparrow )\end{array}$

Thus, the velocity of the collar D is $\underset{\_}{12.25\text{in}\text{./s}(\uparrow )}$.

To determine

Find the angular velocity of the bar BD and velocity of the collar D when $\theta =90°$.

Answer to Problem 15.57P

The angular velocity of the bar BD and velocity of the collar D are $\underset{\_}{42\text{in/s}(\downarrow )}$ and $\underset{\_}{\text{0}\text{rad/s}}$.

Explanation of Solution

Calculation:

Refer to Part (a).

Show the disk as shown in Figure 3.

Refer Figure 3.

Calculate the velocity at B using the relation:

$v_B=\left(AB\right){\omega }_A$

Substitute $2.8\text{in}\text{.}$ for AB and $15\text{rad/s}$ for ${\text{ω}}_{\text{A}}$.

$\begin{array}{c}{v}_B=2.8\times 15\\ =42\text{in}\text{./s}\end{array}$

Consider $\theta =90°$.

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

${\text{v}}_B=42\text{in}\text{./s}(\downarrow )$

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

$\begin{array}{l}\sin \beta =\frac{5.6}{10}\\ \beta ={\sin }^{-1}\left(\frac{5.6}{10}\right)\\ \beta =34.06°\end{array}$

Consider bar BD.

Show the relation between the velocity ${\text{v}}_D$ and ${\text{v}}_B$ as follows:

$\begin{array}{l}{\text{v}}_D={\text{v}}_B+{\text{v}}_{D/B}\\ v_D(\downarrow )=42(\downarrow )+{\text{v}}_{D/B}\left(\text{Acting at }\beta \text{ Clockwise above the horizontal}\right)\end{array}$ (1)

Equate the horizontal component of Equation (1).

$\begin{array}{l}0=0+{\text{v}}_{D/B}\cos \left(\beta \right)(\leftarrow )\\ {\text{v}}_{D/B}=0\end{array}$

Equate the vertical component of Equation (1).

$v_D=42-{\text{v}}_{D/B}\sin \beta$

Substitute 0 for ${\text{v}}_{D/B}$.

$v_D=42\text{in/s}(\downarrow )$

Thus, the velocity of the collar is $\underset{\_}{42\text{in/s}(\downarrow )}$.

Calculate the angular velocity of bar BD using the relation:

$v_{D/B}={\omega }_{D/B}\left(DB\right)$

Substitute $0$ for $v_{D/B}$.

${\omega }_{D/B}=0\text{rad/s}$

Thus, the angular velocity of bar BD is $\underset{\_}{\text{0}\text{rad/s}}$.

To determine

Find the angular velocity of the bar BD and velocity of the collar D when $\theta =180°$.

Answer to Problem 15.57P

The angular velocity of the bar BD and velocity of the collar D when $\theta =180°$ are $\underset{\_}{\text{4}\text{.38}\text{rad/s}\left(\text{Counterclockwise}\right)}$ and $\underset{\_}{12.25\text{in}\text{./s}(\downarrow )}$.

Explanation of Solution

Given information:

The constant angular velocity of the disk is ${\text{ω}}_{\text{A}}=15\text{rad/s}\left(\text{Clockwise}\right)$.

The distance AB is $2.8\text{in}\text{.}$

Calculation:

Show the disk as shown in Figure 4.

Refer Figure 4.

Calculate the velocity at B using the relation:

$v_B=\left(AB\right){\omega }_A$

Substitute $2.8\text{in}\text{.}$ for AB and $15\text{rad/s}$ for ${\text{ω}}_{\text{A}}$.

$\begin{array}{c}{v}_B=2.8\times 15\\ =42\text{in}\text{./s}\end{array}$

Consider $\theta =180°$.

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

${\text{v}}_B=42\text{in}\text{./s}(\leftarrow )$

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

$\begin{array}{l}\sin \beta =\frac{2.8}{10}\\ \beta ={\sin }^{-1}\left(\frac{2.8}{10}\right)\\ \beta =16.260°\end{array}$

Consider bar BD.

Show the velocity diagram as shown in Figure 5.

Refer Figure 5.

Show the relation between the velocity ${\text{v}}_D$ and ${\text{v}}_B$ as follows:

${\text{v}}_D={\text{v}}_B+{\text{v}}_{D/B}$

Calculate the velocity of point D with respect to B using the relation:

$\begin{array}{l}{v}_{D/B}\cos \beta =v_B\\ v_{D/B}=\frac{42}{\cos \beta }\end{array}$

Substitute $16.260°$ for $\beta$.

$\begin{array}{c}{v}_{D/B}\cos \beta =42\\ v_{D/B}=\frac{42}{\cos \left(16.260°\right)}\\ v_{D/B}=43.75\text{in}\text{./s}\end{array}$

Calculate the angular velocity of bar BD using the relation:

$\begin{array}{l}{v}_{D/B}={\omega }_{D/B}\left(DB\right)\\ {\omega }_{D/B}=\frac{v_{D/B}}{DB}\end{array}$

Substitute $43.75\text{in}\text{./s}$ for $v_{D/B}$ and $10\text{in}\text{.}$ for DB.

$\begin{array}{c}{\omega }_{D/B}=\frac{43.75}{10}\\ =4.375\text{rad/s}\\ \approx \text{4}\text{.38}\text{rad/s}\left(\text{Counterclockwise}\right)\end{array}$

Thus, the angular velocity of bar BD is $\underset{\_}{\text{4}\text{.38}\text{rad/s}\left(\text{Counterclockwise}\right)}$.

Calculate the velocity of the collar $\left(v_D\right)$ using the relation:

$v_D=v_B\tan \beta$

Substitute $16.260°$ for $\beta$ and $42\text{in}\text{./s}$ for ${\text{v}}_B$.

$\begin{array}{c}{v}_D=42\tan \left(16.260°\right)\\ =12.25\text{in}\text{./s}(\downarrow )\end{array}$

Thus, the velocity of the collar D is $\underset{\_}{12.25\text{in}\text{./s}(\downarrow )}$.