Step 1

We are given the angular velocity of the wheel ${\omega _B} = 4\;{\rm{rad/s}}$, the radius for point $P$ as ${r_P} = 4\;{\rm{in}}$, and the angle as $\theta = 30^\circ $.

We are asked to determine the angular velocity ${\omega _A}$ and angular acceleration ${\alpha _A}$.

 
Step 2

We will find the radius of wheel $B$.

\[{r_B} = {r_P}\tan \theta \]

Substitute the given value in the above equation.

\[\begin{array}{c} {r_B} = \left( {{\rm{4}}\;{\rm{in}}} \right)\tan 30^\circ \\ {r_B} = 2.31\;{\rm{in}} \end{array}\]
 
Step 3

We will find the velocity of point $P$.

\[{{\bf{v}}_P} = {\omega _B}\left( { - {\bf{k}}} \right) \times {r_P}{\bf{i}}\]

Substitute the given value in the above equation.

\[\begin{array}{c} {{\bf{v}}_P} = \left( {4\;{\rm{rad/s}}} \right){\bf{k}} \times \left( {{\rm{2}}{\rm{.31}}\;{\rm{in}}} \right){\bf{i}}\\ {{\bf{v}}_P} = - \left( {9.24{\bf{j}}} \right)\;{\rm{in/s}} \end{array}\]
 
Step 4

We will find the acceleration of point $P$.

\[{{\bf{a}}_P} = - \left( {{\omega _B}^2{r_B}} \right){\bf{i}}\]

Substitute the given value in the above equation.

\[\begin{array}{c} {{\bf{a}}_P} = - \left( {{{\left( {4\;{\rm{rad/s}}} \right)}^2}\left( {2.31\;{\rm{in}}} \right)} \right){\bf{i}}\\ {{\bf{a}}_P} = - \left( {36.96\;{\rm{in/}}{{\rm{s}}^{\rm{2}}}} \right){\bf{i}} \end{array}\]
 
Step 5

We will find the position vector ${{\bf{r}}_{B/A}}$.

\[{{\bf{r}}_{P/A}} = {r_P}{\bf{j}}\]

Substitute the given value in the above equation.

\[{{\bf{r}}_{P/A}} = \left( {{\rm{4}}\;{\rm{in}}} \right){\bf{j}}\]
 
Step 6

We will find the angular velocity of wheel $A$.

\[{{\bf{v}}_P} = {{\bf{v}}_A} + {\omega _A}{{\bf{r}}_{P/A}} + {\left( {{{\bf{v}}_{P/A}}} \right)_{xyx}}\]

Here, ${{\bf{v}}_A} = 0\;{\rm{in/s}}$ is the velocity of point $A$ and ${\left( {{{\bf{v}}_{P/A}}} \right)_{xyz}} = {v_{P/A}}{\bf{j}}$ is the velocity of point $P$ relative to $A$.

Substitute the given value in the above equation.

\[\begin{array}{c} - \left( {9.24{\bf{j}}} \right)\;{\rm{in/s}} = 0\;{\rm{in/s}} + {\omega _A}\left( { - {\bf{k}}} \right) \times \left( {{\rm{4}}\;{\rm{in}}} \right){\bf{j}} + {v_{P/A}}{\bf{j}}\\ - \left( {9.24{\bf{j}}} \right)\;{\rm{in/s}} = \left( {4{\omega _A}{\bf{i}} + {v_{P/A}}{\bf{j}}} \right)\;{\rm{in/s}} \end{array}\]
 
Step 7

Compare the components ${\bf{i}}$ and ${\bf{j}}$ in the above equation to find the angular velocity of the wheel $A$ and the velocity ${v_{P/A}}$.

\[\begin{array}{c} 4{\omega _A} = 0\\ {\omega _A} = 0\;{\rm{rad/s}} \end{array}\]

And

\[{v_{P/A}} = - 9.24\;{\rm{in/s}}\]
 
Step 8

We will find the angular acceleration of the wheel $A$.

\[{{\bf{a}}_P} = {{\bf{a}}_A} + {\alpha _A}{\bf{k}} \times {{\bf{r}}_{P/A}} + {\omega _A}{\bf{k}} \times \left( {{\omega _A}{\bf{k}} \times {{\bf{r}}_{P/A}}} \right) + 2{\omega _A} \times {\left( {{{\bf{v}}_{P/A}}} \right)_{xyz}} + {\left( {{{\bf{a}}_{P/A}}} \right)_{xyz}}\]

Here, ${{\bf{a}}_A} = 0\;{\rm{in/}}{{\rm{s}}^{\rm{2}}}$ is the velocity of point $A$, ${\left( {{{\bf{v}}_{P/A}}} \right)_{xyz}} = {v_{P/A}}{\bf{j}}$ is the velocity of point $P$ relative to $A$, and ${\left( {{{\bf{a}}_{P/A}}} \right)_{xyz}} = - {a_{P/A}}{\bf{j}}$ is the acceleration of point $P$ relative to $A$,

Substitute the given value in the above equation.

\[\begin{array}{c} - \left( {36.96\;{\rm{in/}}{{\rm{s}}^{\rm{2}}}} \right){\bf{i}} = \left[ \begin{array}{l} 0\;{\rm{in/}}{{\rm{s}}^{\rm{2}}} + {\alpha _A}{\bf{k}} \times \left( {{\rm{4}}\;{\rm{in}}} \right){\bf{j}} + \\ \left( {0\;{\rm{rad/s}}} \right){\bf{k}} \times \left( {\left( {0\;{\rm{rad/s}}} \right){\bf{k}} \times \left( {{\rm{4}}\;{\rm{in}}} \right){\bf{j}}} \right) + \\ 2\left( {0\;{\rm{rad/s}}} \right) \times \left( {{v_{P/A}}} \right){\bf{j}} + \left( { - {a_{P/A}}} \right){\bf{j}} \end{array} \right]\\ - \left( {36.96\;{\rm{in/}}{{\rm{s}}^{\rm{2}}}} \right){\bf{i}} = \left[ { - 4{\alpha _A}{\bf{i}} - {a_{P/A}}{\bf{j}}} \right]\;{\rm{in/}}{{\rm{s}}^{\rm{2}}} \end{array}\]
 
Step 9

Compare the components ${\bf{i}}$ in the above equation to find the angular acceleration of the wheel $A$.

\[\begin{array}{c} - \left( {36.96\;{\rm{in/}}{{\rm{s}}^{\rm{2}}}} \right) = - 4{\alpha _A}\\ {\alpha _A} = 9.24\;{\rm{in/}}{{\rm{s}}^{\rm{2}}} \end{array}\]