Step 1

We are given the weight of the uniform rods is $W = 10\;{\rm{lb}}$ and the angular velocity of the disk is ${\omega _D} = 4\;{\rm{rad/s}}$.

We are asked to determine the angle $\theta $ made by each rod during the motion, and the components of the force and moment developed at the pin A.

 
Step 2

The diagram of the system is shown as:

The free body diagram of the segment AB is shown as:

We have the radius of the rotating disk is $R = 1.75\;{\rm{ft}}$.

We have the length of rod is $l = 2\;{\rm{ft}} + 2\;{\rm{ft}} = 4\;{\rm{ft}}$.

 
Step 3

The expressions for the equations of motion are,

\[{A_x} = m{\bar a_x}\] ... (1) \[{A_y} - W\sin \theta = - mr{\left( {{\omega _D}} \right)^2}\cos \theta \] ... (2) \[{A_z} - W\cos \theta = - mr{\left( {{\omega _D}} \right)^2}\sin \theta \] ... (3)

Here, m is the mass of the rod, ${\bar a_x}$ is the acceleration in x-direction, and r is the distance of point G from the disk axis.

 
Step 4

The expressions for the moment equations about G in x-direction are,

\[\begin{array}{c} \Sigma {M_{Gx}} = 0\\ {A_y}\left( {\frac{l}{2}} \right) = {I_x}{{\dot \omega }_x} - \left( {{I_y} - {I_z}} \right){\omega _y}{\omega _z} \end{array}\] ... (4)
 
Step 5

The expressions for the moment equations about G in y-direction are,

\[\begin{array}{c} \Sigma {M_{Gy}} = 0\\ {M_y} - {A_x}\left( {\frac{l}{2}} \right) = {I_y}{{\dot \omega }_y} - \left( {{I_z} - {I_x}} \right){\omega _z}{\omega _x} \end{array}\] ... (5)
 
Step 6

The expressions for the moment equations about G in z-direction are,

\[\begin{array}{c} \Sigma {M_{Gz}} = 0\\ {M_z} = {I_z}{{\dot \omega }_z} - \left( {{I_x} - {I_y}} \right){\omega _x}{\omega _y} \end{array}\] ... (6)
 
Step 7

The disk is rotating with a constant speed; therefore the following value at different angle of $\theta $ will be zero. i.e, ${\bar a_x},{\omega _x},{\dot \omega _x},{\dot \omega _y}{\rm{ and }}{\dot \omega _z} = 0$.

Substitute the value of ${\bar a_x}$ in equation (1), we get:

\[{A_x} = 0\]
 
Step 8

Substitute the value of ${A_x},{\dot \omega _y}$ and ${\omega _x}$ in equation (5), we get:

\[{M_y} = 0\]
 
Step 9

Substitute the value of ${\omega _y}$ and ${\omega _x}$ in equation (6), we get:

\[{M_z} = 0\]
 
Step 10

The formula to calculate the distance of point G from the disk axis is,

\[r = \frac{l}{2}\sin \theta + R\]
 
Step 11

Substitute the values in the above expression.

\[\begin{array}{l} r = \frac{{4\;{\rm{ft}}}}{2}\sin \theta + 1.75\;{\rm{ft}}\\ r = \left( {1.75 + 2\sin \theta } \right)\;{\rm{ft}} \end{array}\]
 
Step 12

Substitute the value ${\omega _z} = {\omega _D}\cos \theta $, ${\omega _y} = {\omega _D}\sin \theta $ and ${I_y} = \frac{1}{{12}}m{l^2}$ in equation (4), we get:

\[\begin{array}{l} {A_y}\left( {\frac{l}{2}} \right) = {I_x}\left( 0 \right) - \left( {\frac{{m{l^2}}}{{12}} - 0} \right)\left( {{\omega _D}\sin \theta } \right)\left( {{\omega _D}\cos \theta } \right)\\ {A_y} = \frac{2}{l}\left( {\frac{{m{l^2}}}{{12}}} \right){\left( {{\omega _D}} \right)^2}\sin \theta \cos \theta \\ {A_y} = \frac{2}{l}\left( {\frac{{W{l^2}}}{{12g}}} \right){\left( {{\omega _D}} \right)^2}\sin \theta \cos \theta \end{array}\]

Here, g is the gravitational acceleration, having a standard value of $32.2\;{\rm{ft/}}{{\rm{s}}^2}$.

 
Step 13

Substitute the values in the above expression.

\[\begin{array}{l} {A_y} = \frac{2}{{2\;{\rm{ft}}}}\left( {\frac{{10\;{\rm{lb}} \times {{\left( {2\;{\rm{ft}}} \right)}^2}}}{{12 \times 32.2\;{\rm{ft/}}{{\rm{s}}^2}}}} \right){\left( {4\;{\rm{rad/s}}} \right)^2}\sin \theta \cos \theta \\ {A_y} = 3.33\sin \theta \cos \theta \end{array}\] ... (7)
 
Step 14

Substitute the values in the equation (2).

\[\begin{array}{l} 3.33\sin \theta \cos \theta - 10\;{\rm{lb}}\sin \theta = - \frac{{10\;{\rm{lb}}}}{{32.2\;{\rm{ft/}}{{\rm{s}}^2}}}\left( {1.75\;{\rm{ft}} + 2\sin \theta \;{\rm{ft}}} \right){\left( {4\;{\rm{rad/s}}} \right)^2}\cos \theta \\ 3.33\sin \theta \cos \theta - \sin \theta = - \frac{{16}}{{32.2}}\left( {1.75 + 2\sin \theta } \right)\cos \theta \end{array}\]

On dividing with $\cos \theta $, we get:

\[\begin{array}{c} 3.33\sin \theta - \tan \theta = - \frac{{16}}{{32.2}}\left( {1.75 + 2\sin \theta } \right)\\ \tan \theta = 0.875 + 1.33\sin \theta \end{array}\]
 
Step 15

On solving the above equation, we get:

\[\theta = 64^\circ \]
 
Step 16

Substitute the value in the equation (7).

\[\begin{array}{l} {A_y} = 3.33\sin 64^\circ \cos 64^\circ \\ {A_y} = 1.3135\;{\rm{lb}} \end{array}\]
 
Step 17

Substitute the value in the equation (3).

\[\begin{array}{c} {A_z} - \left( {10\;{\rm{lb}}} \right)\cos 64^\circ = - \left( {\frac{{10\;{\rm{lb}}}}{{32.2\;{\rm{ft/}}{{\rm{s}}^2}}}} \right)\left( {1.75 + 2\sin 64^\circ } \right){\left( {4\;{\rm{rad/}}{{\rm{s}}^2}} \right)^2}\left( {\sin 64^\circ } \right)\\ {A_z} = 20.326\;{\rm{lb}} \end{array}\]