Step 1

We are given the mass of link $AB$ and $CD$ as $m = 6\;{\rm{kg}}$, the mass of link $BD$ as ${m_{BD}} = 20\;{\rm{kg}}$, the angular velocity as $\omega = 2\;{\rm{rad/s}}$, the moment as $M = 3{\rm{0}}\;{\rm{N}} \cdot {\rm{m}}$, the angle as $\theta = 45^\circ $, the length $BD$ as ${r_{BD}} = 1.5\;{\rm{m}}$, the length $AB$ as ${r_{AB}} = 1\;{\rm{m}}$, and the length $CD$ as ${r_{CD}} = 1\;{\rm{m}}$.

We are asked to determine the angular velocity.

 
Step 2

We will draw a free body diagram of the system.

Here, ${W_1}$ is the weight of the link $BD$.

 
Step 3

We will find the moment of inertia of the rod $AB$.

\[I = \frac{{mr_{AB}^2}}{{12}} + m{\left( {\frac{{{r_{AB}}}}{2}} \right)^2}\]

Substitute the given value in the above equation.

\[\begin{array}{c} I = \frac{{\left( {6\;{\rm{kg}}} \right){{\left( {1\;{\rm{m}}} \right)}^2}}}{{12}} + \left( {6\;{\rm{kg}}} \right){\left( {\frac{{1\;{\rm{m}}}}{2}} \right)^2}\\ = 2\,{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}} \end{array}\]

Both the rods $AB$ and $CD$ are identical in shape. So, the moment of inertia will be same for both rods.

 
Step 4

We will find the velocity at point $B$.

\[{v_B} = {r_{AB}}\omega \]

Here, $\omega $ is the angular velocity of the rod $AB$.

Substitute the given value in the above equation.

\[\begin{array}{l} {v_B} = \left( {1\;{\rm{m}}} \right)\omega \\ {v_B} = \left( \omega \right)\;{\rm{m/s}} \end{array}\]
 
Step 5

We will find the final kinetic energy of the linkage.

\[K{E_f} = 2\left( {\frac{1}{2}I{\omega ^2}} \right) + \frac{1}{2}{m_{BD}}v_B^2\]

Substitute the given value in the above equation.

\[\begin{array}{c} K{E_f} = 2\left( {\frac{1}{2}\left( {2\,{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}} \right){\omega ^2}} \right) + \frac{1}{2}\left( {20\;{\rm{kg}}} \right){\left( {\left( \omega \right)\;{\rm{m/s}}} \right)^2}\\ = \left( {12{\omega ^2}} \right)\;{\rm{J}} \end{array}\]
 
Step 6

We will apply the energy balance equation to find the initial kinetic energy of the linkage.

\[\begin{array}{l} K{E_i} = K{E_f}\\ K{E_i} = \left( {12{\omega ^2}} \right)\;{\rm{J}} \end{array}\]

Substitute the given value in the above equation.

\[\begin{array}{c} K{E_i} = \left( {12{{\left( 2 \right)}^2}} \right)\;{\rm{J}}\\ = 48\;{\rm{J}} \end{array}\]
 
Step 7

We will find the change vertical distance of the rod.

\[\Delta y = - \left( {{r_{AB}} - {r_{AB}}\cos \theta } \right)\]

Substitute the given value in the above equation.

\[\begin{array}{c} \Delta y = - \left( {1\;{\rm{m}} - \left( {1\;{\rm{m}}} \right)\cos 45^\circ } \right)\\ \approx - 0.293\;{\rm{m}} \end{array}\]
 
Step 8

We will find the wok done due to the weight of the rods.

\[\begin{array}{l} U = - 2W\Delta y - {W_1}\Delta y\\ U = - 2mg\Delta y - {m_{BD}}g\Delta y \end{array}\]

Substitute the given value in the above equation.

\[\begin{array}{c} U = - 2\left( {6\;{\rm{kg}}} \right)\left( {{\rm{9}}{\rm{.81}}\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\left( { - 0.293\;{\rm{m}}} \right) - \left( {20\;{\rm{kg}}} \right)\left( {{\rm{9}}{\rm{.81}}\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\left( { - 0.293\;{\rm{m}}} \right)\\ \approx 91.98\;{\rm{J}} \end{array}\]
 
Step 9

We will find the work done due to couple moment.

\[{U_M} = M\theta \]

Substitute the given value in the above equation.

\[\begin{array}{c} {U_M} = \left( {3{\rm{0}}\;{\rm{N}} \cdot {\rm{m}}} \right)\left( {45^\circ } \right)\\ = \left( {3{\rm{0}}\;{\rm{N}} \cdot {\rm{m}}} \right)\left( {\frac{\pi }{4}} \right)\\ = \left( {7.5\pi } \right)\;{\rm{J}} \end{array}\]
 
Step 10

We will find the angular velocity of the rod at $\theta = 45^\circ $ using the energy equation.

\[K{E_i} + U + {U_M} = K{E_f}\]

Substitute the given value in the above equation.

\[\begin{array}{c} 48\;{\rm{J}} + 91.98\;{\rm{J}} + \left( {7.5\pi } \right)\;{\rm{J}} = \left( {12{\omega ^2}} \right)\;{\rm{J}}\\ \omega = \sqrt {\frac{{48 + 91.98 + 7.5\pi }}{{12}}} \\ \approx 3.{\rm{7}}\;{\rm{rad/s}} \end{array}\]