Step 1

We are given the angular velocity of the shaft is $\omega = \left\{ {2i} \right\}\;{\rm{rad/s}}$, and the mass per unit length of each rod is $m = 5\;{\rm{kg/m}}$.

We are asked to determine the reactions at the bearings, and the shaft's angular acceleration.

 
Step 2

The diagram of the system is shown as:

We have the distance between the journal bearings A and B is ${l_1} = 2\;{\rm{m}} + 1\;{\rm{m}} = 3\;{\rm{m}}$.

We have the length of intermediate shaft segment is ${l_2} = 1\;{\rm{m}}$.

 
Step 3

The formula to calculate the moment of inertia of the system about the x-plane is,

\[{I_x} = \frac{1}{3}\left( {m{l_2}} \right){\left( {{l_2}} \right)^2}\]

Substitute the values in the above expression.

\[\begin{array}{c} {I_x} = \frac{1}{3}\left( {5\;{\rm{kg/m}} \times 1\;{\rm{m}}} \right){\left( {1\;{\rm{m}}} \right)^2}\\ {I_x} = 1.67\;{\rm{kg}} \cdot {{\rm{m}}^2} \end{array}\]
 
Step 4

The formula to calculate the moment of inertia of the system about the y-plane is,

\[{I_y} = \frac{1}{3}\left( {m{l_1}} \right){\left( {{l_1}} \right)^2} + \left( {m{l_2}} \right){\left( {{l_2}} \right)^2}\]

Substitute the values in the above expression.

\[\begin{array}{c} {I_y} = \frac{1}{3}\left( {5\;{\rm{kg/m}} \times 3\;{\rm{m}}} \right){\left( {3\;{\rm{m}}} \right)^2} + \left( {5\;{\rm{kg/m}} \times 1\;{\rm{m}}} \right){\left( {1\;{\rm{m}}} \right)^2}\\ {I_y} = 50\;{\rm{kg}} \cdot {{\rm{m}}^2} \end{array}\]
 
Step 5

The formula to calculate the moment of inertia of the system about the z-plane is,

\[{I_z} = \frac{1}{3}\left( {m{l_1}} \right){\left( {{l_1}} \right)^2} + \left\{ {\frac{1}{{12}}\left( {m{l_2}} \right){{\left( {{l_2}} \right)}^2} + \left( {m{l_2}} \right)\left[ {{{\left( {{l_2}} \right)}^2} + {{\left( {\frac{{{l_2}}}{2}} \right)}^2}} \right]} \right\}\]

Substitute the values in the above expression.

\[\begin{array}{c} {I_z} = \left[ \begin{array}{l} \frac{1}{3}\left( {5\;{\rm{kg/m}} \times 3\;{\rm{m}}} \right){\left( {3\;{\rm{m}}} \right)^2}\\ + \left\{ {\frac{1}{{12}}\left( {5\;{\rm{kg/m}} \times 1\;{\rm{m}}} \right){{\left( {1\;{\rm{m}}} \right)}^2} + \left( {5\;{\rm{kg/m}} \times 1\;{\rm{m}}} \right)\left[ {{{\left( {1\;{\rm{m}}} \right)}^2} + {{\left( {\frac{{1\;{\rm{m}}}}{2}} \right)}^2}} \right]} \right\} \end{array} \right]\\ {I_z} = 51.67\;{\rm{kg}} \cdot {{\rm{m}}^2} \end{array}\]
 
Step 6

The formula to calculate the moment of inertia in x-y-plane is,

\[{I_{xy}} = m{\left( {{l_2}} \right)^2}\left( {\frac{{{l_2}}}{2}} \right)\]

Substitute the values in the above expression.

\[\begin{array}{c} {I_{xy}} = 5\;{\rm{kg/m}}{\left( {1\;{\rm{m}}} \right)^2}\left( {\frac{{1\;{\rm{m}}}}{2}} \right)\\ {I_{xy}} = 2.50\;{\rm{kg}} \cdot {{\rm{m}}^2} \end{array}\]
 
Step 7

The expression to calculate the moment of inertia in y-z and z-x-plane is,

\[{I_{yz}} = {I_{zx}} = 0\]

The expression to calculate the angular acceleration of system from free body diagram is,

\[\begin{array}{c} {\omega _y} = {\omega _z} = 0\\ {{\dot \omega }_y} = {{\dot \omega }_z} = 0 \end{array}\]
 
Step 8

The formula to calculate the moment of the system about x-axis is,

\[\begin{array}{c} {M_x} = {I_x}{{\dot \omega }_x}\\ \left( {m{l_2}} \right)g\left( {\frac{{{l_2}}}{2}} \right) = {I_x}{{\dot \omega }_x} \end{array}\]

Here, g is the acceleration due to gravity and its standard value is $9.81\;{\rm{m/}}{{\rm{s}}^2}$.

Substitute the values in the above expression.

\[\begin{array}{c} \left( {5{\rm{kg/m}} \times 1\;{\rm{m}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {\frac{{1\;{\rm{m}}}}{2}} \right) = 1.67{{\dot \omega }_x}\\ {{\dot \omega }_x} = - 14.7\;{\rm{rad/}}{{\rm{s}}^2} \end{array}\]
 
Step 9

The formula to calculate the moment of the system about y-axis is,

\[\begin{array}{c} {M_y} = {I_{xy}}{{\dot \omega }_y}\\ \left( {m{l_2}} \right)g + 2mg\left( {{l_2} + \frac{{{l_2}}}{2}} \right) - {B_z}\left( {{l_1}} \right) = {I_{xy}}{{\dot \omega }_y} \end{array}\]

Substitute the values in the above expression.

\[\begin{array}{c} \left[ \begin{array}{c} \left( {5{\rm{kg/m}} \times 1\;{\rm{m}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^2}} \right) + \\ \left( {5{\rm{kg/m}} \times {\rm{2}}\;{\rm{m}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {1\;{\rm{m}} + \frac{{1\;{\rm{m}}}}{2}} \right)\\ - {B_z}\left( {3\;{\rm{m}}} \right) \end{array} \right] = \left( { - 2.50\;{\rm{kg}} \cdot {{\rm{m}}^2}} \right)\left( { - 14.715} \right)\\ {B_z} = 77.7\;{\rm{N}} \end{array}\]
 
Step 10

The formula to calculate the moment of the system about z-axis is,

\[\begin{array}{c} {M_z} = - {I_{xy}}{\left( {{\omega _x}} \right)^2}\\ - {B_y}{l_1} = - {I_{xy}}{\left( {{\omega _x}} \right)^2} \end{array}\]

Substitute the values in the above expression.

\[\begin{array}{c} - {B_y}\left( 3 \right) = - 2.50{\left( 2 \right)^2}\\ {B_y} = 3.33\;{\rm{N}} \end{array}\]
 
Step 11

The expression to calculate the reaction forces at bearings in x-axis is,

\[{A_x} = 0\]

The expression to calculate the reaction forces at bearings in y-axis is,

\[ - {A_y} - {B_y} = m\left[ { - {2^2}\left( {\frac{{{l_2}}}{2}} \right)} \right]\]

Substitute the values in the above expression.

\[\begin{array}{c} - {A_y} - 3.33 = 5\left[ { - {2^2}\left( {\frac{1}{2}} \right)} \right]\\ {A_y} = 6.67\;{\rm{N}} \end{array}\]
 
Step 12

The expression to calculate the reaction forces at bearings in z-axis is,

\[{A_z} + {B_z} - mg{l_1} - mg{l_2} = m\left[ {{{\dot \omega }_x}\left( {\frac{{{l_2}}}{2}} \right)} \right]\]

Substitute the values in the above expression.

\[\begin{array}{c} {A_z} + 77.7 - \left( {5 \times 9.81 \times 3} \right) - \left( {5 \times 9.81 \times 1} \right) = \left( 5 \right)\left[ {{{\dot \omega }_x}\left( {\frac{1}{2}} \right)} \right]\\ {A_z} - 118.5 = \left( 5 \right)\left[ {\left( { - 14.7} \right)\left( {\frac{1}{2}} \right)} \right]\\ {A_z} - 118.5 = - 36.75\\ {A_z} = 81.75\;{\rm{N}} \end{array}\]