Step 1

We are given the angular velocity and angular acceleration ${\omega _{AB}} = 3\,{\rm{rad/s}}$, ${\alpha _{AB}} = 6\,{\rm{rad/}}{{\rm{s}}^2}$ and length of link CD, BC and AB is ${r_{CD}} = 0.5\;{\rm{m}}$, ${r_{BC}} = 1\;{\rm{m}}$ and ${r_{AB}} = 1\;{\rm{m}}$.

We are asked to determine the angular acceleration and angular velocity of link CD.

 
Step 2

To find the velocity of link AB we will use the relation,

\[\begin{array}{c} {v_B} = {\omega _{AB}}{r_{AB}}\\ {v_B} = \left( {3\;{\rm{rad/s}}} \right)\left( {1\;{\rm{m}}} \right)\\ {v_B} = 3\;{\rm{m/s}} \end{array}\]

To find the acceleration of AB, we will use the relation,

\[{\vec a_B} = \left( {{{\vec \alpha }_{AB}} \times {{\vec r}_{AB}}} \right) - {\vec \omega _{AB}}^2{\vec r_{AB}}\]

On plugging the values in the above relation, we get,

\[\begin{array}{c} {a_B} = \left( { - 6\;{\rm{rad/}}{{\rm{s}}^2}} \right)\hat k\left( {1\;{\rm{m}}} \right)\hat i - {\left( {3\;{\rm{rad/s}}} \right)^2}\left( {1\;{\rm{m}}} \right)\hat i\\ {a_B} = \left[ {\left( { - 6\hat j} \right) - \left( {9\hat i} \right)} \right]\;{\rm{m/}}{{\rm{s}}^2} \end{array}\]
 
Step 3

To find the acceleration of link CD, we will use the relation,

\[{\vec a_C} = \left( {{{\vec \alpha }_{CD}} \times {{\vec r}_{CD}}} \right) - {\vec \omega _{CD}}^2{\vec r_{CD}}\]

On plugging the values in the above relation, we get,

\[\begin{array}{c} {a_C} = \left( {{\alpha _{CD}}\hat k} \right)\left( { - 0.5\hat j} \right) - {\left( {{\omega _{CD}}} \right)^2}\left( { - 0.5\hat j} \right)\\ {a_C} = 0.5{\alpha _{CD}}\hat i + 0.5{\omega _{CD}}^2\hat j \end{array}\]
 
Step 4

To find the velocity at B, we will use the relation,

\[\begin{array}{c} {v_B} = {\omega _{BC}}{r_{CD}}\\ 3\;{\rm{m/s}} = \left( {0.5} \right){\omega _{BC}}\\ {\omega _{BC}} = 6\,{\rm{rad/s}} \end{array}\]

To find the velocity of link CD, we will use the relation,

\[{v_C} = {\omega _{BC}}{r_{C/IC}}\]

Here, ${r_{C/IC}}$ is the distance of the point C from the IC whose value is $1\,{\rm{m}}$.

On plugging the values in the above relation, we get,

\[\begin{array}{c} {v_C} = \left( {{\rm{6}}\,{\rm{rad/s}}} \right)\left( {1\;{\rm{m}}} \right)\\ {v_C} = {\rm{6}}\;{\rm{m/s}} \end{array}\]
 
Step 5

To find the angular velocity of link CD, we will use the relation,

\[\begin{array}{c} {v_C} = {\omega _{CD}}{r_{CD}}\\ \left( {6\;{\rm{m/s}}} \right) = {\omega _{CD}}\left( {0.5\;{\rm{m}}} \right)\\ {\omega _{CD}} = 12\;{\rm{rad/s}} \end{array}\]

To find the required angular acceleration, we will use the relation,

\[{\vec a_C} = {\vec a_B} + \left( {{{\vec \alpha }_{BC}} \times {{\vec r}_{BC}}} \right) - {\vec \omega _{BC}}^2{\vec r_{BC}}\]

On plugging the values in the above relation, we get,

\[\begin{array}{c} 0.5{\alpha _{CD}}\hat i + 0.5{\omega _{CD}}^2\hat j = \left( { - 6\hat j} \right) - \left( {9\hat i} \right) + \left( {{\alpha _{BC}}\hat k} \right)\left( { - {r_{CD}}\hat i + {r_{BC}}\hat j} \right) - {\omega _{BC}}^2\left( { - {r_{CD}}\hat i + {r_{BC}}\hat j} \right)\\ 0.5{\alpha _{CD}}\hat i + 0.5{\left( {12} \right)^2}\hat j = \left( { - 6\hat j} \right) - \left( {9\hat i} \right) + \left( {{\alpha _{BC}}\hat k} \right)\left( { - 0.5\hat i + 1\hat j} \right) - {\left( 6 \right)^2}\left( { - 0.5\hat i + 1\hat j} \right)\\ 0.5{\alpha _{CD}}\hat i + 72\hat j = \left( {9 + {\alpha _{BC}}} \right)\hat i + \left( {0.5{\alpha _{BC}} - 42} \right)\hat j \end{array}\]

On equating the components of $\hat j$, we get,

\[\begin{array}{c} 72 = 0.5{\alpha _{BC}} - 42\\ {\alpha _{BC}} = 228\;{\rm{rad/}}{{\rm{s}}^2} \end{array}\]

On equating the components of $\hat i$, we get,

\[\begin{array}{c} 0.5{\alpha _{DC}} = 9 + {\alpha _{BC}}\\ 0.5{\alpha _{DC}} = 9 + 228\;{\rm{rad/}}{{\rm{s}}^2}\\ {\alpha _{DC}} = 474\;{\rm{rad/}}{{\rm{s}}^2} \end{array}\]