Falling stick

This problem can be solved by the concept of IAOR. First note that since there is no horizontal force hence COM will fall vertically. Also the lower end of the rod will slide horizontally. So we can locate the IAOR by drawing perpendiculars from their line of velocities and marking their intersection. From the IAOR we can assume the motion to be completely rotational. Applying energy conservation we have :

$\frac { mgl(1-\sin { \theta ) } }{ 2 } =\frac { 1 }{ 2 } (\frac { m{ l }^{ 2 } }{ 12 } +\frac { m{ l }^{ 2 }{ cos }^{ 2 }\theta }{ 4 } ){ \omega }^{ 2 }$

Here moment of inertia is taken about IAOR.(note parellel axes theorom has been used here) Solving $\omega =\sqrt { \frac { 12g(1-sin\theta ) }{ l(1+{ 3cos }^{ 2 }\theta ) } }$

Also ${ V }_{ com }=\frac { \omega lcos\theta }{ 2 }$

Hence ${ V }_{ com }=\sqrt { \frac { 3gl{ cos }^{ 2 }\theta (1-sin\theta ) }{ 1+3{ cos }^{ 2 }\theta } }$

Since $\theta ={ 30 }^{ 0 }$

Putting the values we get ${ V }_{ com }=\sqrt { \frac { 9gl }{ 26 } }$

Hence $a=9,b=26$ So $\boxed{a+b=35}$

Note: 9 can be taken out of square root but we will not because of the from given in the question.