Triangle swinging around a pivot im studying oscilatory motion, and i have a problem that asks me for the angular frequency of a group of sticks,each stick has mass M and length L, that form an equilateral triangle swinging around a pivot located at a vertex. But this triangle has an extra stick that is in the same position of the height  and with the same length (this stick goes from the pivot to the opposite side) The lenght of this stick is $$ \frac{L\sqrt3}{2} $$ and the mass of this stick is M too. This would be the figure. Figure
I know how to solve if the pendulum was only a triangle using the formula
$$  w = \frac{\sqrt(MgD)}{MI} $$ where M is the mass of the object which is 3M if the physical pendulum has the form of a triangle ,  g is the acceleration of gravity and D  , de distance from the center of mass of the triangle to the pivot, this distance would be  $$ \frac{L\sqrt3}{3} $$ and MI is the moment of inertia of the triangle.
So  this figure in the picture  has two objects now , a triangle and a stick. each object has its own center of mass , using the same formula to obtain the angular frequency of this object  , the total mass would be 4M , g is a constant , and D is the distance from pivot to center of mass . 
My question is do i have to sum the distance of each center of mass to the pivot. The distance of the center of mass of the stick to pivot would be   $$ \frac{L\sqrt3}{4} $$ . So summing $$ D = \frac{L\sqrt3}{3} + \frac{L\sqrt3}{4} $$.
Is this correct? Please help me.
 A: No, the distance would not be the addition of the center of masses. As the two sticks have been  joined, they are a rigid body. Thus the whole object(with triangle and stick in the middle will have a common center of mass). You can find this distance,$x$ from the pivot by, where r is the distance from center of mass of each part of rigid body to the reference point(here we're taking it as the pivot point of vertex of triangle) and $m_i$ is the mass of each part:
$$x=\frac{\sum{m_ir}}{\sum{m_i}}$$
$$=\frac{(\sqrt{3}/3)L*3M+(\sqrt{3}/4)L*M}{4M}$$
$$=\frac{5}{16}\sqrt{3}L$$
So thus this is the distance from which torque should be taken for finding the frequency of oscillation. The moment of inertia should on the other hand be taken also for the triangle and stick combined but this time using parallel axis theorem and just adding the individual moment of intertidal of triangle and stick will suffice as moment of inertia is additive for a rigid body.
It's useful to remember that for rigid bodies the distances for center of mass is not additive. And you should have one common center of mass for the body.
