How Do You Calculate the Instantaneous Linear Speed of a T-Shaped Pendulum?

[myko]

1. A T-shaped assembly is made of two identical uniform bars of length d and mass m each. One end of one of the bars is rigidly connected to the midpoint of the other. The assembly is pivoted at the connection point and held in position A. After it is released, the assembly swings down in the vertical plane. Neglect all resistive forces.

Find the instantaneous linear speed of the bottom point of the assembly as it passes through position B
2. I tryed to use energy equation to find the velocity, but moment of inertia is unknown
2mg(d/2)=1/2(2m)v_{cm}^2+1/2I_{cm}\omega^2 so I can't figure out how to continue

 

[myko]

picture

The corresponding image for the situation

 

[dean barry]

The assembly is in the first position.
The length of each bar is L

Im guessing the assembly centre of mass is ( L / 4 ) above the pivot point.
( i considered each bar as a point mass at its individual centre of mass, then assumed the combined centre of mass at a point halfway inbetween )

The distance the combined centre of mass falls is ( L / 4 ) * 2
(call this distance h )
So, the potential energy lost = m * g * h

 

[myko]

The assembly is in the first position.
The length of each bar is L

Im guessing the assembly centre of mass is ( L / 4 ) above the pivot point.
( i considered each bar as a point mass at its individual centre of mass, then assumed the combined centre of mass at a point halfway inbetween )

The distance the combined centre of mass falls is ( L / 4 ) * 2
(call this distance h )
So, the potential energy lost = m * g * h

I agree, that height lost is h=L/2, but the potential energy is 2mgh, because of the combined mass. But this is what I have written in the equation of my question. This leads me nowhere. I have 1 equation 2 unknowns..

 

[haruspex]

1. A T-shaped assembly is made of two identical uniform bars of length and mass each.

That sentence seems incomplete. Have you missed out something?
You will certainly need the length of the bars to answer the question.
I would consider the KE of each bar separately. If each has mass m, length L, what are their moments of inertia about O?

 

[myko]

That sentence seems incomplete. Have you missed out something?
You will certainly need the length of the bars to answer the question.
I would consider the KE of each bar separately. If each has mass m, length L, what are their moments of inertia about O?

I added some letters that where missed, sry.
The momento of inertia of horizontal bar is I=md^2/12 and vertival bar I=md^2/3
so combined kinethik energy at position B is:
K=((1/2)md^2/12+(1/2)md^2/3)\omega^2=2mg(d/2)
is this correct?