# T shaped assembly swings down

1. Aug 16, 2014

### 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

Last edited: Aug 16, 2014
2. Aug 16, 2014

### myko

picture

The corresponding image for the situation

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3. Aug 16, 2014

### 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

4. Aug 16, 2014

### myko

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 writen in the equation of my question. This leads me nowhere. I have 1 equation 2 unknowns..

5. Aug 16, 2014

### haruspex

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?

6. Aug 16, 2014

### myko

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?

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