Why we can't assume centre of mass of chain falling through incline? In a mechanics problem involving finding final velocity of falling chain when half of chain has fallen from incline I am getting two results by applying two different methods , one involving taking centre of mass of whole chain and other by applying centre of mass of two parts of chain (fallen and incline) part.


The  chain  of  length  L  is  released  from  rest  on  a  smooth  fixed  incline  with  $x  =  0$  as  shown  in  the figure.  Determine  velocity  v  of  the  chain  when  a  half  of  the  length  has  fallen  (Take $ L  = 425$ m)  in
$ms^{-1}$. (Neglect edge effect of inclined). 


(When chain has fallen through half length)
In the first method $$ \Delta x_{com} = \frac{L}{2} \sin 30^o-0 \implies \frac{1}{2}mv^2=mg\Delta x_{com}$$
$$ \implies \frac{1}{2}mv^2=mg\frac{L}{2} \sin 30^o$$
In the second method
$$\frac{1}{2}mv^2-m_2g\frac{x}{2}+m_1g\frac{L-x}{2}=mg\frac{L}{2} \sin 30^o$$
$$\frac{1}{2}mv^2-\frac{1}{2}mg\frac{L}{4}+\frac{1}{2}mg\frac{L}{4}=mg\frac{L}{2} \sin 30^o$$
From first and second method $v$ comes out to be $\frac{2}{\sqrt5}$ and $1$.
So my question is that why can't we simply take centre of mass of whole chain rather than breaking the chain into two parts since the centre of mass is the point where whole body mass is supposed to be concentrated .
 A: Most approaches, including that of the OP, will be wrong because they start from the idea that a chain element going 'over the edge' will change direction instantaneously, in a time interval $\Delta t=0$. But that requires an acceleration:
$$\mathbf{a}=\frac{\text{d}\mathbf{v}}{\text{d}t}=\infty$$
and of course a causal force to be infinitely high too. So that's not possible in the real world.
So we can deduce that the chain will bulge to the left, instead of 'adhering' to the vertical part of the incline.
One way to solve the problem without the erroneous assumption is to determine the Equation of Motion of the chain.
The free body diagram of a chain element $\text{d}m$ 'going over the edge' is:

From it, the acceleration components $\mathbf{a_x}$ and $\mathbf{a_y}$ could be determined. However, I have difficulty working out the angle of $T_2$ to the horizontal.
A: In your second sketch and first calculation, you are assuming that the center of mass has moved to the edge of the incline. Actually, once the chain bends, the center of mass is no longer at the center of the chain. (It has moved to the right, and dropped to a vertical position (L/16) below the edge.  The loss of potential energy is (5/16)mgL.)  In your second calculation, the term relating to the lower half of the chain going down the incline should be negative and include a sin($30^o$).  Either approach (done correctly) gives $v^2$ = (5/8)gL.
A: First of all there is a bit of a problem with the problem. A chain slides down an incline. It's velocity has vertical and horizontal components. It slides off the edge, and is shown moving straight down. What happened to the horizontal component? If you are going to use conservation of energy, you need to include it.
The chain starts with the bottom just at the edge. Immediately falls off after very little acceleration. The bottom link travels almost straight down. Links above it have accelerated longer and have acquired more horizontal velocity. There are also tension forces between links. So it gets complicated to say what the force is on any given link, and what direction it is traveling. Whoever wrote the problem probably just missed that point.
The nice thing about conservation of energy is that you can say things about the motion without having to know all the details. At any given time, the sum of the kinetic energies of each link will total the change in potential energy from the start. To calculate the total kinetic energy of the chain, you do need to know the height of each link. The problem is apparently supposed to be one where the heights are easy to see and do calculations. It would certainly be harder figure out the heights if the chain was in a curve. So we will go with the intent of the simplified problem. Each link travels straight down after falling off the edge.
You can use the height of the center of mass to calculate the potential energy. If there are N links of mass $m$ and the height of each is $h_i$, then
$$H = \frac{{\sum{mh_i}}}{{\sum{m}}}$$
You can calculate this by adding up each link.
Or you can get the same answer by adding up the links in two parts of the chain to get the mass and height of the center of mass of each part. Then add up the two parts to get the the height of the center of mass of the entire chain. I leave it to you to prove this, or perhaps find the proof in your textbook.
In this case, the second approach is an easy way to do this problem.  It is easy to calculate the height of the center of mass of a straight chain.
