I was practising some questions until I stumbled upon a question.
A chain of length $l$ and mass $m$ lies on the surface of a smooth sphere of radius $R>l$ with one end tied to the top of the sphere.
(a) Find the gravitational potential energy of the chain with reference level at the center of the sphere.
(b) Suppose the chain is released and slides down the sphere. Find the kinetic energy of the chain, when it has slid through and angle $\theta$.
(c) Find the tangential acceleration $\frac{dv}{dt}$ of the chain when the chain starts sliding down.
(H.C. Verma , Work, Power Energy, Q63)
I am only concerned with part (a) of this question which is trivial to solve. The only thing which I don't understand is why I'm getting different answers when solving through different methods.
Method 1: (Integration)
The problem can be reduced to, $$V=\int_0^{l/r}{\frac{m}{l}gR\cdot R\cos\theta\,d\theta}=\dfrac{mR^2g}{l}\sin\biggl({\dfrac{l}{R}}\biggl)$$ which is stated as the answer in the book.
Method 2: (Center of Mass)
Assuming that the whole mass of the system can be taken on the center of mass, therefore, the center of mass of the chain will subtend an angle $\alpha=\dfrac{l}{2R}$ because $\alpha \propto arc$, i.e $\dfrac{l}{2}$. $$\implies V=mg\cdot R\cos\biggl(\frac{l}{2R}\biggl)$$
To summarize myself, I'm curious as to why the second method doesn't yield the expected answer? Is it wrong to calculate the gravitational potential energy of a body through its center of mass?