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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?

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  • $\begingroup$ Please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead. $\endgroup$ – user191954 Nov 1 '18 at 5:48
  • $\begingroup$ Yes, I also had that in mind initially, but I was mistaken that this question had a figure, so I posted a picture but forgot about it in the process. I will make required edits soon. $\endgroup$ – Utkarsh Verma Nov 1 '18 at 6:11
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You are assuming that the COM lies on the chain. This is not the case. The COM will be closer to the center of the sphere than the chain. The COM of an extended body need not be within the body itself.

Let's look at the height of COM coordinate. $$my_{com}=\int y\ dm=\lambda\int y\ ds=\lambda\int_0^{l/R} R\cos\alpha\cdot R\ d\alpha$$

This is your first integral (without the $mg$ term). Therefore the methods are equivalent.


For a general object of mass $M$ $$Mgy_{com}=\int gy\ dm=\int dU=U$$

So the two integrals are in fact equal in general, assuming $g$ is uniform.

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  • $\begingroup$ OP asked "Is it wrong to calculate the gravitational potential energy of a body through its center of mass?" You replied "If g is not uniform then you just use the center of gravity instead, but the arguments still play out the same." I don't understand. In general potential energy can't be computed using c.o.g. A simple example will prove it. Put a system formed by two equal balls vertically over Earth and compute their potential energy exactly, as the sum of both p.e. You will see that it's different from that of a single ball of twice the mass, located in their c.o.m. $\endgroup$ – Elio Fabri Nov 1 '18 at 10:41
  • $\begingroup$ @ElioFabri I'm not sure I fully understand your example. To me that seems fine. Are you just saying that one statement is incorrect, or my entire answer? $\endgroup$ – Aaron Stevens Nov 1 '18 at 10:58
  • $\begingroup$ Did you keep into account variation of $g$ with height? $\endgroup$ – Elio Fabri Nov 1 '18 at 13:53
  • $\begingroup$ @ElioFabri If you have a varying $g$ then of course you cannot use the COM. I never said you can if $g$ varies. This is the point of my final sentence. $\endgroup$ – Aaron Stevens Nov 1 '18 at 13:54
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    $\begingroup$ @UtkarshVerma Thanks for the clarification. I think Elio became more concerned with the more general statement of use the center of gravity in a non-uniform field. I have removed the statement due to the confusion. $\endgroup$ – Aaron Stevens Nov 2 '18 at 2:05

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