# Tension in the ring

1 vote
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In this I think the equation should be like this $T=N\cos 45° =N\sin 45° =mg$

And how they have commented on option B .

Is the ring horizontal? If so, by symmetry the tension will be constant. Otherwise it is not obvious that the tension is constant. Perhaps the ring is rigid, so it does not deform like string or rope. If it is a rigid circle, then it must be horizontal, because all sections parallel to the base are circular, whereas all other conic sections are ellipses/parabolas/hyperbolas.
Yes it is horizontal
You are correct that $N\sin45^{\circ}=mg$ but $N\cos45^{\circ}$ is the *radial stress* in the ring. The question is asking for the *hoop stress*.
I think hoop stress would be T/a . and according to option C , $T=mg/2\pi$
@sammygerbil I couldn't get what is not understandable in my answer.can you be more specific
Check edit to my answer.I was busy at that time and just got carried away.
@Green My comment above was addressed to koolman's attempt.

1 vote

If the ring does not extend very much it will remain circular. The only conic section which is circular is one which is sliced parallel to the base of the cone. The ring remains horizontal, with all points at the same height, so by symmetry the tension will the same at all points in it (A).

The radial force $F$ is related to the tension $T$ in the ring by $F=2\pi T$. See my answer to Interaction between an ideal pulley and an ideal rope.

The horizontal component of normal reaction $N$ must equal the radial force $F$, while the vertical component of $N$ must equal the weight of the ring :
$N\cos\alpha=F=2\pi T$
$N\sin\alpha=mg$
$T=\frac{mg}{2\pi}\cot\alpha=\frac{mg}{2\pi}$
because $\alpha=45^{\circ}$.

Thus the tension is independent of radius (B).

The extension $\Delta x$ of the perimeter $x=2\pi R$ of the ring is given by
$\frac{T}{a}=Y\frac{\Delta x}{2\pi R}$
$\Delta x=\frac{2\pi R}{Y}\frac{mg}{2\pi a}=\frac{mgR}{aY}$.
This is (C).

The elastic PE stored in the ring is
$U=\frac12T\Delta x=\frac12 \frac{mg}{2\pi}\frac{mgR}{aY}=\frac{(mg)^2 R}{4\pi aY}$.
So (D) is incorrect.

answered May 3, 2017 by (27,948 points)
selected May 4, 2017 by koolman
Your answer is similar to mine except that you have taken the whole ring where the angle subtended is $2\pi$ whereas I took only a small portion of ring where angle subtended is $2a$
1 vote

Take a small elementary portion of ring .then if the angle subtended by the portion of ring at its centre is 2a then $$2Ta=N_x$$ also $N_x=N_y=\text{mass of the small portion}$ which gives $$2Ta=\frac{mRg2a}{2\pi R}$$ and the result follows.

N is the normal reaction to the ring .we should find it by taking only a small portion of ring and not the whole ring as tension is only defined for a part of ring.

answered May 3, 2017 by (564 points)
edited May 4, 2017 by Green