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 .

1 vote

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 .

1 vote

Best answer

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.

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.

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