Tension in A Rotating Loop?

Here in the figure, we have a loop and we took an elemental part magnified its view as the above picture.

You can see that Tension $T$ is acting on the arc, and it is making an angle $\theta$ with the center.

So breaking up the rectangular components of the tension, we have

$T \cos \theta = T \cos \theta$ which cancel out each other.

We are left with

$2T \sin \theta$ which will balance the centrepetal force $\dfrac{mv^2}{r}$ .

$\implies 2T \sin \theta = \dfrac{mv^2}{r}...(i)$

where $m$ is the mass of that small portion of the loop.

So, $m= \dfrac{M}{2 \pi r} \times dl \implies \dfrac{M.dl}{2 \pi r}$

Here $dl$ is the arc length of the small portion.

We know $dl=R 2\theta$

Putting $dl$ , We get

$m= \dfrac{M.r. 2\theta}{2 \pi r} \implies \dfrac{M \theta}{\pi}$

Putting values of m in (i)

We get

$2T \sin \theta = \dfrac{mv^2}{r} \implies 2T \sin \theta = \dfrac{M .\theta . v^2}{r. \pi}$

As $\theta$ is small, Using small angle approximation we have

$\sin \theta \approx \theta$ [Actually not exactly equal]

We have

$\implies 2T .\theta = \dfrac{M .\theta . v^2}{r. \pi}$

$\implies 2T = \dfrac{Mv^2}{r. \pi}$

$\implies T = \dfrac{Mv^2}{2r. \pi}$

So we have our Tension as $T = \dfrac{Mv^2}{2r. \pi}$

Putting the values of the question, we get $\boxed{9 N}$