Circular Motion #13

Initially, the string is horizontal. When it is released, it will perform oscillatory motion. When the string makes an angle of $\theta$ with the vertical, we see that the tension in the string is $T = 3 mg\cos \theta$ . (Proof given below).

We can see that tension is maximum and is equal to $3mg$ when $\theta = 0$ , i.e. the tension in the string is maximum when the string is vertical.

If the tension in the string at that point is less than or equal to 12, than the string will never break at any point during the motion. We obtain the following inequality:

$T \leq 12 \implies 3mg \leq 12 \implies \boxed{m \leq 0.4} ~~ _\square$

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Proof: We will draw the free body diagram of the bob when the angle of the string is $\theta$ with the vertical.

Let the speed of the bob at this point be $v$ , and the length of the string be $l$ . Net force towards center is equal to $mv^{2}/l$ .

$T - mg \cos \theta = \frac{mv^2}{ l} \quad \quad \quad(*)$

Using conservation of energy,

$\frac{ 1 } {2} m v^{2} = mgl \cos \theta \implies \frac{mv^2}{ l} = 2mg \cos \theta$

Substituting this in the $(*)$ equation gives

$T = mg\cos \theta + 2mg \cos \theta \implies T = 3 mg \cos \theta ~~~ _\square$