A classical mechanics problem by Miraj Shah

After the instant, when tension in the string becomes zero, the body does not perform circular motion but instead follows a parabolic path. Let the velocity of the body at the instant tension in the string becomes zero, be $u$ as shown in the figure. Let the time taken from this instant to reach the point of suspension be $t$ .

Therefore,

$\large \boxed{t = \dfrac{2u}{g\cos\theta}}$ .... $(1)$

Also, at the exact instant when the tension in the string becomes zero, we can write the equation of motion as follows:

Net Force towards Centre = $\large \dfrac{mu^2}{r}$

$\Rightarrow mg\sin\theta = \large \dfrac{mu^2}{r}$

$\Rightarrow \boxed{u^2 = gr\sin\theta}$ .... $(2)$

Now, we also know that the horizontal component of $u$ is $u_x=u\sin\theta$

Therefore,

$(u_x)(t)$ = Net displacement in the horizontal direction.

$(u\sin\theta)(t) = r\cos\theta$

$\Rightarrow 2u^2\sin\theta = gr\cos^2\theta$ ....(from equation $(1)$ )

$\Rightarrow 2gr\sin^2\theta = gr\cos^2\theta$ ....(from equation $(2)$ )

$\Rightarrow \tan\theta = \dfrac{1}{\sqrt{2}} \Rightarrow \sin\theta = \dfrac{1}{\sqrt{3}} \Rightarrow \boxed{\large u^2 = \dfrac{gr}{\sqrt{3}}}$

Now that we know the value of $u^2$ we can use the Law of Conservation of Energy to find $v^2$

On calculating we get:

$v^2 = gr(2+\sqrt 3)$

Therefore, $\boxed{ab = 6}$

let us consider the string to become slackened at an angular position \(\theta) measured from horizontal in anticlockwise direction.

at this point, tension in string is zero.

thus \(mgsin\theta=\frac{mv^{'2}}{r}\) where $v^{'}$ is the velocity at point of slackening.

thus $v^{'}=\sqrt{grsin\theta}$

from this point onwards, the path of ball is parabolic with general equation being,

$y = xcot\theta- \frac{gx^{2}}{2v^{'2} sin^{2} \theta}$

plugging in the values in this equation, $-2sin^{4}\theta=2cos^{2}\theta sin^{2}\theta-cos^{2}\theta$

convert everything in terms of sine and obtain $sin\theta=\frac{1}{\sqrt{3}}$

now by energy conservation the relation between $v and v^{'}$ is found to be $v^{2}=2gr+\sqrt{3}gr$

therefore a is 2 and b is 3.

It's an hc verma example question