So Much Tension!

We analyse this problem with respect to the car(moving along, In the reference frame of car, the net force(excluding tension) is $m \sqrt{a^2+g^2}$ . Clearly vertical will make an angle $\theta_{0} = \tan^{-1} \frac{a}{g}$ with the equilibrium position where tension is opposite to then net force excluding tension .

At an instant when angle between string and the position of string in equilibrium is $\phi$ , centripetal force is provided by tension and a component of $m \sqrt{a^2+g^2}$ ,hence,

$T - m\sqrt{a^2+g^2} \cos \phi = \frac{mv^2}{r} \Rightarrow T = m\sqrt{a^2+g^2} \cos \phi + \frac{mv^2}{r}$ ,

From above expression , it is clear that tension is maximum at equilibrium and minimum at initial position.

Say the velocity at equilibrium is $v$ . Apply conservation of energy in this frame to get:

$\frac{1}{2} m v^2 = m\sqrt{g^2 + a^2} l (1 - \cos \theta_{0})$

Hence, $\frac{mv^2}{l} = 2m\sqrt{a^2+g^2} (1 -\cos \theta_{0})$

Hence, $T_{min} = m \sqrt{ a^2+g^2} \cos \theta_{0}$ , (since initial velocity is 0)

also, $T_{max} = \frac{mv^2}{l} + m \sqrt{a^2+g^2} = m\sqrt{a^2+g^2} (3 - 2\cos \theta_{0})$

Clearly, $r = \frac{T_{max}}{T_{min}} = \boxed{3 \sec \theta_{0} - 2}$

In our case, $\theta_{0} = \tan^{-1} \frac{a}{g} = {30}^{\circ}$ , Hence, $r = \boxed{1.464}$

First of all, it is pretty obvious what the minimum tension in the string is going to be..it is Mg and it occurs at t=0 .

Let A and B represent the maximum and minimum tensions respectively.

Hence, we have,

$\boxed{\Rightarrow B=Mg}$

Now, Let us think about when will there be a maximum tension in the string.

Clearly, it occurs when the velocity of the pendulum is maximum .

(1)We use Work Energy Theorem for this and let $\theta$ , $\ell$ , M and $\upsilon$ represent the angle made by the string with the vertical at any instant of time, the length of the string, the mass of the bob of the pendulum and the maximum velocity of the bob respectively.

(2)Also, observe that when there is a maximum velocity, the tangential components of the pseudo force and the gravitational force must cancel out, because at that moment, the tangential acceleration must be zero , otherwise it wouldn't be called a maximum.

From the application of the above concepts, we have the 2 equations:

$\boxed{\Rightarrow Mg\sin\theta = \frac {Mg}{\sqrt{3}}\cos\theta}$

and, Work done by pseudo force + Work done by the gravity = $\Delta$ Kinetic Energy

$\boxed{\Rightarrow \frac {Mg}{\sqrt{3}}\ell\sin\theta - Mg\ell\left(1-\cos\theta\right) = \frac {1}{2}M\upsilon^2}$

From the first equation, we have $\tan\theta = \frac {1}{\sqrt{3}}$

Applying this result in our second equation, we get,

$\upsilon^2 = 2g\ell\left(\frac {2-\sqrt{3}}{\sqrt{3}}\right)$

Now, analyzing the Free Body Diagram of the bob, we get that,

A = $Mg\cos\theta + \frac {Mg}{\sqrt{3}}\sin\theta + \frac {M\upsilon^2}{\ell}$

Substituting the value of $\theta$ and $\upsilon^2$ , we get,

$\boxed{\Rightarrow A = 2Mg\left(\sqrt{3}-1\right)}$

In the Problem, we are required to evaluate $\frac {A}{B}$

Hence, $\frac {A}{B} = r = 2\left(\sqrt{3}-1\right) \approx 1.464$

$\Rightarrow 100r \approx 146.4$

$\Rightarrow \lfloor 100r \rfloor = \boxed{146}$ , is the final solution.