Looping

For bob 1 (B) to complete a full circle, at its highest point, the centripetal force on it must be balanced by the gravitational force on it. That is $\dfrac {mv_1^2}{l_1} = mg$ and its velocity $v_1 = \sqrt{gl_1}$ . Similarly, the velocity of bob 2 (D) at its highest point is $v_2 = \sqrt {gl_2}$ .

At point D, just before the collision bob 1 is moving at $v_1$ , while bob 2 is at rest. After the elastic collision, bob 1 is at rest and bob 2 of the same mass is moving at $v_1$ .

Considering the conservation of energy of bob 2 by equating the kinetic energy loss being converted to potential energy, we have:

$\begin{aligned} \frac 12mv_1^2 - \frac 12mv_2^2 & = 2mgl_2 \\ \frac 12mv_1^2 & = 2mgl_2 + \frac 12mv_2^2 \\ \frac 12 mgl_1 & = 2 mg l_2 + \frac 12 m g l_2 \\ \frac 12 l_1 & = \frac 52 l_2 \\ \implies \frac {l_1}{l_2} & = \boxed{5} \end{aligned}$

How fast must the bob initially be going in order to make it to the top? The most obvious first thought is to suppose that the initial kinetic energy must equal the change in gravitational potential energy that occurs when the bob moves upward by $2 l_1$ .

$\frac{1}{2} m v_0^{2} = mg(2l_1) \\ v_0^2 = 4gl_1$

This gives us an absolute lower bound for the initial speed. However, there is a more subtle condition that derives from the fact that a rope can pull on an object but cannot push on an object. As the bob moves on its circular trajectory, the rope pulls it toward the center of the circle with variable tension $T$ . $T$ , thus defined, must never be negative, because that would correspond to a push.

The other force acting in the radial direction (besides the string tension) is the component of the gravitational force in the (inward) radial direction. This component can be expressed as: $-mg cos\theta$

$\theta$ is the angle between the string and the vertical, and $(\theta = 0)$ corresponds to the initial position. The force / acceleration relationship in the radial direction is thus:

$T - mg cos\theta = m\frac{v^2}{l_1}$

We need to apply conservation of energy in order to determine an expression for $v$ in terms of $v_0$ and $\theta$ .

$\frac{1}{2}m v_0^2 = \frac{1}{2}mv^2 + mg l_1(1-cos\theta) \\ v^2 = v_0^2 - 2g l_1(1-cos\theta)$

Plugging into the previous equation gives: $T - mg cos\theta = \frac{m}{l_1}(v_0^2 - 2g l_1(1-cos\theta)) \\ T - mg cos\theta = \frac{m}{l_1}v_0^2 - 2mg + 2mg cos\theta \\ T = \frac{m}{l_1}v_0^2 - 2mg + 3mg cos\theta$

For $T$ to be positive for all $\theta$ , we thus have the condition:

$\frac{m}{l_1}v_0^2 \geq 5mg \\ v_0^2 \geq 5gl_1$

Thus, the "positive tension" requirement is more demanding than the "conservation of energy" requirement. If the bob begins with initial velocity $(v_0^2 = 5gl_1)$ , the $v^2$ value at the top will be $(v_f^2 = 5gl_1 - 4gl_1 = gl_1)$ .

Since the collision with the second bob is perfectly elastic, the second bob begins with speed $(v_{0_2}^2 = gl_1)$ . Then we apply the tension condition again to get:

$gl_1 = 5gl_2 \\ \implies \large{\boxed{\frac{l_1}{l_2} = 5}}$