Animated Solution for Physics - Work, Energy, and Power: A particle is suspended vertically from a point $O$ by an inextensible massless string of length $L$. A vertical line $AB$ is at a distance $L/8$ from $O$ as shown in figure. The object is given a horizontal velocity $u$. At some point, its motion ceases to be circular and eventually the object passes through the line $AB$. At the instant of crossing $AB$, its velocity is horizontal. Find $u$.

This problem is a masterpiece of classical mechanics, seamlessly blending two fundamental concepts: vertical circular motion and projectile motion. When a particle is given an initial velocity at the bottom of a suspended string, its fate depends entirely on that initial kinetic energy. If the energy is too low, it oscillates. If it's very high, it completes the circle. But in this specific scenario, the energy is just enough to carry it above the horizontal, but not enough to reach the top.

This causes the string to slack, and the particle transitions from a constrained circular path into a free-falling parabolic trajectory. Let's break down this beautiful transition step by step.

The Circular Journey

Let the particle start at the lowest point $P$ with an initial velocity $u$. It travels in a circular arc of radius $L$. As it climbs, it loses kinetic energy and gains gravitational potential energy. At some point $Q$, located at an angle $\theta$ above the horizontal, the tension in the string drops exactly to zero.

At this critical instant, the string slacks. The only force providing the necessary centripetal acceleration is the component of gravity acting along the radial direction. We can write the dynamic equation as:

This simplifies to a direct relationship between the velocity at $Q$ and the angle $\theta$:

Simultaneously, we must respect the Conservation of Mechanical Energy. The particle has climbed a total vertical height of $h=L+L\sin \theta$ from its starting point. Applying the work-energy theorem between $P$ and $Q$:

Rearranging this gives us our master energy equation:

The Parabolic Arc

Once the string slacks at $Q$, the particle is no longer constrained by the string. It becomes a projectile launched with velocity $v$. Since the string was at an angle $\theta$ to the horizontal, the velocity vector (which is tangential to the circle) is launched at an angle of $90^{\circ }-\theta$ to the horizontal.

The problem provides a crucial geometric constraint: the particle crosses the vertical line $AB$ (located at $x=L/8$) with a horizontal velocity. In projectile motion, the velocity is purely horizontal only at the highest point (the vertex) of the parabolic trajectory.

This means the horizontal distance from the launch point $Q$ to the line $AB$ is exactly half the range of the projectile.

Let's calculate this geometric distance. The x-coordinate of $Q$ is $L\cos \theta$, and the x-coordinate of the line $AB$ is $L/8$. Therefore, the horizontal distance traveled is:

Now, we use the standard formula for the range of a projectile. Since the launch angle is $90^{\circ }-\theta$, the range is $\frac{v^2\sin (2(90^{\circ }-\theta ))}{g}=\frac{v^2\sin (180^{\circ }-2\theta )}{g}=\frac{v^2\sin (2\theta )}{g}$. Equating our geometric distance to half the range:

We now have a system of equations. We can elegantly eliminate $v^2$ by substituting $v^2=gL\sin \theta$ from our centripetal equation into the range equation:

Canceling $g$ and $L$, and simplifying the right side:

Using the identity ${\sin }^2\theta =1-{\cos }^2\theta$, we get:

The $\cos \theta$ terms cancel out beautifully, leaving:

Taking the cube root yields $\cos \theta =\frac{1}{2}$, which means the string slacks exactly at $\theta =60^{\circ }$ above the horizontal!

With $\theta =60^{\circ }$ unlocked, the rest of the problem collapses easily. First, we find the velocity squared at $Q$:

Finally, we substitute $v^2$ and $\theta$ back into our master energy equation to find the initial velocity $u$:

Taking the square root gives us our final, elegant answer: