Animated Solution for Physics - Laws of Motion: A particle of mass $m$ is suspended from a ceiling through a string of length $L$. The particle moves in a horizontal circle of radius $r$ such that $r=\frac{L}{\sqrt{2}}$. The speed of particle will be

Imagine you are holding a string with a small stone tied to its end, and you start whirling it around so that the stone moves in a perfect horizontal circle while your hand stays still. The string traces out a cone in the air. This beautiful and classic physics setup is known as a Conical Pendulum.

In this problem, we are given a particle of mass $m$ suspended from a ceiling by a string of length $L$. It is moving in a horizontal circle of radius $r$, and we are given a special geometric constraint: $r=\frac{L}{\sqrt{2}}$. Our mission is to find the speed $v$ of this particle.

To solve any mechanics problem, our first and most powerful tool is the Free Body Diagram (FBD). Let's isolate the particle and see what forces are acting on it.

1. Gravity: The Earth is pulling the particle straight down with a force equal to its weight, $mg$. 2. Tension: The string is pulling the particle up and towards the center of suspension with a force $T$.

Since the particle is moving in a horizontal circle, it is not moving up or down. This means the forces in the vertical direction must perfectly balance each other. But the tension $T$ is acting at an angle $\theta$ with the vertical.

Let's resolve this tension into two mutually perpendicular components: - A vertical component pointing upwards: $T\cos \theta$ - A horizontal component pointing towards the center of the circle: $T\sin \theta$

Now, let's apply Newton's Laws of Motion.

Vertical Equilibrium: Because there is no vertical acceleration, the upward force must equal the downward force.

Horizontal Dynamics: For any object to move in a circle, it requires a net force directed towards the center, known as the centripetal force. In our setup, the only force pointing towards the center is the horizontal component of the tension. Therefore, this component provides the necessary centripetal force.

We have two equations, but we don't know the tension $T$. The most elegant way to eliminate $T$ is to divide equation (ii) by equation (i):

Notice how beautifully the tension $T$ and the mass $m$ cancel out! This tells us that the angle and speed do not depend on how heavy the particle is.

Rearranging this to solve for $v$, we get our master formula for the speed of a conical pendulum:

We are almost there, but we still need the value of $\theta$. This is where the geometric constraint given in the problem comes into play. Look at the right-angled triangle formed by the string $L$, the radius $r$, and the vertical axis.

From basic trigonometry, we know that:

The problem states that $r=\frac{L}{\sqrt{2}}$. Let's substitute this into our sine ratio:

What angle has a sine of $\frac{1}{\sqrt{2}}$? Exactly, $\theta =45^{\circ }$!

Now, we just plug this angle back into our master formula for speed:

Since $\tan 45^{\circ }=1$, the equation simplifies beautifully to:

$$v=\sqrt{rg}$$

And there we have it! The speed of the particle is simply $\sqrt{rg}$. This problem is a fantastic demonstration of how combining free body diagrams, Newton's laws, and a little bit of geometry leads us straight to the solution.