Animated Solution for Physics - Laws of Motion: A bead of mass $m$ stays at point $P(a,b)$ on a wire bent in the shape of a parabola $y=4C{x}^2$ and rotating with angular speed $\omega$ (see figure). The value of $\omega$ is (neglect friction)

A Shift in Perspective

Imagine you are the tiny bead, resting on a smooth, frictionless parabolic wire. Suddenly, the entire wire begins to spin around its vertical axis. What do you experience?

From the perspective of an observer standing on the ground, you are moving in a circle. But let's step into the rotating frame of reference—a perspective where the wire appears stationary. In this frame, you feel a mysterious outward push. This is the centrifugal force, a pseudo force that arises because your frame of reference is accelerating.

To stay perfectly balanced at point $P(a,b)$, the forces acting on you must perfectly cancel out. Let's identify these forces. First, there is the relentless pull of gravity, $mg$, acting downwards. Second, there is the outward centrifugal force, $ma{\omega }^2$, pushing you horizontally away from the axis of rotation. Finally, there is the normal force, $N$, exerted by the wire, pushing perpendicular to the surface.

For the bead to remain stationary in this rotating frame, the net force must be zero. We can achieve this by resolving the normal force into its vertical and horizontal components.

The vertical component of the normal force must balance gravity to prevent the bead from sliding down. Therefore, we write:

The horizontal component of the normal force must balance the outward centrifugal push to prevent the bead from flying off. Thus, we have:

Here, $\theta$ is the angle the tangent to the parabola makes with the horizontal. By dividing these two equations, we elegantly eliminate the unknown normal force $N$:

This equation is a profound statement: it tells us that the required slope of the wire at any point is directly proportional to the centrifugal force and inversely proportional to gravity.

Now, we must connect this physical requirement to the actual shape of the wire. The wire is bent into a parabola described by the equation:

In calculus, the slope of a curve at any point is given by its derivative, $\frac{dy}{dx}$. Geometrically, this derivative is exactly equal to $\tan \theta$. Let's differentiate our parabolic equation with respect to $x$:

At our specific point $P(a,b)$, the $x$-coordinate is $a$. Substituting this into our derivative, we find the slope at point $P$:

We now have two distinct expressions for $\tan \theta$: one derived from the physics of balancing forces, and the other derived from the pure geometry of the parabola. Let's equate them:

Look closely at this equation. The variable $a$, representing the horizontal position of the bead, appears on both sides. As long as $a eq 0$, we can cancel it out!

This is a remarkable result. It implies that the required angular speed $\omega$ is completely independent of the bead's position on the parabola. Whether the bead is close to the bottom or high up on the curve, it requires the exact same angular speed to stay in place. This is a unique property of the parabolic shape!

Finally, we solve for the angular speed $\omega$:

And there we have it. The exact angular speed required to keep the bead dancing in perfect equilibrium on the rotating parabola.