A piece of wire is bent in the shape of a parabola y = kx^2 (y-axis vertical) with a bead of mass m

Q: Wait, why do we use a pseudo force here? Can't we just solve it from the ground frame?

You absolutely can solve it from the ground frame! In the ground frame, the bead is accelerating horizontally with acceleration $a$ (since it's at rest relative to the accelerating wire). The net horizontal force would be $N \sin\theta = ma$, and the net vertical force would be $N \cos\theta = mg$. Dividing these gives $\tan\theta = \frac{a}{g}$. However, using the non-inertial frame and a pseudo force often makes it easier to visualize the 'uphill' push balancing the 'downhill' pull of gravity along the tangent.

Q: Why did we resolve forces along the tangent instead of just using horizontal and vertical axes?

Resolving along the tangent is a classic physics 'cheat code'. The normal force $N$ is always perpendicular to the tangent. By resolving forces *along* the tangent, the component of $N$ is zero ($N \cos 90^\circ = 0$). This instantly removes an unknown variable from our equations, allowing us to directly relate the pseudo force and gravity.

Q: Why does the mass $m$ of the bead cancel out? Does a heavier bead not go as high?

It's a beautiful consequence of the equivalence principle! Both gravity ($mg$) and the pseudo force ($ma$) are directly proportional to the mass of the object. When you equate their tangential components ($ma \cos\theta = mg \sin\theta$), the mass $m$ appears on both sides and cancels out. A 1 kg bead and a 100 kg bead will settle at the exact same height!

Q: I don't get why $\tan\theta$ is equal to $\frac{dy}{dx}$. Can you explain that?

Imagine a tiny right-angled triangle drawn along the tangent line. The vertical side is a tiny change in $y$ ($dy$), and the horizontal side is a tiny change in $x$ ($dx$). By basic trigonometry, the tangent of the angle $\theta$ this line makes with the horizontal is the 'opposite' over 'adjacent', which is exactly $\frac{dy}{dx}$. This is the fundamental geometric definition of a derivative!

Q: What would happen if the wire was a straight line $y = mx$ instead of a parabola?

If the wire were a straight line, its slope $\frac{dy}{dx}$ would be a constant $m$ everywhere. The equilibrium condition $\tan\theta = \frac{a}{g}$ would mean that equilibrium is *only* possible if the wire's slope $m$ exactly equals $\frac{a}{g}$. If $m \neq \frac{a}{g}$, the bead would either slide all the way down or fly all the way off the top; there would be no stable resting point!

Question

A piece of wire is bent in the shape of a parabola $y = kx^2$ ($y$-axis vertical) with a bead of mass $m$ on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the $x$-axis with a constant acceleration $a$. The distance of the new equilibrium position from the $y$-axis of the bead, where the bead can stay at rest with respect to the wire,

Options:

(A) $\frac{a}{gk}$

(B) $\frac{a}{2gk}$

(C) $\frac{2a}{gk}$

(D) $\frac{a}{4gk}$

Sigma Percentile · Accepted Answer

### Riding the Parabola: When Pseudo Force Meets Calculus

Imagine a smooth wire bent into a perfect parabola, described by the mathematical equation $y = kx^2$. A small bead of mass $m$ is threaded onto this wire. When everything is perfectly still, gravity does what it does best—it pulls the bead down to the lowest possible point, the origin $(0,0)$.

But physics problems rarely stay still for long. Suddenly, the entire parabolic wire is given a constant horizontal acceleration, $a$. The bead, possessing inertia, resists this sudden change in motion. To analyze what happens next, we need to shift our perspective.

### The Frame Shift: Enter the Pseudo Force

Analyzing the bead's motion from the ground (an inertial frame) is complicated because the wire itself is moving. Instead, let's "jump onto the wire" and observe the bead from this accelerating, non-inertial frame of reference.

Whenever we step into an accelerating frame, we must pay a toll: the **pseudo force**. This fictitious force acts on every object in the frame, pointing in the exact opposite direction of the frame's acceleration. If the wire accelerates to the left with $a$, the bead experiences a pseudo force $ma$ pushing it to the right.

Now, let's draw the Free Body Diagram for the bead in this new frame. We have three forces at play:
1.  **Gravity ($mg$)**: Pulling straight down.
2.  **Pseudo Force ($ma$)**: Pushing horizontally.
3.  **Normal Force ($N$)**: The wire pushing back on the bead, acting strictly perpendicular to the curve of the parabola.

### The Geometry of Equilibrium

Pushed by the pseudo force, the bead slides up the side of the parabola until it reaches a new equilibrium position where it stays at rest relative to the wire. Let's call the coordinates of this new position $(x, y)$.

At this point, if we draw a tangent line to the parabola, let's say it makes an angle $	heta$ with the horizontal. Because the bead is constrained to move *only* along the wire, the most intelligent way to analyze the forces is to resolve them along this tangent line. Why? Because the normal force $N$ is perpendicular to the tangent, meaning it has absolutely zero component along the wire. This clever choice of axes eliminates an unknown variable immediately!

Let's resolve the remaining forces along the tangent:
*   The pseudo force $ma$ is horizontal. Its component pushing the bead *up* along the tangent is $ma \cos	heta$.
*   Gravity $mg$ is vertical. Its component pulling the bead *down* along the tangent is $mg \sin	heta$.

For the bead to remain perfectly at rest in the wire's frame, these two opposing tangential forces must perfectly balance each other. This gives us our master equilibrium equation:

$$ma \cos	heta = mg \sin	heta$$

By rearranging this equation—dividing both sides by $mg \cos	heta$—we notice something beautiful. The mass $m$ cancels out entirely! The equilibrium position doesn't depend on how heavy the bead is. We are left with a simple trigonometric ratio:

$$	an	heta = \frac{a}{g}$$

### The Calculus Bridge

We have found $	an	heta$ using pure Newtonian mechanics. But what is $	an	heta$ geometrically? It is the slope of the tangent line to the curve at the point $(x, y)$.

This is where we cross the bridge from physics into calculus. The slope of any curve $y = f(x)$ at a given point is given by its first derivative, $\frac{dy}{dx}$. For our parabola $y = kx^2$, taking the derivative with respect to $x$ is straightforward:

$$\frac{dy}{dx} = \frac{d}{dx}(kx^2) = 2kx$$

So, calculus tells us that the slope of the tangent line is $2kx$.

### The Grand Finale

We now have two distinct expressions for the exact same physical quantity—the slope of the tangent line at the equilibrium position. Physics dictates that the slope must be $\frac{a}{g}$ to balance the forces. Calculus dictates that the slope must be $2kx$ based on the geometry of the parabola.

By equating these two expressions, we unite the physical reality with the mathematical framework:

$$2kx = \frac{a}{g}$$

Our goal is to find the horizontal distance from the y-axis, which is simply the coordinate $x$. By isolating $x$, we arrive at our final, elegant solution:

$$x = \frac{a}{2gk}$$

This result tells us exactly where the bead will settle. The harder you accelerate the wire (larger $a$), the further up the bead rides. The steeper the parabola (larger $k$) or the stronger the gravity (larger $g$), the closer the bead stays to the center.

Select Answer:

Visualized Solution

The Parabolic Wire

The Non-Inertial Frame

Free Body Diagram

The Tangent Line

Resolving Forces

Tangential Components

Equilibrium Condition

Finding $tan θ$

The Calculus Connection

Differentiating the Parabola

Equating the Slopes

Final Position

The Sigma Insight: Pseudo Force

Riding the Parabola

The Frame Shift

The Geometry of Equilibrium

The Calculus Bridge

The Grand Finale

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Select Answer:

Visualized Solution

The Parabolic Wire

The Non-Inertial Frame

Free Body Diagram

The Tangent Line

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Equilibrium Condition

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