A piece of wire is bent in the shape of a parabola `y = kx^2` (y axis vertical) with a bead of mass mon 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 paralle to the x-axis with a constant acceleration a. The distance of the the new equilibrium position of the bead. Find the x-coordinate where the bead can stay at rest with respect to the wire ?

Answer

To solve the problem, we need to analyze the forces acting on the bead when the wire is accelerated. Let's break it down step by step.

### Step 1: Understand the Setup
The wire is shaped like a parabola given by the equation \( y = kx^2 \). The bead can slide along this wire without friction. Initially, when the wire is at rest, the bead is at the lowest point, which is at the origin (0,0).

### Step 2: Analyze the Forces Acting on the Bead
When the wire is accelerated with a constant acceleration \( a \) in the x-direction, the bead experiences two forces:
1. The gravitational force acting downwards: \( \vec{F_g} = m \vec{g} \)
2. A pseudo force acting in the opposite direction of the acceleration due to the wire's motion: \( \vec{F_p} = -m \vec{a} \)

### Step 3: Break Down the Forces
At the new equilibrium position, the normal force \( \vec{N} \) acting on the bead can be resolved into two components:
- \( N \cos(\theta) \) balancing the gravitational force \( mg \)
- \( N \sin(\theta) \) balancing the pseudo force \( ma \)

### Step 4: Set Up the Equations
From the balance of forces, we have:
1. \( N \cos(\theta) = mg \) (1)
2. \( N \sin(\theta) = ma \) (2)

### Step 5: Divide the Equations
Dividing equation (2) by equation (1):
\[ \frac{N \sin(\theta)}{N \cos(\theta)} = \frac{ma}{mg} \]
This simplifies to:
\[ \tan(\theta) = \frac{a}{g} \]

### Step 6: Relate the Angle to the Parabola
The slope of the parabola at any point \( x \) is given by the derivative of \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d(kx^2)}{dx} = 2kx \]
Thus, at the new equilibrium position, we have:
\[ \tan(\theta) = 2kx \]

### Step 7: Equate the Two Expressions for \( \tan(\theta) \)
From the previous steps, we have:
\[ 2kx = \frac{a}{g} \]
Solving for \( x \):
\[ x = \frac{a}{2kg} \]

### Conclusion
The x-coordinate where the bead can stay at rest with respect to the wire when it is accelerated is:
\[ x = \frac{a}{2kg} \]