## Homework Statement

An accelerometer is made of a piece of wire with a bead on it that can slide on the wire with no friction. The wire is formed as a parabola y = kx2, as shown in the drawing. The bead rests at the lowest point of the parabola when it is at rest. When accelerated parallel to the x-axis the bead will climb up some on the wire. Find the relationship between the acceleration a of the wire and the bead’s maximum horizontal displacement x relative to the wire.

F=ma

## The Attempt at a Solution

Suppose the wire is accelerating and the bead is at rest at some point P. Then in the frame of the bead, we have a gravitational force downward, a fictitious force to the left, and a normal force perpendicular to the wire.
Consider the tangent line to the wire at the point where the bead is resting. The slope of this is given by

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

Then the angle between a line parallel to the point where the bead is at rest, and the horizontal satisfies:

$$tan \theta = 2kx$$

Look at the components of the forces parallel to this line, and they must be equal. Then we have

$$ma cos \theta = mg sin \theta$$

then dividing by sin theta and plugging in tan theta = 2kx we can get

$$x = - \frac{a}{2kg}$$

However the answer should be only

$$x = - \frac{a}{kg}$$

Do you see where I have gone wrong?

Last edited:

tiny-tim
Homework Helper
Hi Alex! … Look at the components of the forces parallel to this line, and they must be equal …
No, that's for the equilibrium point, where the bead could remain at rest …

you need the point where it's temporarily at rest! (one way would be to use the potential energy of the "fictitious gravity")

Hi Alex! No, that's for the equilibrium point, where the bead could remain at rest …

you need the point where it's temporarily at rest! (one way would be to use the potential energy of the "fictitious gravity")
Hmm... I see. I was unaware that you could associate a potential energy with a fictitious force. Would I find that in this way?

$$U = \int F dx = m \int \frac{d}{dt} \frac{dx}{dt} dx = m \int d \left( \frac{dx}{dt} ^2 \right) = m ( \dot{x} )^2$$

Also, I don't quite see how the bead would be temporarily at rest anywhere other than the equilibrium position. Of course this is just a comment, and not a question. But if you could provide more insight into this I would be much obliged.

Actually, I've made a mistake, it should be

$$U = - m( \dot{x} )^2$$

tiny-tim
Homework Helper
Hi Alex! (just got up :zzz: …)

Sorry, but your equation is rubbish. Are you trying to say dv/dt = dv/dx dx/dt = v dv/dx = 1/2 d(v2) ?

Even so, all you're doing is proving the work-energy theorem.

To get the "fictitious potential", find the magnitude and direction of the total acceleration (including fictitious), and use that instead of g.
Also, I don't quite see how the bead would be temporarily at rest anywhere other than the equilibrium position. Of course this is just a comment, and not a question. But if you could provide more insight into this I would be much obliged.
If the dish is stationary, the equilibrium position will be at the bottom, but the bead will be temporarily at rest at positions either side. Hi Alex! Sorry, but your equation is rubbish. Are you trying to say dv/dt = dv/dx dx/dt = v dv/dx = 1/2 d(v2) ?

Even so, all you're doing is proving the work-energy theorem.
Ah yes you are right. That was quite wrong The a is taken to be constant so I suppose it should have just been

$$\int ma dx = max$$

So how about this. I consider the work done by each of the forces in the frame of reference of the wire

$$W = - mgy + max = \frac{1}{2} m v^2$$

then we find

$$v^2 = 2x (-gkx+a)$$

and for v to be zero we have either x=0 or

$$x = \frac{a}{kg}$$

where x is really just the magnitude of the horizontal distance traveled.

tiny-tim
Yes, that's fine. (Another way of doing it would be to say that the "fictitious gravity" is g down and a horizontally, so lines of equal potential would be parallel to gy = ax, and so the point at the same potential as the origin is gkx2 = ax, or x = a/kg. )
Yes, that's fine. (Another way of doing it would be to say that the "fictitious gravity" is g down and a horizontally, so lines of equal potential would be parallel to gy = ax, and so the point at the same potential as the origin is gkx2 = ax, or x = a/kg. )