Bead Parabola Accelerometer

[AlexChandler]

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.

Homework Equations

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?

 

[tiny-tim]

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")

 

[AlexChandler]

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.

 

[AlexChandler]

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

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

 

[tiny-tim]

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.

 

[AlexChandler]

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. )

 

[AlexChandler]

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. )

Nice! I like this way. Very good!