Accelerometer - device on a parabolic track

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  • #1
Shinaolord
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Accelerometer -- device on a parabolic track

EDIT: Picture of problem in its entirety farther down.
A device on a parabolic trac moves on the track described as y=x^2. Both in meters
A$ find an expression for acceleration using position.
B) what is a if x=20cm?

My solution.
First, I took dy/dx(x^2)=2x for velocity. We know the tangent line of a point on velocity is the acceleration. The tangent line is defined as the reciprocal of the function, the slope of the velocity is x/2. At .2m this equals .1 m/s^2 of acceleration.
Am I right?
 
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Answers and Replies

  • #2
Shinaolord
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Please emit slope of velocity is. My apologies
 
  • #3
berkeman
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A device on a parabolic trac moves on the track described as y=x^2. Both in meters
A$ find an expression for acceleration via position.
B) what is a if x=20cm?

My solution.
First, I took dy/dx(x^2)=2x for velocity. We know the tangent line of a point on velocity is the acceleration. The tangent line is defined as the reciprocal of the function, the slope of the velocity is x/2. At .2m this equals .1 m/s^2 of acceleration.
Am I right?

Please emit slope of velocity is. My apologies

Is there more to the problem statement? How does the device move? Is the track vertical, and gravity is causing the device to move down the slope? Or is the track horizontal, and something else is driving the device? Is it supposed to be moving with a constant velocity in some direction?
 
  • #4
Shinaolord
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It just says derive a formula showin (edit) acceleration from position (edit) if I remember correctly. That's the whole problem. I'll confirm if what I've said is right when I have the book with me. It won't let me upload a picture. I can email it though.
 
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  • #6
nasu
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For a given acceleration the ball is in equilibrium in some position. There is no velocity to worry about. For a given acceleration the equilibrium position is determined by the condition that the net force (or total force) is equal to the mass times the acceleration.

The forces acting on the ball are two: weight and normal force. The weight is the same in every point. However the normal force depends on the position.
 
  • #7
Shinaolord
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I don't quite understand, could you going to a little more detail? I understand what you're saying but the mathematical translation of it I don't quite get.
 
  • #8
berkeman
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I don't quite understand, could you going to a little more detail? I understand what you're saying but the mathematical translation of it I don't quite get.

We don't do your homework for you here. Please show some effort and draw the free body diagram for the ball, showing the forces on it in equilibrium.
 
  • #9
Shinaolord
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I didn't want a solution, all you would have had to said is take a closer look at the fbd. I take offense to the way you stated that. I was not looking for a solution.
 
  • #10
nasu
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Do you take offence when a police officer tells you that you broke the l
rules and went over the speed limit? :smile:

Berkeman just pointed one of the forum rules: you have to show some work before you get more help. He (or other of the admins) have said this same words many times, when a new subscriber do not follow the usual rules.

"Taking offense" has been abused too much lately. This is my personal opinion. And I feel like "taking offense" when I see people throwing this left and right.
What if someone "takes offense" for the sloppy (and hard to follow) way you wrote the OP?(no offense):wink:

But maybe is just a problem of language. I am pretty sure nobody intended to offend here.
 
  • #11
berkeman
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I didn't want a solution, all you would have had to said is take a closer look at the fbd. I take offense to the way you stated that. I was not looking for a solution.

No offense meant. You will now be willing to show us your FBD, so we can keep trying to help you?
 
  • #13
nasu
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Why do you need to integrate? This is an equilibrium problem. The ball is at rest in respect with the device, for a given acceleration. It is moving in respect to the ground but this is irrelevant.

You just need to find the acceleration for a given position of the ball relative to that parabolic shape.
Take the y axis vertical and x axis horizontal. And write the equation for each direction as you did.
Use the y=x^2 to find the angle between the tangent to the parabola and the x axis, for a given position.
 
  • #14
Shinaolord
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Okay so with that methodology would just be the correct way to go about this problem?

http://s12.postimg.org/f7oye4wnh/image.jpg

My problem-solving skills are not very good so I apologize for any immaturity
 
  • #15
nasu
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Well, the result looks OK to me.
However the equations you wrote seem to indicate that there is still some confusion about the physics, about what is going on.
I don't know what you mean by these Δx and Δy. Are these displacements along the x and y axes?
And how do you associate accelerations with these displacement?

You don't have a displacement Δy corresponding to acceleration g, for example.
Why don't you stick to the forces and FBD?

Imagine that you have a small object on a frictionless inclined plane of angle θ which is moving with acceleration a. What should be a (as a function of θ) to have the object at rest in respect to the plane?
 
  • #16
Shinaolord
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I wrote it badly, I meant the acceleration in the x and y direction. Does that help clarify?
So you're saying use the forces and sine function?
And yes the physics does confuse me slightly, I'd be able to figure it out IMO if no one had invented math yet :)
 
  • #17
Shinaolord
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Also, I have a question on my method. Why would I take the line perpendicular to the derivative of F[x], what does that represent as a physical quantity?
 
  • #18
nasu
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I wrote it badly, I meant the acceleration in the x and y direction. Does that help clarify?
So you're saying use the forces and sine function?
And yes the physics does confuse me slightly, I'd be able to figure it out IMO if no one had invented math yet :)

There is no acceleration in the y direction.
For the vertical direction you have that the y component of the normal force (N) balances the weight.
Ny-mg=0
For the horizontal direction you have Nx=ma.
These are the equations. You just have to express the components in terms of the angle of the local tangent and eliminate N.
Do you understand so far?
 
  • #19
Shinaolord
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So, I'm looking for the line tangent to what? That confused me a bit...
 
  • #20
Shinaolord
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Would Nx=Nsinx , and N=mgsinx=ma
gsinx=a
sinx=a/g
Sin^-1[a/g]=x ?

This result completely makes me feel the problem is unsolvable.

Or do you mean
Nx=gcosx and Ny=gsinx

so y/x => (gsinx/gcosx) => tanx

Where to go?
 
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  • #21
nasu
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The line tangent to the parabola.

For the second part, what do you call "x"? In the problem x is the horizontal coordinate. It looks like you have mixed up notations.
Before you write sin of some angle, it help if you mention what is that angle. Between what lines?
The components of a force (N) along some direction can be something like Nsinθ or N cosθ where θ is a relevant angle.
But not Nx=gsinθ. The right hand side is not a force,
 
  • #22
Shinaolord
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I was using Nx as the magnitude of the normal force in the x directdion, sorry I should have used subscripts.
The Normal line of the tangent to the parabola is the vector of the normal force correct? If so, I'm using the angle θ from the vertical. Does that help?

PS: I've turned in my assignment, but I would like to continue discussion about this problem so I can properly understand it.
 
  • #23
nasu
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OK, if θ is the angle between the normal and the vertical (which you call y), you will have
Ny=Ncosθ and Nx=Nsinθ.
And Newton's second law for each direction will be like in post 18.
Solve these to find a as a function of θ.
 

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