Parabolic track dynamics + calculus (Hard)

In summary, the equation that describes the motion of a ball rolling on a track is y=x2, where x and y are in meters. A scale along the bottom is used to measure the ball’s horizontal position x. The main forces that act on the ball are Fg=-mgj, n, and the acceleration of the car in any direction. If the ball has a positive x coordinate, then the acceleration of the car is in the -x direction.
  • #1
Ruinmaniac
5
0

Homework Statement


The problem is in the photo below:

https://www.physicsforums.com/attachment.php?attachmentid=71928&stc=1&d=1407288108

Text: Figure CP6.74 shows an accelerometer, a device for measuring the horizontal acceleration of cars and airplanes. A ball is free to roll on a parabolic track described by the equation y = x2, where both x and y are in meters. A scale along the bottom is used to measure the ball’s horizontal position x.

Homework Equations



F = ma
Possibly other kinematics equations.

The Attempt at a Solution



I can imagine how the problem would play out well in my head, and understand how the acceleration has an effect on the ball.
At any "x", slope is y' = 2x
I do not understand how to go from this slope to the angle (theta, see diagram) that represents the rotation of the tangent to the horizontal.

[EDIT] theta = tan-1(2x) (just realized, man I am stupid sometimes lol)

The main forces that I think are acting on the ball are Fg, n, and force due to acceleration of vehicle in any direction, which i think points horizontally, the pic below shows my understanding of the problem:

https://www.physicsforums.com/attachment.php?attachmentid=71929&stc=1&d=1407288854
 

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  • #2
[EDIT] theta = tan-1(2x) (just realized, man I am stupid sometimes lol)
... this relationship is why the trig function is called "tangent" ;)
Good on you for leaving the mistake in though - that helps others.

Note: The acceleration of the car (what car?) does not produce a force.
The ball is "free to roll" - so it will roll back and forth along the track.
The forces on the ball are what produces the acceleration.
You have done problems with ramps before.

Note: $$\sin\left(\arctan x\right)=\frac{x}{\sqrt{1+x^2}}$$
http://en.wikipedia.org/wiki/List_o...mpositions_of_trig_and_inverse_trig_functions
 
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  • #3
Sorry, that's my imagination talking, I imagined the ball and track inside a car and the car is moving to the left in this case. You are right there is no horizontal force acting on the ball but, when the "car" reaches constant acceleration, the ball should be stuck at one point, where the Fnet = 0. So there must be some force keeping the ball in that location right? So i just made a vector to help me out. I no longer have any idea on how to approach this. Sigh.
 
  • #4
Oh I see what it is now.
OK - the situation is for a non-inertial frame so there will be a pseudoforce from the cars acceleration.
The vector sum of the forces should end up as zero for a constant acceleration.
 
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  • #5
Yeah I did that however the unfortunate thing is that this has no answer key so I cannot even check. Regardless, you were extremely helpful, thank you!
 
  • #6
No worries - real life does not have an answer key either... well, apart from actually doing it for real.
You need to get used to what the right answer "feels" like rather than relying on someone telling you.

You will probably be fine neglecting the moment of inertia of the ball - treat it as a block sliding on a frictionless track.
Gravity is Fg=-mgj and the normal force N has to cancel the net force perpendicular to the slope as usual.

Take care: if the ball has a positive x coordinate, then the acceleration of the car is in the -x direction.
So when a<0, x>0, and Fa points in the +x direction too ... it's easy to mess up the signs.

You will need to take sines and cosines of the arctangent ... the link I gave you will give you the shortcuts for that.
I cannot help you further because you have yet to show me the maths ;)
 

Related to Parabolic track dynamics + calculus (Hard)

1. What is a parabolic track?

A parabolic track is a curved path that follows the shape of a parabola. It is commonly seen in physics and engineering, such as in the trajectory of a projectile or the shape of a satellite's orbit.

2. How is calculus used in studying parabolic track dynamics?

Calculus is used to analyze the motion of objects along a parabolic track. It helps in determining the position, velocity, and acceleration of the object at any given point on the track. It also allows for the calculation of the maximum height, range, and time of flight of the object.

3. What are the key equations used in studying parabolic track dynamics?

The key equations used in studying parabolic track dynamics are the equation of motion (s = ut + 1/2at^2), the kinematic equations (v = u + at, v^2 = u^2 + 2as), and the equation for a parabola (y = ax^2 + bx + c). These equations can be derived from the principles of calculus.

4. What factors affect the trajectory of an object along a parabolic track?

The trajectory of an object along a parabolic track is affected by several factors, including the initial velocity, initial angle of launch, air resistance, and gravity. These factors can be manipulated to change the shape and distance of the parabolic track.

5. How is the calculus concept of optimization used in parabolic track dynamics?

The concept of optimization in calculus is used in parabolic track dynamics to find the ideal launch angle and initial velocity for an object to travel the maximum distance or reach the maximum height. This is important in applications such as sports, where athletes need to optimize their performance by understanding the dynamics of a parabolic track.

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