Dynamics Question: What is the velocity of a car traveling on a known curvature

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SUMMARY

The discussion focuses on simulating the velocity of a car on a known curvature using Matlab. The user PN seeks guidance on integrating physics concepts, particularly the relationship between kinetic energy and motion constraints. AN provides a mathematical framework involving a cubic function for the ramp and derives the kinetic energy equation, emphasizing the roll without slip condition. The key takeaway is the need to express the motion in terms of x(t) and v_x(t) to solve the problem effectively.

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  • Understanding of classical mechanics, specifically kinetic energy and motion constraints.
  • Familiarity with Matlab for simulation purposes.
  • Knowledge of calculus for integrating functions and deriving equations.
  • Basic understanding of coordinate systems and cubic functions.
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  • Study the principles of kinetic energy and motion in classical mechanics.
  • Learn how to implement simulations in Matlab, focusing on physics-based modeling.
  • Explore the concept of rolling motion and the conditions for rolling without slipping.
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This discussion is beneficial for physics students, mechanical engineers, and software developers working on simulations involving motion dynamics and vehicle dynamics modeling.

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Hi all,

I'm trying to write a Matlab simulation that determines the velocity of a car of known mass and moment of inertia which travels on a track whose curvature is also known.

To say the least, I'm at a loss as to what approach I should take to create my simulation. I'm finding it somewhat difficult to grasp the physics of the situation. The free body diagram is easy enough, but I can't recall how to couple this information with the constraint that the cart is confined to roll on the path.

If someone could provide some hints or a resource that I can read, I'd really appreciate it.


Thanks,

-PN
 
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I've given this some thought, was wondering if some higher up could check my reasoning/math:

Coordinate system: x,y conventional.

Say we have a ramp given by the cubic function: [itex]y(x)[/itex] = a[itex]x^{3}[/itex] + b[itex]x^{2}[/itex] + [itex]cx[/itex] + d

If we start the cart at x = 0 and y = [itex]y_{0}[/itex], where [itex]y_{0}[/itex], is a maximum, the kinetic energy at any point is

Kinetic Energy T = [itex]\frac{1}{2}[/itex]M[itex]v_{x}(t)^{2}[/itex] + [itex]\frac{1}{2}[/itex]I[itex]ω^{2}[/itex] = mg([itex]y_{0}[/itex] - [itex]y(t)[/itex])

Assuming roll without slip condition: ω = [itex]\frac{v_{x}}{R}[/itex] and some simplification we get:
T = C[itex]v_{x}(t)^{2}[/itex] = mg([itex]y_{0}[/itex] - [itex]y(t)[/itex]) where C is some constant.

In order to get the equation solely in terms of [itex]x(t)[/itex] and [itex]v_{x}(t)[/itex], we can substitute [itex]y(t)[/itex] for the cubic function [itex]y(x)[/itex], which is implicitly a function of time through x.

By integrating both sides of the equation now, we can get a function for [itex]x(t)[/itex], which we can plug back into [itex]y(x)[/itex] to get [itex]y(t)[/itex].Alright, how far off am I?
Thanks in advance,

-AN
 

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