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

AI Thread Summary
The discussion centers on simulating the velocity of a car on a curved track using Matlab, focusing on the physics of motion and constraints. The original poster seeks guidance on how to couple the free body diagram with the rolling constraint of the car. A participant proposes a mathematical approach using a cubic function to describe the ramp and derives an equation for kinetic energy that incorporates both translational and rotational motion. They suggest substituting the cubic function into the kinetic energy equation to derive the car's position and velocity over time. The conversation highlights the complexities of modeling dynamics in a constrained environment and the need for clarity in the underlying physics.
physicsnewblol
Messages
9
Reaction score
0
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
 
Physics news on Phys.org
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: y(x) = ax^{3} + bx^{2} + cx + d

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

Kinetic Energy T = \frac{1}{2}Mv_{x}(t)^{2} + \frac{1}{2}Iω^{2} = mg(y_{0} - y(t))

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

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

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

-AN
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top