Kinematics, curvilinear motion (a couple of questions)

Thread starter srnixo
Start date Jan 27, 2024
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Position vector

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SUMMARY

The discussion centers on the existence of the position vector in the tangential and normal coordinate system during curvilinear motion. It is established that the position vector does not exist in normal-tangential coordinates because these coordinates are attached to a moving particle, thus lacking a fixed origin. The conversation also clarifies the use of tangential and normal coordinates versus Cartesian coordinates; the former is preferred for known curved paths, while the latter is used for general cases. Additionally, curvilinear motion is defined as one-dimensional motion along a curved path, distinct from rectilinear motion.

PREREQUISITES

Understanding of curvilinear motion and its definitions
Familiarity with normal-tangential (n-t) coordinate systems
Knowledge of Cartesian coordinate systems
Basic principles of kinematics and motion analysis

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Study the mathematical representation of normal-tangential coordinates in curvilinear motion
Explore the differences between curvilinear motion and rectilinear motion
Learn about the applications of curvilinear coordinates in physics problems
Investigate the role of position vectors in various coordinate systems

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Students of physics, educators teaching kinematics, and professionals in engineering fields focusing on motion analysis will benefit from this discussion.

Jan 28, 2024

haruspex

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srnixo said:

the tangential and normal coordinate systems do not have a fixed origin (the origin moves with the object). Therefore, the position vector of the object will constantly change

Ok, so in eqn 1, what does ##\vec r## mean? It is not the location of the object moving along the path s since that is by definition (0,0) in n-t. It must be the location of some other point expressed in n-t coordinates relative to the object, but what point?

##\rho \hat n## is the location of the centre of curvature, but what is ##R(t)##? I can find no online reference to "radius of curvature in tangent direction", and @Delta2's proposed interpretation in post #19 makes it the same as ##\rho##.

Here's another interpretation: ##R## and ##\rho## (or perhaps it is ##P##, not ##\rho##) have nothing to do with radii of curvature. All the eqn is saying is that given some arbitrary point ##\vec r## and a point on a curve as origin, ##\vec r## can be expressed in curvilinear coordinates as ##(R, P)##, meaning ##R\hat t+P\hat n##.