Jaan Kalda kinematics -- Radius of Curvature of a Cycloid

In summary, "Jaan Kalda kinematics -- Radius of Curvature of a Cycloid" explores the geometric properties of cycloids, specifically focusing on calculating the radius of curvature at various points along the curve. The study highlights the mathematical relationships and derivations involved in determining the curvature, emphasizing its applications in physics and engineering. It provides detailed equations and graphical representations to illustrate the behavior of the cycloid as it relates to motion and forces.
  • #1
imbumb
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Cycloid is a curve which can be defined as a trajectory of a point marked on the rim of a rolling wheel or radius R. Determine the curvature radius of such curve at its highest point.

what you need to do is to equate 4v²/r with v²/R and to get that r=4R but i dont understand why the answer shpuldnt just be 2r
 
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  • #2
imbumb said:
Cycloid is a curve which can be defined as a trajectory of a point marked on the rim of a rolling wheel or radius R. Determine the curvature radius of such curve at its highest point.

what you need to do is to equate 4v²/r with v²/R and to get that r=4R but i dont understand why the answer shpuldnt just be 2r
2r is the distance between the highest point and the ground. What they are looking for is the radius of the arc that is made when the point on the wheel is at its highest point. If you look at a picture of it, you will see that the curve made by the point is much flatter than the curve of the wheel.

-Dan
 
  • #3
topsquark said:
2r is the distance between the highest point and the ground. What they are looking for is the radius of the arc that is made when the point on the wheel is at its highest point. If you look at a picture of it, you will see that the curve made by the point is much flatter than the curve of the wheel.

-Dan
why would the curve of the point be flatter than the curve of the wheel?
edit: i think i get it now
 
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  • #4
imbumb said:
why would the curve of the point be flatter than the curve of the wheel?
edit: i think i get it now
Because the center of curvature is not a fixed location. It moves as the wheel rotates.
 
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  • #5
imbumb said:
Cycloid is a curve which can be defined as a trajectory of a point marked on the rim of a rolling wheel or radius R. Determine the curvature radius of such curve at its highest point.

what you need to do is to equate 4v²/r with v²/R and to get that r=4R but i dont understand why the answer shpuldnt just be 2r
2r is the distance to the instantaneous center of rotation, which equals the curvature radius, if that center of rotation is a static point, but not necessarily if it translates.
 
  • #6
Hm, just do the calculation. Start by parametrizing the cycloid and then analyze it with the usual (plane) Fresnet analysis.
 
  • #7
vanhees71 said:
Hm, just do the calculation. Start by parametrizing the cycloid and then analyze it with the usual (plane) Fresnet analysis.
The OP already states a simpler way to calculate it, based on acceleration being the same in the ground frame and the wheel center frame. The question was why the result doesn't match the value intuitively expected by the OP.
 

FAQ: Jaan Kalda kinematics -- Radius of Curvature of a Cycloid

What is the basic definition of the radius of curvature in the context of a cycloid?

The radius of curvature at a particular point on a cycloid is the radius of the osculating circle at that point, which is the circle that best approximates the curve near that point. It provides a measure of how sharply the cycloid curve bends at that specific point.

How is the radius of curvature of a cycloid derived?

The radius of curvature \( R \) of a cycloid can be derived using the formula \( R = \frac{(x'^2 + y'^2)^{3/2}}{|x'y'' - y'x''|} \), where \( x \) and \( y \) are the parametric equations of the cycloid, and the primes denote derivatives with respect to the parameter \( \theta \). For a cycloid given by \( x = r(\theta - \sin\theta) \) and \( y = r(1 - \cos\theta) \), the radius of curvature simplifies to \( R = 4r \sin(\theta/2) \).

What role does the parameter \( \theta \) play in the radius of curvature of a cycloid?

The parameter \( \theta \) represents the angle through which the generating circle has rotated to form the cycloid. It directly influences the radius of curvature, as the radius of curvature \( R \) is a function of \( \theta \). Specifically, \( R = 4r \sin(\theta/2) \) shows that the curvature varies with \( \theta \), being zero at the cusps (where \( \theta \) is a multiple of \( 2\pi \)) and maximal midway between the cusps.

Why is understanding the radius of curvature important in the study of cycloids?

Understanding the radius of curvature is crucial because it provides insights into the geometric and physical properties of cycloids. For example, in physics, the radius of curvature is related to the centripetal force experienced by a particle moving along the cycloid. In engineering, this knowledge is useful in designing cycloidal gears and other mechanical systems where smooth motion is critical.

Can you provide an example of where the radius of curvature of a cycloid is applied in real-world problems?

One real-world application of the radius of curvature of a cycloid is in the design of roller coasters. The cycloid is the solution to the brachistochrone problem, where it represents the curve of fastest descent under gravity. Engineers use this property to design roller coaster tracks that provide thrilling yet smooth rides. The radius of curvature helps ensure that the transitions between different sections of the track are safe and comfortable for riders.

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