Cylindrical and Spherical Coordinate systems

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SUMMARY

The discussion centers on the mechanics of cylindrical and spherical coordinate systems, specifically the equations for velocity and acceleration. The velocity vector in cylindrical coordinates is expressed as \vec v = \dot r\hat e_r + r\dot\theta\hat e_\theta + \dot z\hat k, while the acceleration vector is \vec a = (\ddot r - r\dot\theta^2)\hat e_r + (r\ddot\theta + 2\dot r\dot\theta)\hat e_\theta + \ddot z\hat k. A key point raised is the physical meaning of \ddot r - r\dot\theta^2 when r is constant, highlighting that even with \dot r = 0, there can still be radial acceleration due to angular velocity \dot\theta. The discussion clarifies that the coordinate systems are fixed and do not move with the object, allowing for acceleration in various directions.

PREREQUISITES
  • Understanding of cylindrical coordinate systems
  • Familiarity with vector calculus
  • Knowledge of angular velocity and acceleration concepts
  • Basic principles of kinematics
NEXT STEPS
  • Study the derivation of velocity and acceleration in cylindrical coordinates
  • Learn about the relationship between angular velocity and radial acceleration
  • Explore applications of spherical coordinate systems in physics
  • Investigate the differences between fixed and moving coordinate systems
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Students and professionals in physics, engineering, and mathematics who are studying motion in cylindrical and spherical coordinate systems, particularly those interested in kinematics and dynamics.

Warr
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I have a question about the equation mechanics of cylindrical and spherical coordinate systems

This is basically about the velocity and acceleration equations of both

Let me just give an example from cylindrical

[tex]\vec v = \dot r\hat e_r + r\dot\theta\hat e_\theta + \dot z\hat k[/tex]

and

[tex]\vec a = (\ddot r - r\dot\theta^2)\hat e_r + (r\ddot\theta + 2\dot r\dot\theta)\hat e_\theta + \ddot z\hat k[/tex]

My question is, what is the physical meaning of [tex]\ddot r - r\dot\theta^2[/tex] if r is not changing. I thought that the coordinate system moved with the object you are measuring, and if so..how can there be an acceleration in the direction of [tex]\hat e_r[/tex] if [tex]\dot r = \ddot r = 0[/tex]

:confused:
 
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Warr said:
how can there be an acceleration in the direction of [tex]\hat e_r[/tex] if [tex]\dot r = \ddot r = 0[/tex]

Think about what [itex]\dot{\theta}[/itex] is: the angular velocity. So think about a record playing at a constant 45 rpm. The [itex]r[/itex]-coordinate of each point on the record is constant, but since [itex]\dot{\theta}\neq0[/itex], the direction of the velocity vector of each point on the record is constantly changing, which means there is an acceleration. And in what direction must that acceleration point? Well, if it points in any direction other than the radial direction then the record will speed up, which is contrary to our assumption of constant [itex]\dot{\theta}[/itex]. Therefore, there must be a nonzero acceleration pointing in the radial direction.
 
Warr said:
I have a question about the equation mechanics of cylindrical and spherical coordinate systems
...
My question is, what is the physical meaning of [tex]\ddot r - r\dot\theta^2[/tex] if r is not changing. I thought that the coordinate system moved with the object you are measuring, and if so..how can there be an acceleration in the direction of [tex]\hat e_r[/tex] if [tex]\dot r = \ddot r = 0[/tex]

:confused:


These cylindrical and spherical coordinate systems do not move together with the body: they are steady, just like the cartesian system of coordinates. You set up the system with its axes, and describe the motion of the body with respect the fixed axes of the system. The body can have acceleration in any direction, x, y, z. In the same way, it can have acceleration along the circles, surrounding the z axis of a cylindrical system, and also normal to these circles in radial direction, and along the z axis as well. Try to describe the velocity and acceleration of a body that moves along a straight line, using polar coordinates!
You might have mixed them with the coordinate system that moves together with the body and changes coordinate axes according to its orbit. In that system, one axis is parallel to the velocity vector and the other one is normal to the velocity in the plane of motion, and the third axis is normal to the plane. In that system, the acceleration has got a tangential component, the time derivative of the speed, and a radial component, the centripetal acceleration, v^2/R , where R is the radius of the curvature. You just have solved such a problem here in the Forum.

ehild
 

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