# Kinetic energy in polar coordinates

• teclo
In summary: In the following derivation, I will first show you the kinetic energy term in rectangular coordinates, and then show you how to convert that to polar coordinates. The polar coordinates are just the coordinates of the particle after the transformation is complete.The kinetic energy of a point-particle in rectangular coordinates is given by:where r is the distance from the origin, a is the position vector, and v is the velocity of the point-particle.To convert this to polar coordinates, we use the following equation:where r is the distance from the origin, theta is the angle between the position vector and the positive x-axis, and v is the velocity of the point-particle in polar coordinates
teclo
if i have a system that I'm describing using polar coordinates, do i need to have an additional term for rotational kinetic energy? it would seem like this is covered since my velocity is in terms of the r and theta basis vectors. (i.e. i will have a term that covers the rotational movement ala theta). I'm not really sure about this, though. thanks if anyone is able to help.

More information would be helpful, because in a general way you're wrong (example, if your velocity just describes linear motion). What's it rotating about? Is it just spinning, etc.

Office_Shredder said:
More information would be helpful, because in a general way you're wrong (example, if your velocity just describes linear motion). What's it rotating about? Is it just spinning, etc.

two particles constrained to each other, free to rotate. each one is described by the position vector a(rhat) + b(theta hat). my velocity would have two components, r and theta. the velocity for each one would be in terms of r and theta, which would (i would think) would cover translational and rotational movement. the last sentence sums up my question fairly well. thanks for your input!-

I assume you are trying to write a lagrangian though I may be incorrect. You will have two kinetic energy terms in your lagrangian in this case, one proportional to the time derivative of your r position, and one proportional to the time derivative of the theta direction.

teclo said:
if i have a system that I'm describing using polar coordinates, do i need to have an additional term for rotational kinetic energy?

YES

marlon

marlon said:
YES

marlon

Write the kinetic energy of each point-particle in rectangular coordinates, then convert to polar coordinates. If there are relations among your point-particles [e.g. extended rigid bodies], you may be able to group terms and reinterpret.

robphy said:
Write the kinetic energy of each point-particle in rectangular coordinates, then convert to polar coordinates. If there are relations among your point-particles [e.g. extended rigid bodies], you may be able to group terms and reinterpret.

yes, this is definitely the best way to see what is going on if you are confused. Start with just a 1/2mv^2 term for each direction and then sub in the substitutions into polar coordinates, yuo should much of it will cancel out and you will be left with a simple expression with a kinetic energy term for each of your coordinates, r and theta.

http://web.me.com/dmwilliams/photo.jpg
Here is the derivation.

## 1. What is kinetic energy in polar coordinates?

Kinetic energy in polar coordinates is the energy an object possesses due to its motion in a polar coordinate system. It takes into account both the velocity of the object and its position in the polar coordinate system.

## 2. How is kinetic energy in polar coordinates calculated?

The formula for calculating kinetic energy in polar coordinates is KE = 1/2 * m * (r^2 * theta'^2 + 2 * r * theta' * r' * theta + r'^2), where m is the mass of the object, r is the distance from the origin, theta is the angle from a reference line, and the primes denote derivatives with respect to time.

## 3. What is the difference between kinetic energy in polar coordinates and Cartesian coordinates?

In Cartesian coordinates, kinetic energy is calculated using the formula KE = 1/2 * m * (x'^2 + y'^2 + z'^2), where x, y, and z are the coordinates in a 3D space. In polar coordinates, the formula takes into account the angular velocity and the radial velocity of the object, making it more specific to the polar coordinate system.

## 4. How does kinetic energy in polar coordinates relate to rotational motion?

Kinetic energy in polar coordinates is closely related to rotational motion because it takes into account the angular velocity of the object. In rotational motion, an object rotates around an axis, and its kinetic energy is dependent on its angular velocity.

## 5. Can kinetic energy in polar coordinates be converted to kinetic energy in Cartesian coordinates?

Yes, kinetic energy in polar coordinates can be converted to Cartesian coordinates using the transformation equations: x = r * cos(theta) and y = r * sin(theta). This allows for the calculation of kinetic energy in both coordinate systems for a more comprehensive understanding of the object's motion.

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