Reference frame for uniform circular motion

[daudaudaudau]

Hi.

Say a particle is moving around in uniform circular motion. The way my book attacks this problem is by placing the coordinate system such that one axis is in the radial direction and the other axis is in the tangential direction. Then we have the following for the radial acceleration
<br /> a_r=\frac{v^2}{R}<br />

But I am just wondering: Why is this an inertial reference frame? Is it not rotating along with the particle?

 

[sganesh88]

Radial and tangential? I don't think its right since the "radial" and "tangential" changes directions continuously in circular motion.
The better way would be to choose the intersection of the two coordinate axes(that lies in the plane of rotation) to be at the center.

 

[vin300]

But I am just wondering: Why is this an inertial reference frame? Is it not rotating along with the particle?

It is not.

 

[Doc Al]

But I am just wondering: Why is this an inertial reference frame? Is it not rotating along with the particle?

No, it's not. In the non-inertial frame that rotates with the particle, the speed and acceleration of the particle would be zero.

 

[BobG]

Hi.

Say a particle is moving around in uniform circular motion. The way my book attacks this problem is by placing the coordinate system such that one axis is in the radial direction and the other axis is in the tangential direction. Then we have the following for the radial acceleration
<br /> a_r=\frac{v^2}{R}<br />

But I am just wondering: Why is this an inertial reference frame? Is it not rotating along with the particle?

Radial and tangential? I don't think its right since the "radial" and "tangential" changes directions continuously in circular motion.
The better way would be to choose the intersection of the two coordinate axes(that lies in the plane of rotation) to be at the center.

It's a non-inertial reference frame.

Using radial and tangential is also some sloppy terminology, which sganesh noticed. Works fine for circular motion, but if you expand that to elliptical motion, your "tangential axis" isn't going to lie tangent to your ellipse. Technically, it's (radial and cross-radial) or (tangential and normal).

Even though sloppy, it's terminology that's also used a lot. Once you've started with "radial" and "tangential" in circular motion, it's hard to stop doing that once you talk about elliptical motion.

 

[Doc Al]

It's a non-inertial reference frame.

No, it's an inertial frame. (See my last post.)

And there's nothing wrong with using the radial and tangential directions (at some instant) to define a coordinate system for describing things.

But you're right--that trick only works well for simple circular motion. But that's all we're dealing with here.

 

[rcgldr]

Say a particle is moving around in uniform circular motion. The way my book attacks this problem is by placing the coordinate system such that one axis is in the radial direction and the other axis is in the tangential direction.

Essentially from the particles viewpoint looing inwards at the center of rotation.

Then we have the following for the radial acceleration
<br /> a_r=\frac{v^2}{R}<br />

This isn't true if v is velocity with respect to the coordinate system you just defined. With v=0 with respect to this coordinate system, which is a non-inertial reference frame, the particle experiences centrifugal force. If V is in the radial direction, then centrifugal force at any point in time is related to R₀ / R, where R₀ is the radial distance from "0" to the center of rotation. If V is in the tangental direction, then centrifugal force is related to (V-V₀)² / V₀²where V₀ = ω₀ R₀, where ω₀ is the rate of rotation of this coordinate system.

 

[James98765]

I have a quote from my physics textbook that may assist you.

"You might wonder if the rtz-coordinate system is an inertial reference frame. It is, and Newton's laws apply, although the reason is rather subtle. We're using the rtz-coordinates to establish directions for decomposing vectors, but we're not making measurements in the rtz-system. That is, velocities and accelerations are measured in the laboratory reference frame. The particle would always be at rest (v = 0) if we measured velocities in a reference frame attached to the particle. Thus the analysis [is really from the laboratory's inertial reference frame]." (Physics for Scientists and Engineers with Modern Physics, 2nd Edition, Randall D. Knight)

I hope this clears up the thought for you. Basically the rtz-system states directions that couldn't easily be stated with circular motion in an xy-coordinate system. The measurements are actually taken from the inertial reference frame of the observer.

-James

 

[Doc Al]

Basically the rtz-system states directions that couldn't easily be stated with circular motion in an xy-coordinate system. The measurements are actually taken from the inertial reference frame of the observer.

Exactly. This is standard practice. (Thanks, James.)

 

[jmb]

As the other posters have already said, it is an inertial, non-rotating frame. I think part of your confusion may be due to the fact that the directions of the unit vectors in this coordinate system (i.e. the radial direction and the tangential direction) do 'rotate' as the particle moves.

The important conceptual point to grasp is that it is perfectly possible for these unit vectors to change direction in time, while the position coordinates are still measured with respect to a non-rotating frame.

In the case of simple polar coordinates like this, the position coordinates are (r,\theta). Where r is measured as a distance from the origin of the non-rotating coordinate system, and \theta is the angle (the bearing if you like) made between a line joining the particle to the origin and some other fixed line in the non-rotating reference frame (many choices are possible, a common one is the 'x-axis'). However the unit vectors in the radial and tangential directions (sometimes written \vec{e_r}\;\text{and}\;\vec{e_{\theta}}) are functions of \theta and so vary with the particle's motion.

In fact one way to derive the apparent centrifugal and Coriolis forces felt by the rotating particle is to consider the derivatives of these unit vectors.