How Can This Angular and Linear Vector Concept Be Simplified?

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    Angular Linear Vectors
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Discussion Overview

The discussion revolves around the relationship between angular and linear vectors, specifically focusing on the expressions for tangential and radial velocities and accelerations. Participants explore the implications of these vector relationships in both polar and rectangular coordinates, examining how to simplify and interpret various terms in the context of rotational motion.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that the expression v = w x r represents tangential velocity, while questioning the absence of radial velocity in this formulation.
  • Others propose that the velocity can be resolved into radial and tangential components, suggesting the use of polar coordinates for clarity.
  • One participant expresses confusion about the perpendicularity of the velocity vector to the radial vector and angular velocity, particularly in elliptical motion.
  • There is a suggestion to differentiate the position vector in polar coordinates to derive the velocity and acceleration components, with some participants advocating for this approach despite challenges in switching coordinate systems.
  • Concerns are raised about the accuracy of applying the chain rule in calculations, with calls for careful re-evaluation of previous steps.
  • Some participants discuss the implications of differentiating the expression v = w x r, noting that additional terms may appear in polar coordinates that are not present in the original expression.
  • There is a debate over whether the angular velocity always points along the axis of rotation, with some participants agreeing that this is a valid definition.
  • One participant attempts to convert expressions into rectangular coordinates, leading to a disagreement about the interpretation of the resulting velocity components.
  • Corrections are made regarding the signs and terms in the vector equations, with participants refining their arguments based on these adjustments.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of radial and tangential components of velocity, the application of coordinate transformations, and the implications of angular velocity. There is no consensus on the simplification of the expressions or the correctness of certain calculations.

Contextual Notes

Participants acknowledge limitations in their understanding of vector properties, the need for careful application of mathematical rules, and the challenges of transitioning between coordinate systems. Some assumptions about the nature of motion and the definitions of terms remain unresolved.

StephenPrivitera
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The "x" means cross product

v=w x r
This is the tangential velocity.
So, a=(dw/dt) x r+w x (dr/dt)
(dw/dt) x r is the tangential acceleration
w x (dr/dt) is the radial (centripetal acceleration)

I learned this from class.

But where is the radial velocity?

If we write s=T x r, then
v=w x r+T x (dr/dt)
Can T x (dr/dt) be interpreted as radial velocity?

Now a=(dw/dt) x r + w x (dr/dt) + w x (dr/dt) + T x (d2r/dt2)
=(dw/dt) x r + 2w x (dr/dt) + T x (d2r/dt2)
Through a little geometry, I can show that dr/dt=w x r
So, a=(dw/dt) x r + 2w x (w x r) + T x (d(w x r/dt)
=(dw/dt) x r + 2w x (w x r) + T x ((dw/dt) x r + w x (w x r))

First question: have I violated physics and mathematics at once?
Second question: can I simplify this? (I don't know the distributive/associative properties of "x")
Third question: What do each of these terms signify? (ie, cent. accel, tang. accel, etc)
Fourth question: What exactly is this vector T?
 
Last edited:
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Originally posted by StephenPrivitera
The "x" means cross product

v=w x r
This is the tangential velocity.
So, a=(dw/dt) x r+w x (dr/dt)
(dw/dt) x r is the tangential acceleration
w x (dr/dt) is the radial (centripetal acceleration)

I learned this from class.

But where is the radial velocity?

I don't understand your question. v is velocity and can be resolved into whatever components you like. For example: if the motion is an ellipse in a plane then you can write the velocity in terms of polar coordinates as

v = vrer + vthetaetheta

where

er = unit vector in the radial direction
etheta = unit vector in tangential direction

vr = radial velocity
vtheta = tangential velocity

Pete
 
Ok, pmb, it's possible that I'm talking nonsense, so bear with me. But according to what I have written v=w x r the velocity must be perpendicular to the radial vector r. It must also be perpendicular to w. If you are considering an ellipse, the velocity is not always perpendicular to the radial vector. There is an additional component. Also, I am trying very hard to stay in rectangular coordinates. I have already seen this development in polar.
 
Originally posted by StephenPrivitera
Ok, pmb, it's possible that I'm talking nonsense, so bear with me. But according to what I have written v=w x r the velocity must be perpendicular to the radial vector r. It must also be perpendicular to w. If you are considering an ellipse, the velocity is not always perpendicular to the radial vector. There is an additional component. Also, I am trying very hard to stay in rectangular coordinates. I have already seen this development in polar.

Yes. The velocity vector v is perpendicular to the postion vector r. I think that you're confusing the radial component of the velocity vector with the radial component of the position vector.

Try this. Write r as

r = rer

Now take the time derivative of this equation and simplify it.

Try to use polar coordinates. I know that its hard switching since you're used to Cartesian coordinates. But its well worth the effort. To aid you in this notice that

er = cos(theta)i + sin(theta) j
etheta = -sin(theta)i + cos(theta) j

I think that if you do this as an exercise then you'll find the answer to some of your questions.

Pete
 
s=r=rer
ds/dt=retheta+vrer
I agree that this velocity expresses both tangential and radial components.
But...
Ok, one step at a time.
Is it true that the angular velocity always points along the axis of rotation?
 
You could use the generic formula for generating a component along another vector:

r<v,r>/|r|

Where <,> is the dot product and || is the norm.

For a particle traveling in a circle (where your other formula holds) this is always zero since <v,r> is zero when v and r are perpendicular to each other.
 
Ok, if we can answer this, I'll be happy. How can we show that ds/dt=retheta+vrer is equivalent to ds/dt=w x r?
EDIT: To support my reasoning: When we continue with the polar coordinates by differentiating to find acceleration, a "Coriolis" force term appears, in addition to some other term that I don't understand. When we differentiate v=w x r we get no such terms.
 
Last edited:
I agree that this velocity expresses both tangential and radial components.
You made an error. Try doing it again. This time be more careful when you apply the chain rule.
Is it true that the angular velocity always points along the axis of rotation?
Actually that is one way the term axis of rotation can be defined.

Pete
 
s=r=rer
ds/dt=rwetheta+vrer
Even better.
I'm NOT crazy (Don't make yourself look stupid Stephen)...
v=w x r
Suppose w=wk and r=xi+yj+zk.
v=w x r =-w(yi+xj)
If we convert ds/dt=rwetheta+vrer into rectangular coordinates, then we get (let me use T for theta)
v=rw(-sinTi+cosTj)+vr(cosTi+sinTj)
If for the first term you factor out a negative and distribute the "r" and keep in mind that x=rcosT and y=rsinT, then
v=-w(yi+xj)+vr(cosTi+sinTj)
Now this disagrees with v=w x r. This verifies my belief that w x r is only the tangential velocity.

Any mistakes?
 
  • #10
Any mistakes?
Yes.

You have

w = wk and r = xi + yj + zk

wxr = (wk)x(xi + yj + zk)

= wx(kxi) + wy(kxj) + wz(kxk)

Note that -

kxi = j
kxj = -i
kxk = 0

Thus

wxr = wxj- wyi = w(-yi + xj)


Pete
 
  • #11
Correction
____________________
v=w x r = -w(yi-xj)

v=rw(-sinTi+cosTj)+vr(cosTi+sinTj)
If for the first term you factor out a negative and distribute the "r" and keep in mind that x=rcosT and y=rsinT, then
v= -w(yi-xj)+vr(cosTi+sinTj)
___________________



Yes, yes. I had that written down. I must have transferred it wrong. In any event, I made another mistake. The sign in front of "xj" is negative. Thus, my argument stands. By converting to rectangular we find,
v=rwetheta+vrer=w x r + vrer
 

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