Angular and linear vectors

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?

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:
Related Other Physics Topics News on Phys.org

pmb

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

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.

pmb

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

StephenPrivitera

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?

NateTG

Homework Helper
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.

StephenPrivitera

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:

pmb

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

StephenPrivitera

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?

pmb

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

StephenPrivitera

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 transfered 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

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving