Solving for the Unknown in a Math Equation

[Panda1175]

Homework Statement

Given an angle for the what is the tangential velocity, A, respecting the ground.

Relevant Equations

v_a = v_cm + w x R
Center of mass velocity in respects to the ground + tangential velocity of a in respects to the center of mass velocity = A in respects to the ground

I don't understand the extra - in the second part of the equation, before the 2.
From my understanding, the squaring of the first part of the first equation is so that the equations consist of magnitudes. I'm just not quite sure where the negative came from. If I had to guess, it would be, w, harboring the negative (because it's pointing into the page)???

Thank you in advance!

 

[haruspex]

Try writing the three vectors, ##\vec v_{cm}, \vec\omega, \vec R##, in ##\hat i, \hat j, \hat k## notation and performing the multiplications.

 

[kuruman]

Or you can dot ##\vec v_A## with itself and see what you get. You may have to use the scalar triple product property of invariance under circular shift for the cross term.

 

[Panda1175]

Try writing the three vectors, ##\vec v_{cm}, \vec\omega, \vec R##, in ##\hat i, \hat j, \hat k## notation and performing the multiplications.

Thank you so much!

Oh! So the ##\vec\omega## and ##\vec R## are cross-multiplied while the ##\vec v_{A}## is dot produced with itself? If that is the case, then
##\vec\omega## x ##\vec R = <r, 0, 0>## x ## <0, 0, \omega> = <0-0, -wr+0, 0-0> = <0, -wr, 0>##

Letting ##v_{cm}## = <a, b, 0>
##\vec v_{A}## (dot) ##\vec v_{A} ## =<0, -wr, 0>² +( ##\vec v_{cm}##)² + ##\vec v_{cm} ## x（<0, -wr, 0>）= w²r² + (##\vec v_{cm}##)² - 2wr ##v_{cm} cos \theta##???

*I apologize for the messy text. I still haven't gotten the hang of this syntax.

 

[Panda1175]

Or you can dot ##\vec v_A## with itself and see what you get. You may have to use the scalar triple product property of invariance under circular shift for the cross term.

Thank you!
I think that is what I did above?

 

[kuruman]

I think that is what I did above?

Not quite.
$$\vec v_A\cdot \vec v_A=(\vec v_{cm}+\vec \omega\times \vec R)\cdot (\vec v_{cm}+\vec \omega\times \vec R)=v_{cm}^2+\omega^2R^2+2~\vec v_{cm}\cdot(\vec \omega\times \vec R).$$The first two terms are easy to see where they come from. Consider now the cyclic invariance $$\vec v_{cm}\cdot(\vec \omega\times \vec R)=\vec R \cdot(\vec v_{cm}\times \vec\omega)=\vec \omega \cdot(\vec R\times \vec v_{cm}).$$One of these three terms (which one?) gives you a vector cross product that is in the plane of the wheel. Draw it and see what you get when you dot it with the third vector.

 

[haruspex]

##\vec\omega## x ##\vec R = <r, 0, 0>## x ## <0, 0, \omega> = <0-0, -wr+0, 0-0> = <0, -wr, 0>##

Letting ##v_{cm}## = <a, b, 0>

First, why did you change the order of the cross product? The ##\omega## was first.
Second, if ##\hat i## is the ##v_{cm}## direction, the ##\vec r## coordinates should involve the angle ##\theta##.
Third, why does ##\vec v_{cm}## have a vertical coordinate?
Fourth, you can use \times for the cross product.

 

[haruspex]

One of these three terms (which one?) gives you a vector cross product that is in the plane of the wheel

Did you mean, one gives a product normal to the wheel?
But the question in post #1 is why the minus sign. The only ways I see to answer that are via a handedness rule or by doing the cross multiplication of coordinates.

 

[kuruman]

Did you mean, one gives a product normal to the wheel?

Yes.

But the question in post #1 is why the minus sign. The only ways I see to answer that are via a handedness rule or by doing the cross multiplication of coordinates.

Or by a drawing. The handedness rule must always be invoked because of the cross product between the vectors themselves of the unit vectors in terms of which one writes them.

Consider the cross term and its two cyclic equivalents $$\vec v_{cm}\cdot(\vec \omega\times \vec R)=\vec R \cdot(\vec v_{cm}\times \vec\omega)=\vec \omega \cdot(\vec R\times \vec v_{cm}).$$ In cases (A) and (B) in the figure below, the angle between the vectors to be dotted is greater than ##\frac{\pi}{2}## so the dot product is negative. In case (C) the vectors to be dotted are antiparallel. Cases (B) and (C) are easier than (A) for finding the entire term using ##\vec A\cdot \vec B=|\vec A||\vec B|\cos\theta## because the cross product involved is along one of the principal axes. The derivations are shown below the figure.

Case (B) ##[\vec R \cdot(\vec v_{cm}\times \vec\omega)##]
##|\vec v_{cm}\times \vec\omega|=v_{cm}~\omega## so that ##\vec R \cdot(\vec v_{cm}\times \vec\omega)=R~v_{cm}~\omega~\cos(\pi-\theta)=-\omega R v_{cm} \cos\theta.##

Case (C) ##[\vec \omega \cdot(\vec R\times \vec v_{cm})]##
##|\vec R\times \vec v_{cm}|=v_{cm}R\sin(\frac{\pi}{2}-\theta)=v_{cm}R\cos\theta.## The direction of this cross product is ##-(\hat {\omega})## so that ##\vec \omega \cdot(\vec R\times \vec v_{cm})=-\omega R v_{cm} \cos\theta.##

 

[haruspex]

The handedness rule must always be invoked because of the cross product between the vectors themselves of the unit vectors in terms of which one writes them.

… in terms of the order in which one writes them.
Yes, but @Panda1175 had that in the solution quoted, and perhaps knew the vector formula anyway. The question being asked was how to get from that to the scalar form with the correct sign.

 

[kuruman]

If it's only a question of sign, then I think diagram (C) is the best way to figure it out. It shows that the sign is positive/negative if ##\vec R## is above/below that horizontal.