Rolling Motion direction of Velocity center of mass

[physwiz222]

TL;DR Summary

Shouldnt V_cm be equal to -wR

Hi I have come across something confusing in rolling motion. If an object moves with a positive V_cm meaning to the right its angular velocity will be clockwise or negative. The formula is V_cm=wR but for a positive V_cm you get a negative w as it moves clockwise if V_cm is to the right. Shouldnt the formula be V_cm = -wR or is this just the Magnitude of V_cm the formula refers to.

 

[BvU]

Hi I have come across something confusing in rolling motion.

Hi,
Isn't there a context for this 'something' that can eliminate your confusion?
Some explanation, typography, examples, etcetera?

## \ ##

 

[physwiz222]

Hi,
Isn't there a context for this 'something' that can eliminate your confusion?
Some explanation, typography, examples, etcetera?

## \ ##

I already explained for a rolling body with positive V_cm the angular velocity is clockwise which is negative. But V_cm = wR so a negative w means negative Vcm but Vcm is positive so its seems like a contradiction. Maybe that equation is only for the Magnitude

 

[jbriggs444]

I already explained for a rolling body with positive V_cm the angular velocity is clockwise which is negative. But V_cm = wR so a negative w means negative Vcm but Vcm is positive so its seems like a contradiction. Maybe that equation is only for the Magnitude

There are several sign conventions lurking here that you may not have considered.

Consider that product of ##\vec{\omega}## and ##\vec{R}##. It is a vector cross product. Technically, it is the cross product of the pseudo-vector ##\vec{\omega}## and the signed displacement ##\vec{R}##.

Vector cross products are not commutative. The sign of the result depends on the order of the operands: $$\vec{\omega} \times \vec{R} = - \vec{R} \times \vec{\omega}$$In addition, the sign of ##\vec{R}## depends on whether you are measuring it from the road to the axle or from the axle to the road.

The sign of ##\vec{\omega}## can be ambiguous. In three dimensions, it depends on whether we chose the right hand rule or the left hand rule when we defined our direction convention for rotations.

If we are careful with all of our sign conventions we could properly infer the relative velocity between two body-fixed points (in three dimensions even) directly from the formula. If one were programming a simulation, that level of care would be called for.

In practice, working simple problems, we are (or I am anyway) often much sloppier than that. We will consider the physical situation, whether we are using a clockwise positive or clockwise negative convention, a rightward positive or rightward negative convention and whether the wheel is rolling on the ceiling or on the floor to figure out whether the resulting motion will be positive or negative. Then we stick a minus sign into our equation if needed so that things work out.

 

[BvU]

I already explained for a rolling body with positive V_cm the angular velocity is clockwise which is negative. But V_cm = wR so a negative w means negative Vcm but Vcm is positive so its seems like a contradiction. Maybe that equation is only for the Magnitude

No need to repeat. But an answer might be useful

Some explanation, typography, examples, etcetera?

notation ? What textbook, webpage, other ?

##\ ##

 

[vanhees71]

The main problem with such confusion is that many books don't use vectors, and I can only guess what is it that's to be described. I assume it's a cylinder rolling without slipping along the direction ##\vec{e}_x## of a Cartesian coordinate system with its axis pointing in the direction ##\vec{e}_z##. Then ##\vec{v}_{\text{CM}}=v_{\text{CM}} \vec{e}_x## and the angular velocity of the cylinder is ##\vec{\omega}=-\omega \vec{e}_3## with ##\omega>0##.

 

[kuruman]

As @jbriggs444 pointed out, the key relation that links the position of an off-center point P ##\vec{r}##, inear velocity of point P ##\vec{v}## and the angular velocity ##\vec{\omega}## about the center of the wheel O is $$\vec{v}=\vec{\omega}\times\vec{r}.$$Now consider a wheel rolling to the right in the plane of the screen and a right-handed coordinate system such that
##\hat x =~## unit vector to the right
##\hat y =~## unit vector down (from the center to the point of contact)
##\hat z =~## unit vector into the screen

Let point P be at the 12 o'clock position on the rim at distance ##R##. Clearly, ##\vec{r}=-R\hat y##. Then $$\vec{v}_P=\vec{\omega}\times\vec{r}_P=\omega\hat{k}\times(-R\hat y)=-\omega R(\hat k\times \hat y)=+\omega R(\hat y\times \hat k)=\omega R \hat{x}.$$This says that point P moves to the right relative to the center of the wheel.

Now let point Q be at the 6 o'clock position at distance ##R##. Clearly, its position vector is the negative of the position vector of P. Then $$\vec{v}_Q=\vec{\omega}\times\vec{r}_Q=\vec{\omega}\times(-\vec{r_P})=-\vec{\omega}\times\vec{r_P}=-\vec{v}_P.$$ The 12 o' clock point is moving to the right whilst the 6 o' clock position is moving to the left. That's a clockwise rotation.

For the rolling (without slipping) part of the motion you shift the reference point to the point of contact Q ##(\vec{r}_Q=0.)## You can easily show by the same method that

##\vec{v}_Q=0##, the point of contact is instantaneously at rest.
##\vec{v}_O=\omega R \hat{x}##, the center moves to the right with speed ##\omega R.##
##\vec{v}_P=2\omega R \hat{x}##, point P moves to the right with speed ##2\omega R.##