Angular Velocity of Thin Wheel/Axle System

[asdf60]

If I have a thin wheel of radius R attached to an axle of length R. The wheel rotates around an axis at the other end of the axle, parallel to the diameter of the wheel. that is, i rotate the wheel around one end of the axle so that the wheel goes around the end of the axle with frequency w, what is the angular velocity vector of the system at any instant.

 

[gothicpie]

Is the speed constant? Would it be possible to use (angular velocity) = (radians [represented by theta]) / (time) ?

 

[Tide]

The angular velocity vector is [itex]\omega[/itex] times a unit vector parallel to the axis of rotation with the sign (direction) given by the right hand rule.

 

[asdf60]

Tide, that doesn't make sense.
If that were the case, then consider the part of the wheel touching the ground. Instantaneously, that wheel has NO velocity, since it rolls without slipping. since WxR=v, an angular velocity going through the z axis would say that the point touching the ground has velocity. Which isn't correct.

in case my first explanation of the situation wasn't clear:
The system consists of an axle of length R with a wheel attached to it, also of radius R. (The normal of the wheel is parallel to the axle). No if the ground is the xy plane, then one end of the axle is constrained to the z axis, and then the wheel axle system rotates around the z-axis. The wheel touches the ground (and is perpendicular to the ground, so to speak), and rotates without slipping.

 

[Tide]

asdf,

Yes, your first description did not make mention of the wheel also rotating about its own principal axis (i.e. rolling along the ground). In that case you'll simply add the two angular velocity vectors.

 

[asdf60]

So essentially, the angular velocity vector points in the same direction as the line connecting the axle at the z-axis to the point touching the ground?

 

[asdf60]

And what would the angular momentum vector be? Wouldn't I be allowed to calculate the angular momentum treating the wheel like a point mass, and then adding the angular momentum of the wheel itself?

 

[Tide]

If I understand correctly, you are looking for something like this:

Suppose [itex]\Omega[/itex] is the angular velocity around the z-axis. If the length of axle is R and the radius of the wheel is r (you'll set them equal in the end but this general) then the angular velocity of the wheel about its principal axis is

[tex]\vec \omega = \Omega \frac {R}{r} \left(\cos(\Omega t) \hat i + \sin(\Omega t) \hat j\right)[/tex]

and the angular velocity about the axle is

[tex]\vec \Omega = \Omega \hat k[/tex]

The total angular velocity is the sum of those two.

 

[asdf60]

Yes, that's basically what I got. However, I think it the angular velocity of the axle should be negative, because of the right hand rule. Your formula shows that angular velocity points outwards, from the origin to the wheel.
I think the easier way to see this is, once again, that the point touching the ground has instantaneous velocity zero.

 

[Tide]

I wasn't paying too much attention to the signs - I figured you could handle the detail.

I'm not sure I see your point about the zero speed at the point of contact.

 

[asdf60]

No problem. Actually, you're correct about the signs either way, since it depends if it is moving clockwise or counterclockwise.

About the zero speed, let me know if I'm wrong. But I figure that since instantaneous v = w x r, then for a rotating object, all of the points that are instantaneously at rest must lie along the line defined by the w vector (and origin). That ensures that those points are moving at zero velocity.