Does a Rolling Cylinder Have One or Multiple Angular Rotations?

[SpartanG345]

If a cylinder is rolling without slipping, C is the centre of zero velocity for a moment and O is the centre

Does the angular rotation about O equal to the angular rotation about C, or is there only one angular rotation when a cylinder is rolling, that is the rotation about the point of contact?

The Attempt at a Solution

This is a question that came to me, not a assignment question or anything, but anyway

I think in the ordinary frame of reference a rolling cylinder only has an angular rotation about the point of contact, not about the centre

Where in a frame of reference where you are following the cylinder, you should see the cylinder rotating about the centre and the surface is moving linearly without sliding.

I know angular rotation is always about an a line, so a single motion can have many angular velocities with respect to many axis's.

Is it possible to evaluate the angular velocity with respect to O in the normal frame of reference when the ground is stationary?

i am not really sure i guess the angular velocity for each point on the shape would vary if you measure it from O since the whole object is kind of translating... and since C has a zero velocity

 

[tiny-tim]

Hi SpartanG345!

(btw, better to say "instant" rather than "moment", so as not to confuse with other types of moment )

Angular velocity (unlike angular momentum) is the same about any point.

Angular velocity is a "free" vector (strictly, a "free" pseudovector), so (unlike force) it has a direction, but not a specific line in that direction.

Changing to a different inertial frame will, of course, alter the velocity, but will not alter the angular velocity.

Formulas that combine I and ω use the same ω, no matter whether I (the moment of inertia) is about the centre of rotation or the centre of mass (btw, they don't generally work about any other point).

Is it possible to evaluate the angular velocity with respect to O in the normal frame of reference when the ground is stationary?

Yes, you get τ = I_Oω, instead of 0 = I_Cω - rmv, which is the same since I_C = I_O + mr².

(I've used your notation, but usually we use C for centre of mass, and O for centre of rotation )

 

[Doc Al]

Just to add to what tiny-tim has already explained...

I think in the ordinary frame of reference a rolling cylinder only has an angular rotation about the point of contact, not about the centre

Since the cylinder rolls without slipping, its instantaneous axis of pure rotation is the point of contact. So you can describe the motion in two ways:
(1) As a pure rotation about the point of contact.
(2) As a combination of rotation about the center of mass plus translation of the center of mass.

(Same ω in both cases, of course.)