Why does tangential velocity equal translational velocity when rolling?

[jonahsaltzman]

Homework Statement

Not actually a problem, I'm just curious why: when calculating total kinetic energy during rolling, you have the translational kinetic energy=0.5mv^2, and the rotational KE=0.5Iω^2. But then ω=v/r, so rotational KE=0.5I(v/r)^2. And for some reason, the v in both equations are equal when the wheel is rolling. I thought the tangential velocity changes depending on where on the wheel the spot is - i.e., on the top, the velocity of a point is 2x the velocity of the center of mass.

Homework Equations

KE_translation=0.5mv^2
KE_rotation=0.5Iω^2

The Attempt at a Solution

Does it have something to do with the fact that the arc length θ traveled by the wheel equals the translational distance traveled?

Thanks!

 

[Greg-ulate]

One is the velocity of the center of mass with respect to the stationary reference frame and the other is the velocity of the stationary ground with respect to the moving frame of reference of the wheel. They are the same because both reference frames are not rotating nor accelerating nor moving at relativistic speed with respect to the other.

The velocity that would be 0 or 2x would be the instantaneous velocity of a wheel segment in the laboratory frame of reference, not the tangential velocity.

 

[Chestermiller]

There is no slip between the wheel and the ground, and from the frame of reference of the wheel, the ground is moving backwards at velocity v. So at the contact point, the tangential velocity of the wheel must be v. Since the wheel is solid, the magnitude of the tangential velocity at all points around the rim must be independent of location around the rim.