So a tire rolls with radius r. The CM of the wheel travels at a velocity of vcom.
What is the acceleration of the top of the wheel and the bottom of the wheel relative to a passenger in the car and to a street sign on the street?
a = v^2/r
The Attempt at a Solution
Relative to the passenger, the top and bottom of the wheel are traveling at a velocity of vcom and -vcom. So the magnitudes of both accelerations are vcom^2/r.
Relative to the street sign, the top of the wheel is traveling at 2vcom. The bottom of the wheel is traveling at 0. I know that these are the wrong velocities to plug into the equation a = v^2/r since the answer for the accelerations are the same as the accelerations in the previous situation.
A friend told me that centripetal acceleration is always relative to the axle. I thought that perhaps since the translational acceleration of the top and bottom of the wheel are 0, and since the acceleration of the wheel when it is rotating is vcom^2/r for the top and bottom, the magnitudes of the accelerations of both the top and bottom could be Arotation + Atranslational = ARolling,
which would be vcom^2/r.
Are any of these explanations correct?