I love it when we get arguments like this I use to bet my wallet on being correct but I lost everything over the years
Yup the top of the tyre is going forward at the same speed that the bottom is going aft aft so it nets out as zero as long as it's attached to the aircraft,
But if a piece now leaves the aircraft it simply maintains the last vector that it was on when it was release (Forward, Aft or Up). That vector is made up of ground speed against the surface speed of the released fragment.
No bets until you work out the numbers. Of course one can always add in a moving runway like a conveyor belt or maybe an aircraft carrier to work your brain