Anyone like physics problems? Not plane on treadmill, dot on tire.

[RaisedByWolves]

Going to try hard not to Eff this up.


There is a car traveling at 100mph. On the sidewall of the front tire of the car is a large red dot that touches the road when at the 6 O'clock position.

What is the difference in speed of the dot at the 12 O'clock position verses the 6 O'clock position relative to the ground?

Go!

 

[Just for fun]

It's only going half the speed..... This is just a wild a** guess, I have no idea.

 

[FOMOGO]

The 6 o'clock is always going to be faster because that old 12 o'clock gets tired a chasing his raggedy a** at 100mph. Sorry to have gotten all technical on ya, but ya just gotsta to tell it like it is. Mike

 

[Inferno]

Easy physics.
MPH is zero. Top of tire is considered 2xMPH
The question is, however, whether or not the surface speed changes.

P.S. The red dot and yellow dot have a purpose. They tell the installer where to mount the tire in relation to the valve stem.

 

[B2]

A little more physics....

@Inferno is correct however, maybe it is easier to understand by considering the speed relative to the car rather than the ground, get the numbers and then shift these by the actual MPH of the car. So consider the car, and a person in the car, to be stationary for a moment.

The person in the car observes the tire to be spinning so he sees the 6 O'clock position moving at -MPH and the 12 O'clock position moving at +MPH "relative" to the car. This is especially obvious if the tire positions are observed from the axle. But the car, and person, are moving at +MPH "relative" to the ground so we simply add this to the two positions to get the numbers "relative" to the ground. This yields -MPH+MPH=0 at the 6 O'clock position and it yields +MPN+MPH=2xMPH at the 12 O'clock position.

When the initial statement asked the question "relative to the ground," what is really meant is what would a person standing on the ground measure the speed of the two dot positions, of course the answer is 0 and 2xMPH.

These are "relative" perceptions. So as the car and passenger moves much faster and approaches the speed of light the passenger still sees the tire rotating in the same manner: - and + the speed of light.

What does the person on the ground observe the speed of the two positions to be?

 

[Inferno]

And a little more physics.
Technically, the tire speed of zero in relation to the road isn't quantifiable as the tire is not a true solid. As it's constantly moving somewhere on the tire even when it's contacting the pavement.
It could be argued that the tire does, in fact, meet the 2xMPH on the top but the fact is that the tire is in a continual state of acceleration and deceleration, simultaneously, relative to the ground.

 

[B2]

Practically speaking, we understand the tire material is "rubbery" and so is deforming at the point of contact during acceleration and due to the car weight. However, if it is moving uniformly at MPH maybe we should consider it not to be accelerating and so the only deformation is due to the car weight...which was not specified. Technically, is the tire material not considered a glass? (glass = no crystal structure, amorphous)

 

[Inferno]

Practically speaking, we understand the tire material is "rubbery" and so is deforming at the point of contact during acceleration and due to the car weight. However, if it is moving uniformly at MPH maybe we should consider it not to be accelerating and so the only deformation is due to the car weight...which was not specified. Technically, is the tire material not considered a glass? (glass = no crystal structure, amorphous)

Depends on the age of the tire.

 

[bulgie]

Going to try hard not to Eff this up.


There is a car traveling at 100mph. On the sidewall of the front tire of the car is a large red dot that touches the road when at the 6 O'clock position.

What is the difference in speed of the dot at the 12 O'clock position verses the 6 O'clock position relative to the ground?

Go!

I see what you did there — trick question!
If the sidewall of the tire is meeting the ground, then the car is on its side Driver probably stepped on the brake when the car fell on its side, so the wheels would be locked up, not rotating.

What do I win?

 

[stupoty]

This sounds a bit like the contact point being "instantaneously at rest" thing.

Stu