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

[mac1911]

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.

No they do not always mark a mounting location.
I ask what to do with the white dot, what to do with the the location when you have more than one dot in different locations and color combinations or no dot.
Unless manufactures specify its generally of little use
Forgot to add that unless you know the runout of the rim. Where the heavy and high spot of the rim is the dots are a bit “useless”
Used to be rumored the valve stem hole would be located at the “low” spot on the rim. After years of spinning up tired on a hunter road force balancer that can do bare rim run out , then locate the high and low spots of a tires and calculated the best mounting orientation …..the dots and valve stems dont align very often.
Plus some manufactured dont use any marks.

 

[RaisedByWolves]

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?

Cant look at it from a passengers point of view as at 6 & 12 the tire would be either advancing or retreating.

Not a trick question, nor does the elasticity of the tire come into play. The tire being squished at the point it touches the road at the dot is mute as it is at zero speed in this condition, otherwise it would be sliding/skidding.

 

[rabler]

Question: what reference frame are you defining speed in relation to? Real question is do you want Newtonian physics, or general relativity? Obviously this is an accelerated point since the tire is in circular motion, so special relativity is not applicable. Unless of course the 100mph is relative to another passing vehicle and the angular velocity of the tire is zero.

Not saying I can do more than recognize that special relativity is not applicable, much less actually crunch the math ...

 

[homebrewed]

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?

Just to add to the above correct reasoning, the relative speed of the tire at the 6 o'clock position has to be zero, otherwise the contact point will be moving w/respect to the ground -- it will be in a skid condition. Given the zero relative velocity there, it's a small jump to conclude that the 12 o'clock position is moving at 2*indicated MPH.

I always find it good to get to the same answer in two different ways.

 

[rabler]

Note that the dot is on the sidewall, not necessarily in contact with the road. If the sidewall extended all the way to the axle, then at the axle the dot would move at 100 MPH. Therefore the velocity of the dot is going to be something like 100 MPH * (1 + (R1/R2)*sin(tire angle)), where R1 is the radius of the dot, and R2 is the effective radius of the tire.

edited to add: "tire angle" is function of effective radius and car's velocity, so you can make some substitutions there if you're worried about the actual frequency and acceleration of the dot's velocity, and express the entire thing as a function of t ...
The difference (to address the actual question as asked) is going to be 100*(1+R1/R2) - 100*(1-R1/R2), since Sine varies between +1 and -1
that simplifies to 100 + 100 * R1/R2 -100 + 100*R1/R2,
so the difference is 200 * R1/R2, assuming I didn't fumble my algebra. As a sanity check, that works for the degenerative case of R1=R2

 

[RaisedByWolves]

Note that the dot is on the sidewall, not necessarily in contact with the road. If the sidewall extended all the way to the axle, then at the axle the dot would move at 100 MPH. Therefore the velocity of the dot is going to be something like 100 MPH * (1 + (R1/R2)*sin(tire angle)), where R1 is the radius of the dot, and R2 is the effective radius of the tire.

edited to add: "tire angle" is function of effective radius and car's velocity, so you can make some substitutions there if you're worried about the actual frequency and acceleration of the dot's velocity, and express the entire thing as a function of t ...
The difference (to address the actual question as asked) is going to be 100*(1+R1/R2) - 100*(1-R1/R2), since Sine varies between +1 and -1
that simplifies to 100 + 100 * R1/R2 -100 + 100*R1/R2,
so the difference is 200 * R1/R2, assuming I didn't fumble my algebra. As a sanity check, that works for the degenerative case of R1=R2

Reread the question, basically both are covered.

You can say the dot is large and I can claim it is infinitesimally small, which takes us off into the realm of semantics.

Some people when at first mistaken want to extrapolate the question outside of how simple it is stated, and it is stated simply for a good reason......

Because in fact on a grander scale the answers given to this point are all wrong when looked at from the scale of the universe.

The car in total, taken as a whole, is rotating upon the earths axis at 1000 miles per hour. But if you look at it from the perspective of how fast it is moving in regards to the sun in our solar system, it is moving at 67,000 miles per hour, and all of this is moving through our galaxy from the perspective of its center at 490,000 miles per hour.

In this grander scenario what is the speed of the dot at the two positions?


I have no idea, and wouldn't even know where to start with all of the variables, but I would think it would need to be nailed down with a time and date and involve calculus.

This is why the question is framed the way it is and references the ground as the point of reference.

Other times I've posed this question I've tried this using the fender as a point of reference, and a passenger in a car, both of which were problematic in their own way.

I'm starting to wonder if it isn't as much of a reading comprehension/retention problem as well as a physics problem.

Its a simple problem and looked at simply has a somewhat simple answer.

 

[rabler]

You did say you were trying to word it carefully, but normally the "sidewall" does not come in contact with the road. So your question had some ambiguity. Of course you can also question whether a "dot" is equivalent to a mathematical point. Of which size is meaningless. But otherwise the whole concept of speed of a non-zero sized area on a flexible rotating tire is pretty vague. Especially if it is a front wheel drive car with some minute slippage as is usual in the drive wheels. Are we having fun yet? Oh, lets add speed verses velocity in physics speak. And then ponder whether the car is turning or not, and how that impacts the answer.

BTW, I meant R1 is the radius of the dot's center relative to the axle, not the size of the dot, so that is a failing/ambiguity in my answer.

ecause in fact on a grander scale the answers given to this point are all wrong when looked at from the scale of the universe.

I earlier pointed out that frame of reference AND general relativity could be considered applicable, lol. That pretty much sums up the scale of the universe perspective, and involves more than basic calculus.

I'm starting to wonder if it isn't as much of a reading comprehension/retention problem as well as a physics problem.

Having graded both college engineering and physics tests, I know how hard it is to write a good question. Having been a flight instructor, teaching people to land an airplane, I know how hard it is to make a concise statement that doesn't have the potential to confuse someone (with drastic consequences). Learning to be a flight instructor is all about communication and fundamentals of learning. I did come up with the basic answer as a degenerative case, so are you looking for an "answer", or a methodology? (I could have added, for the case of R1 = R2, it simplifies to 200).

I'll move it from physics, to philosophy: this comes down to "What answer did you, as the author, want?". That frame of reference may be more fluid/murky than general relativity.

 

[Inferno]

I just read all this and now I'm tired

 

[RJSakowski]

How about this one. You have two gears, one with 20 teeth and one with 60 teeth. The gears a meshed with the smaller gear at 12o'clock. The larger gear is stationary and smaller gear revolves around it.

How many revolutions does the small gear make to get back to 12 o'clock?

 

[Inferno]

How about this one. You have two gears, one with 20 teeth and one with 60 teeth. The gears a meshed with the smaller gear at 12o'clock. The larger gear is stationary and smaller gear revolves around it.

How many revolutions does the small gear make to get back to 12 o'clock?

ONE/