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

[bulgie]

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?

4

 

[toprecyler]

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 will need more information to begin the math. radial tire or Bias ply? Diameter of tire? Biggest info is Air pressure in tire. Weight of vehicle actually putting on the tire. What is the distance of the dot to center axis of the tire?

My initial thoughts as radial tires usually look flat, or low on air due to bellowing out the bottom part in contact on the road. The sides and top are held in normal shape due to pressure of air in tire, relatively to atmospheric air pressure.

If the tire is at proper inflation, the center of the tire will be closer to the ground because the tire flattens out to provide road traction. If you increase the tire pressure, the flatten part will decrease, raising the center up. Vice versa if you decrease tire pressure.

So my initial answer would be what is the radius of the circle created when the dot was touching the ground, vs when it was at 12 o’clock.


Then when you throw the speed of 100 miles into the equation, it will make a difference in the Outside diameter of the tire. If we are taking a 12” diameter tire, that will be rotating very fast, vs a 36” diameter tire, that will be rotating 3 times slower, the centrifugal force of the tires rotating will be different, based all the other criteria I asked for in the first paragraph.

A good thinking question when you get down to it.


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[rwm]

This question is crazy. At 12 o'clock the driver is likely having lunch so, not very fast. At 6 o'clock he is stuck in rush hour traffic trying to get home from work. Otherwise it would be 200.

 

[higgite]

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

Depends on your point of reference. OP's riddle is stated as relative to the ground. Relative to the ground, nowhere on the tire ever retreats. At 6:00, the dot velocity is zero. Anywhere else in the tire's rotation, the dot has a forward velocity between zero and 200 mph. It never retreats.

Tom

 

[rwm]

One interesting this about this is moving a race car. You may see them push on the top of the tire to move the car. That gives them a 2:1 advantage. For each inch you move the tire, the car moves 1/2"

 

[higgite]

One interesting this about this is moving a race car. You may see them push on the top of the tire to move the car. That gives them a 2:1 advantage. For each inch you move the tire, the car moves 1/2"

It also keeps their hands from being run over.

Tom

 

[RaisedByWolves]

I will need more information to begin the math. radial tire or Bias ply? Diameter of tire? Biggest info is Air pressure in tire. Weight of vehicle actually putting on the tire. What is the distance of the dot to center axis of the tire?

My initial thoughts as radial tires usually look flat, or low on air due to bellowing out the bottom part in contact on the road. The sides and top are held in normal shape due to pressure of air in tire, relatively to atmospheric air pressure.

If the tire is at proper inflation, the center of the tire will be closer to the ground because the tire flattens out to provide road traction. If you increase the tire pressure, the flatten part will decrease, raising the center up. Vice versa if you decrease tire pressure.

So my initial answer would be what is the radius of the circle created when the dot was touching the ground, vs when it was at 12 o’clock.


Then when you throw the speed of 100 miles into the equation, it will make a difference in the Outside diameter of the tire. If we are taking a 12” diameter tire, that will be rotating very fast, vs a 36” diameter tire, that will be rotating 3 times slower, the centrifugal force of the tires rotating will be different, based all the other criteria I asked for in the first paragraph.

A good thinking question when you get down to it.


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The tire being squished at 6 Oclock is meaningless as it is momentarily stationary at that point.

The centrifugal force of the tire increasing the diameter could be an issue, but that is outside of the scope of the problem.

Add enough metrics and any problem becomes impossible to solve.

 

[RaisedByWolves]

One interesting this about this is moving a race car. You may see them push on the top of the tire to move the car. That gives them a 2:1 advantage. For each inch you move the tire, the car moves 1/2"

I learned this pushing tractors, but it is impossible to get across to some people.