# Centrifugal Force on Rotating 4" tire / wheel @ 100mph

• B
nitsuj
TL;DR Summary
Centrifugal Force on Rotating tire / wheel
Heya PhysicsForums!

Remote Control Car toy tires and wheels.

a 4" tire/wheel rotates at 8400rpm at 100mph.

Am wondering how many "g's" the tire "experiences" at that rpm; I imagine it being hundreds of times (if below is accurate am WAY off with my guess)

Using a centrifugal force calculator and the following values (sorry about units)

Mass (of just tire) = 250 grams
radius = 2 inches
tangential velocity = 100mph

equals a force of 9835 Newtons and centrifugal acceleration (I guess this is the value am interested in?) of 39,340m/s/s

so for G units is that the centrifugal acceleration of 39,340 / 9.81 = 4,010g?

meaning the tire experiences 4000g's of force perpendicular to the axel? (if so omg that is some strong glue holding the tire to the wheel lol)

Lnewqban

## Answers and Replies

Gold Member
Summary:: Centrifugal Force on Rotating tire / wheel

meaning the tire experiences 4000g's of force
Yes, about 4000g, but g is a unit of acceleration, not force.
Your force figure assumes the entire mass of the tire is a point object traveling around the edge, where in fact it is distributed evenly around what is assumed to be an infinitely thin perimeter.
The force on this tire is thus the tension resulting from the spin, not any hypothetical centripetal force exerted by a string with a mass twirling at the end of it.

Lnewqban and nitsuj
Staff Emeritus
2022 Award
You've used the full mass of the tire, but we're only concerned about a tiny 'slice' of the wheel since the mass is spread out throughout the wheel. Additionally, the mass is not concentrated at the outer edge of the wheel either, so the calculation is much more complicated. But we can simplify things and get a reasonable guess.

Let's divide the mass by 360 to get the mass of a single 1 degree slice. Dividing 250 grams by 360 gives us about 0.694 grams per slice. Now let's assume that all the mass is concentrated at the edge. That gives as a force on this slice of about 27 Newtons. If we make the slices thinner we will get a lower force, which is just as it should be since smaller slices have less mass and require less force for the same acceleration.

The acceleration is, of course, just as you calculated, nearly 40,000 m/s2, or roughly 4,000 g's at the edge of the wheel. This will drop as you move inward toward the center.

Lnewqban and nitsuj
nitsuj
You've used the full mass of the tire, but we're only concerned about a tiny 'slice' of the wheel since the mass is spread out throughout the wheel. Additionally, the mass is not concentrated at the outer edge of the wheel either, so the calculation is much more complicated. But we can simplify things and get a reasonable guess.

Let's divide the mass by 360 to get the mass of a single 1 degree slice. Dividing 250 grams by 360 gives us about 0.694 grams per slice. Now let's assume that all the mass is concentrated at the edge. That gives as a force on this slice of about 27 Newtons. If we make the slices thinner we will get a lower force, which is just as it should be since smaller slices have less mass and require less force for the same acceleration.

The acceleration is, of course, just as you calculated, nearly 40,000 m/s2, or roughly 4,000 g's at the edge of the wheel. This will drop as you move inward toward the center.
DOH!

thank you for clearing up my err in conceptualizing and providing a MUCH better approach!

Lnewqban
nitsuj
Yes, about 4000g, but g is a unit of acceleration, not force.
Your force figure assumes the entire mass of the tire is a point object traveling around the edge, where in fact it is distributed evenly around what is assumed to be an infinitely thin perimeter.
The force on this tire is thus the tension resulting from the spin, not any hypothetical centripetal force exerted by a string with a mass twirling at the end of it.
Thanks! same comment for you as Drakkith.

Fantastic explanations imo

Lnewqban
Homework Helper
Gold Member
That is the reason behind the need to carefully balance these little rotating parts, just like RC airplane propellers.
Also, one of the reasons for the speed limitations provided by manufacturers of real tires.
https://en.wikipedia.org/wiki/Tire_code#Speed_rating

nitsuj
Gold Member
(if so omg that is some strong glue holding the tire to the wheel lol)

To expand on what other have said, 250 g under 40 000 m/s² gives you a force of 10 000 N. If we assume we have a glue that can hold 3800 lb/in² (26x106 N/m²) and that your wheel has a diameter of 2.5 in (i.e. a perimeter of 0.050π m), the width of glue needed is:
$$\frac{10000 \text{ N}}{(26\times 10^6\text{ N/m²}) (0.050\pi \text{ m})} = 0.0025 \text{ m}$$
So a 2.5 mm stream of glue (or 2 times 1.25 mm on each edge) should be enough to hold your tire on the wheel.

nitsuj and Lnewqban
nitsuj
holy snappers! I think you guys see the physical world much more clearly than me.

is the force needed to turn the wheels, while they're rotating at the 8500rpm VERY high compared to when the wheels are not rotating? What I imagine is the speed with which the servo could change the angle of the rotating wheel would be the only difference. (I guess the "resistance" is changing the angle of momentum of the wheel (sorry if messing up wording / terms)

For example even a low torque servo could change the angle of the wheels; just more slowly than a higher toque servo.

Lnewqban
nitsuj
That is the reason behind the need to carefully balance these little rotating parts, just like RC airplane propellers.
Also, one of the reasons for the speed limitations provided by manufacturers of real tires.
https://en.wikipedia.org/wiki/Tire_code#Speed_rating
Yuppers, even at MUCH lower speeds the balance matters, at 8500rpm it is absolutely CRUCIAL mechanically, more so than the mass of the wheel / tire combo imo.

The tires am using are rated for right around 100mph. I believe some have gone 120+ with them.

they're "speed ratings" for RC tires too; of course not safety standards level of certification but yea there is a distinct difference in tires meant for normal RC speeds and "speed run" type tires. (and for decades have compound ratings as well; depending on manufacturer) Generally speaking the grouping is intended for high speed on-road use or not. Not like cars with s / z / or which ever indicating incremental increase in speed capability.

Lnewqban
Staff Emeritus
2022 Award
is the force needed to turn the wheels, while they're rotating at the 8500rpm VERY high compared to when the wheels are not rotating?
No, any amount of force, at any speed, will accelerate the wheels as long as it's enough to overcome friction. A very small force applied over a long enough time can spin the wheels up to a speed at which they simply come apart under the stress.

nitsuj
Gold Member
My post assumed the tire was not mounted, and was a simple rotating ring held together by tension, which I figured was around 1560 N for the specs you gave. 4 inches is a big skateboard wheel, and the stresses on such a wheel (effectively a solid cylinder) at those speeds would be quite different than my ring, and different again from the computation others gave for a weak tire glued to a rim that takes all the stresses.

Read car tires often have steel belts which take up the tension of high speed so that the glue isn't what prevents the tire from exiting the rim.

Changing speed of a tire at non-relativistic speeds is not a function of the speed at which it is rotating. A certain torque will result in a certain angular acceleration whether or not the tire/wheel is already rotating or stopped. If the bearings are good, there's not much friction slowing a 4 inch wheel, but a vacuum would still be better.
Yes, they show videos of various wheels (car tires, fidget spinners, etc) coming apart when the RPM gets too high.

Turning it (getting it to rotate in a different direction, such as a car turning a corner) does involve significant stresses if the wheel is already rotating due to gyroscopic effects, absent for a stationary wheel.

So it will take noticeable effort to reorient the spin axis of say your 4 inch wheel spinning on a gimball.

nitsuj
No, any amount of force, at any speed, will accelerate the wheels as long as it's enough to overcome friction. A very small force applied over a long enough time can spin the wheels up to a speed at which they simply come apart under the stress.
Opps, sorry about my wording. by "turning the wheels" my meaning is the steering action; changing the angle of the wheel/tire that is spinning so fast. The thing that powers the steering is called a servo.

Like the classic bicycle wheel demonstration where there is resistance when changing the angle of the spinning wheel.

I cannot picture / intuit if at 4000g this start to become too much for a servo to be able to steer the wheels.

Staff Emeritus
2022 Award
Opps, sorry about my wording. by "turning the wheels" my meaning is the steering action; changing the angle of the wheel/tire that is spinning so fast. The thing that powers the steering is called a servo.
Hmmm. Honestly I don't know. Offhand I don't think so, but my only experience is with my own vehicles. Perhaps there's some gyroscopic effect that I'm unfamiliar with that resists turning the wheel.

nitsuj
nitsuj
Hmmm. Honestly I don't know. Offhand I don't think so, but my only experience is with my own vehicles. Perhaps there's some gyroscopic effect that I'm unfamiliar with that resists turning the wheel.
oh wow you have remote control car toys also?!

I guess that's best way for me to know, is to try it out myself. I could put the car on a stand-secured and get the wheels up to speed and see for my self...I guess even disconnect servo arm and feel it for myself. :D yay "science" lol

Staff Emeritus