Center Cap vs Tire Diameter Rotation

[element80]

Hello all,

Just trying to find a formula and need some help.

If a tire is 22 inches in diameter, what are the RPMs of the outer diameter of a 22 inch tire versus the RPMs of the outer diameter of a 2.5 inch center cap, on a car traveling 30 MPH?

Are they the same, or are the rotations of the diameter of center cap different than the diameter of the tire?

Any help would be appreciated.

 

[DragonPetter]

Hello all,

Just trying to find a formula and need some help.

If a tire is 22 inches in diameter, what are the RPMs of the outer diameter of a 22 inch tire versus the RPMs of the outer diameter of a 2.5 inch center cap, on a car traveling 30 MPH?

Are they the same, or are the rotations of the diameter of center cap different than the diameter of the tire?

Any help would be appreciated.

If a point somewhere on the circumference of the cap revolved a different amount of times than a point somewhere on the circumference of the tire during a minute of rotations, then they would no longer be aligned, and your tire would probably have something wrong with it (if the tire slipped around the hub or if the hub twisted). So you can assume the RPMs are the same, which is a scalar quantity. What is not the same is the distance traveled by each point and the tangential velocity.

 

[DragonPetter]

Here is a little more help:

Tangential velocity = V_{tangent}
Angular frequency = \omega in revolutions/minute, this must be converted to radians/minute for units to work correctly, which 1 rpm = 6.28 rads/min.
Radius = r in inches
speed of car = 30mph, so to make units agree, 30mph = 31680 inches/minute

And we use the formula:
V_{tangent} = \omega * r

If you consider the tangential velocity of the point on the tire that contacts the road, this velocity vector will tell you how fast the car is moving. So,

V_{tangent} = 31680 \frac{inches}{minute} = 11in\;*\;\omega

So now, solve for \omega:

\omega = 31680\frac{inches}{minute} * \frac{1}{11\;inches} = 2880 \frac{rad}{min}

This is in units of radians/minute, so convert to RPM now:

\frac{2880\;rads}{min}\;*\;\frac{1\;revolution}{6.28\;rads} = 458.6 RPMThere is also a much simpler way of computing this, if you consider that when a tire travels distance of its circumference, it has completed 1 revolution, and then you can calculate how many of those circumferences (revolutions) must be traveled in an hour to get 30 miles, and then convert to minutes.

 

[element80]

Thank you for your reply DragonPeter. So the rotations are the same for both , it is just the distance that is effected by the radius.

 

[Naty1]

...it is just the distance that is effected by the radius.

the circumference gets bigger as the radius gets bigger...so an outer point moves thru
a greater circumference during one rotation than does an inner point...hence it has to go faster [v = wr] than an inner point...