OT: Theoretical question on acceleration (deceleration)

[Jon K]

Ok so I was having a talk about wheel size vs. weight with bahnstormer. we came upon an argument over braking.

Assuming the following is the same in both scenarios:
1) Coefficient of friction applied to both wheels
2) Force applied for negative acceleration
3) Initial centrifugal (outward) velocity

Now, he believes that "if u apply the same amount of stopping force to 2 wheels, the lighter one will stop first, regardless of diameter." So he is discounting centrifugal kenetic energy.

I bellieve that a 15" diameter wheel @ 20lbs vs. a 20" wheel @ 20lbs, will accelerate in the negative direction (decelerate) faster than the 20" scenario. It makes theoretical sense since a smaller diameter wheel equates to more torque, and since torque equates to acceleration, it therefore equates to deceleration which is acceleration but in the negative vector. If the smaller diameter is proportional to the lighter larger wheel (1:1, weight:diameter), this should be true.

We just want to settle and see which is true! Input!!!

[Paul in NZ]

the 20 inch wheel would ake longer to stop i reckon cos even though it is the same weight it carries it s weight further outward thereby has more inertia???

[Javier]

in rotating scenarios, momentum has to do with angular speed and weight distribution along the radius. The problem is not quite the total weight but how is it distributed along the radius (the torque concept is quite similar). If all the weight concentrates toward the center of a 20 inches wheel, and to the outwards on the 15 inches, you may have more momentum in the 15 inches. But if you consider homogeneous material (with total weight identical), then actually the 20 inches well will have higher "Moment of Inertia" leading to more kinetic energy at the same angular speed, leading to more decelerating time under same decelerating forces circumstance.

This is a nice link about the subject:

http://online.cctt.org/physicslab/content/PhyAPB/lessonnotes/angularmomentum/lessonangularmomentum.asp

Javier

[George M]

Jon,
Javier said it correctly. The nomenclature you used to define the dynamic isn't correct so hard to respond to your specific questions but will try:
You wrote:
I bellieve that a 15" diameter wheel @ 20lbs vs. a 20" wheel @ 20lbs, will accelerate in the negative direction (decelerate) faster than the 20" scenario.
Ans: No such thing as accelerating in the negative direction. A wheel is either accelerating, static, at a constant speed or decelerating. The moment of inertia or angular momentum is affected by distance from the centroid or center of the wheel.
An example is a figure skater that sticks out his arms to slow down his spin. His overall mass is conserved, i.e. the same, but he increases angular momentum by sticking out his arms analogous to increasing wheel size that is the same weight and due to conservation of momentum, his angular spin velocity decreases.

You wrote:
It makes theoretical sense since a smaller diameter wheel equates to more torque.
Ans: Nope...a smaller wheel diameter does not equate to more torque.

You wrote:
and since torque equates to acceleration, it therefore equates to deceleration which is acceleration but in the negative vector. If the smaller diameter is proportional to the lighter larger wheel (1:1, weight:diameter), this should be true.
Ans: Torque doesn't relate to acceleration or deceleration. In summary, to answer your question....a larger wheel with the same overall mass and mass distribution will be
both harder to accelerate and harder to decelerate or take more braking force to decelerate compared to an equivalent yet smaller diameter wheel.
HTH,
George

[bofh]

I bellieve that a 15" diameter wheel @ 20lbs vs. a 20" wheel @ 20lbs, will accelerate in the negative direction (decelerate) faster than the 20" scenario.

Not that I want to confuse this more, but assuming that the 15" and 20" refer total size, and not just the wheel.

Assuming the that weight is evenly distributed (which makes the math easier) And solveing for rotational energy, which is the amount of energy it takes to spin or unspin something. And assuming the cars with the said wheels are going the same speed. Hmm. This seems actually doable with my meger math ability.

First we figure out the rotational inertial fo each wheel. From here, my units get ugly. Since we started in pounds and inches, we stay in pound and inches.

the formula is I= 1/2 M R^2 for uniform discs. (Yes, I had to look it up. :)

so for the 20" wheel, we get 1000 lbs in^2
and the for the 15" one, we get 562.5 lbs in^2

Inertial for rotating things is by angle, not by distance covered. so to make things a lot easier, I'm going to say we're meassuring 1 rpm^2 (it doesn't matter if we choose one or 313 rpm^2 or radian per sec^2. But 1 rpm^2 makes the math easy, for those who read this far, feel free to use how ever many radians are in a full rotation.)

So 1 rpm^2 moves the 15" tire, um 47.1" and the 20" moves 62.8" ish.

since we were dealing with diameter earlier, we have to solve for actual road speed, so we find out the circumference of each tire. I used 3.14, but feel free to use as many digits as you want. It cancels out anyway.

It takes the 15" tire 1.333 rpm^2 to keep up groundspeed-wise with the 20"

(47.1x1.333 = 62.8x1 for those who are keeping me honest with my math. )

so we take those numbers, and multiply by the corrected rpm^2 for each tire size and we get how much energy it takes to spin the tires to the same ground speed

15" 562.5 x 1.333 = 750 ish. (I'm rounding)
20" 1000 x 1 = 1000

Hmm. some how that doesn't look right. But the math says the 15" will take less force to accelerate to the same ground speed. in fact 3/4 of the force. which is the same ratio to the wheel sizes.

Can somebody better than me, double check that I did that right?

Shaun

/not good at physics.

[Jon K]

based on everything i've learned in physics over the past 2 years, I believe you're close to correct. Although "total overall size" will be closer because a 15" wheel would have a 60 profile tire and a 20" wheel would have a like 25 or 30 profile tire, assumingly. However, this was without tires, in which case your logic, like mine, seems to be correct.

[Jon K]

GeorgeM: When I had physics one of the catch 22's was that there is no such thing as a decelerating object, but an accelerating object with negative vector. And velocity is absolute always greater than zero, even if it were going backwards.

The general RULE OF THUMB is:

If an object is slowing down, then its acceleration is in the opposite direction of its motion.

This RULE OF THUMB can be applied to determine whether the sign of the acceleration of an object is positive or negative, right or left, up or down, etc. Consider the two data tables below.

From this link (http://www.physicsclassroom.com/Class/1DKin/U1L1e.html)


based on that one rather large detail i didn't continue to read your other arguments, as there is relation between torque and angular acceleration, and in that law it explains the smaller the radius of the object subject to angular acceleration, the great the actual torque transfered (when compared to same torque on larger radius).

Another easy visual example is in a RC Car. The cheap RC cars aren't that light, but their motors are cheap and not powerful. If you ever opened one up you'd see a rather large gear driving a smaller gear. This kills top end but increased the torque AND acceleration, while knocking the top speed.