Rotational Mass and Energy Required

In summary: A lighter tire will almost always require more energy to maintain a constant speed on these types of surfaces.In summary, a heavier tire will generally require more energy to maintain a constant speed on a variety of surfaces.
  • #1
maximiliano
43
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Thinking about rotating mass (for example a bike tire) and the disadvantages / advantages that come with differing weights/masses. Most everybody you speak to are programmed to think that a lighter tire will require less energy, thus, everyone tries to find the lightest tire possible. BUT...I'm not so sure about that conclusion, based on different situations.

Of course, due to Newton's first two laws, the higher the mass of a rotating object, the more energy will be required to CHANGE its speed. So, from a stop or from a lower rotating speed to a higher rotating speed (or higher to lower too I guess), the more massive the item that must be accelerated, more energy will be necessary to accomplish this. That's fine...

Q 1- However, what about holding a steady speed? Assuming no other outside factors such as friction of all types, if I wanted to HOLD a steady 10mph on level ground, would a heavier tire require LESS, MORE or THE SAME amount of energy ?? (let's assume for this question that the only variable in terms of mass/weight is the TIRE...thus the extra weight is at a distance of 13" from the axis).

Q 2- Next... can anyone speak to the relationship between the distance from the axis/axle and the weight of the mass, in terms energy required to make it rotate (change in speed), and also KEEP it rotating (constant speed)? This obviously has something to do with angular velocity since a weight positioned 1" from the hub would be moving slower (all things equal) than one at 13" from the hub. Thus...this more energy would be required to accelerate the one FURTHER from the hub. But again...what about maintaining a constant speed? Same? More energy...less as the weight moves from closer to the axis to further away from the axis?
 
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  • #2
Q2: Both the distance from the axis and the weight make a difference in the tire's moment of inertia. The higher the moment of inertia, the harder it is to change the object's speed of rotation. Increasing the weight increases the moment of inertia linearly, and moving the weight further from the axis will increase the moment of inertia exponentially. Basically, twice the weight = 2x larger moment of inertia, or same weight but at twice the distance = 4x larger moment of inertia.

Q1: A larger moment of inertia will indeed make no difference in maintaining speed (ignoring friction).

However, there is always friction. Most of the power consumed by a bicycle, for example, is due to air resistance. Moment of inertia doesn't matter in this regard. If you run into wind or water or hit the brakes, it will indeed be harder to slow the bicycle, but then it will be equally harder to get back up to your original speed.

However, heavier tires are, well, heavier. This has less, but some effect on the efficiency of the bicycle. The heavier it is, the more friction in bearings and the tires themselves, resulting in a slighly slower bike. Of course, going uphill will also be harder, whether accelerating or not. There are also other effects, such unsprung weight, which might cause issues.

So yeah, if you're riding the bike in space then perhaps the moment of inertia won't matter for maintaining speed. But everywhere else, it WILL matter. It will never help you out in the efficiency department, but iw will almost always hurt you. Basically, the less weight the better.
 
  • #3
maximiliano said:
Q 1- However, what about holding a steady speed? Assuming no other outside factors such as friction of all types, if I wanted to HOLD a steady 10mph on level ground, would a heavier tire require LESS, MORE or THE SAME amount of energy ?? (let's assume for this question that the only variable in terms of mass/weight is the TIRE...thus the extra weight is at a distance of 13" from the axis).

If we neglect friction and assume a perfect wheel, then it requires no energy at all to keep turning at the same speed.
 
  • #4
hmmm, thanks for the responses.

First, part of Lsos' answer "So yeah, if you're riding the bike in space then perhaps the moment of inertia won't matter for maintaining speed. But everywhere else, it WILL matter. It will never help you out in the efficiency department, but iw will almost always hurt you. Basically, the less weight the better."

I see what you're saying. I guess at the end of the day, the extra energy input to get the heavier tire spinning could be partially re-deposited over time riding the bike. The specific example I'm thinking of (and what got me on this), is mountain biking, where you spend maybe 1-2 hours doing nothing but climbing at 3-5mph. In that case, there is nearly zero wind resistance. Almost everybody is buying flimsy little fly-weight tires, because they think it helps them. (but it hurts on the way down, because they are not strong and you end up getting a torn tire). I think it could either HURT them or not matter at all in terms of the CLIMB. Yes, it takes some energy to get the larger/heavier tire moving...but once it is moving, it seems that it would be no worse, and possibly slightly better, than the light tire. (but now as I think further, I guess it really can't be better...because that would require creation of energy from nothing...I think?).

Another example based question- Consider 2 bicycles, that will travel 10 miles in exactly 1 hours, on flat, paved ground (identical bike, wheels, tire tread and width, course and wind). What can you say about the energy input required to do that work, if the only variable is tire weights...listed below?
  • Bike with 3 Kg tire on a 26" wheel
  • Bike with 1/2 Kg tire on a 26" wheel

Same?? Heavier tire require more?? If so, is it a significant amount or miniscule and thus worth having the more durable (if heavier equals durable) tire?

Michael C- yes...I screwed that up by saying "ignore friction". That was dumb. I just meant to ignore the wind and rolling resistance.
 
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  • #5
maximiliano said:
Another example based question- Consider 2 bicycles, that will travel 10 miles in exactly 1 hours, on flat, paved ground (identical bike, wheels, tire tread and width, course and wind). What can you say about the energy input required to do that work, if the only variable is tire weights...listed below?
  • Bike with 3 Kg tire on a 26" wheel
  • Bike with 1/2 Kg tire on a 26" wheel

Same?? Heavier tire require more?? If so, is it a significant amount or miniscule and thus worth having the more durable (if heavier equals durable) tire?

Michael C- yes...I screwed that up by saying "ignore friction". That was dumb. I just meant to ignore the wind and rolling resistance.

The heavier tire WILL take more energy in your scenario, but not much. Like I said, it's mostly air resistance that slows a bike down (on a decent road). In fact, I live in the Netherlands where the whole country is basically the scenario you described: flat, paved, good roads for bikes. I bike 10km to work. Indeed the bikes here tend to be heavier because people ride them daily for transportation and not so much recreation. They have mudguards, chain guards, drum brakes, enclosed gears, everything enclosed and protected and armored...so that you could wear a suit and still ride your bike to work. In the rain. But it doesn't matter much because, again, it's all flat and the performance penalty is minimum.

On a hill though, especially when mountain biking, the extra weight WILL hurt you. YOU have to work against gravity to lift it. Extra momentum might help you for an extra few inches but only if you plan for it and get extra speed on purpose. Realistically how often does that happen when mountain biking? (I honestly don't know). I do know that they try to make mountain bikes as light as possible, and add weight only when they need it for strength.

A heavier tire is less efficient, but as you think the extra momentum MIGHT help in some very limited instances. But at the end of the day, the extra weight is simply an energy-storage device, and not only do YOU have to provide the extra energy for it to store, but in the rare instance when it does help it also has its own extra weight to work against.
 
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  • #6
I'll expand my point:

1. No energy is required to keep a perfect wheel turning at a constant speed.
2. Therefore, if all resistive forces are equal (friction in bearings, wind, rolling resistance...), the energy required to keep a heavy wheel turning at constant speed is the same as that required to keep a light wheel turning at constant speed.

So, on a flat road, once you have got up to speed x, assuming that resistive forces are equal, you will need the same amount of energy, whatever the weight of your wheels.

On an incline you are also working against gravity. If you are climbing at constant speed, any extra weight, in the wheels or elsewhere, will need more energy to lift it.
 
  • #7
The reason for having light wheels is for roadholding. If you have a low 'unsprung weight' then it is easier for the suspension to keep wheels in contact with the road when you hit a bump. It is possible, when the unsprung mass (brakes / hubs / suspension arms) to oscillate - which is the reason for having a damped suspension (it's not only for comfort).
 
  • #8
Lsos and Michael C- thanks again...good answer. I see it works the way I thought. The only thing I'll add is that a heavier tire, for my example (slow-speed mountain CLIMBING) won't matter except to a calculator, at least due to the extra mass being lifted via the climb. Yes, it's extra mass fighting the acceleration of gravity...but it's so small a portion of the pie. For example, a "very heavy" tire might weigh 1.3 Kg; while a "cross country" tire may weight .6 Kg. So, I'm looking at an extra .7 Kg by going with the heavy tire (which will not pinch, tear and will grip like crazy on the way down, plus be more stable). That .7 Kg is part of the total mass being accelerated against gravity. Since I weigh 111 Kg. My bike weighs about 13 Kg. My water and tools go maybe 3 Kg. So, the total mass I'm moving is 127Kg. The extra .7 Kg is less than 1/2 of 1% increase in mass, and would only matter to a calculator, stopwatch or over very very long distances.

So, is the bottom line: lighter tires will always require less energy when traveling (all else being equal); BUT, on level ground, the extra energy to maintain a constant speed is either minimal, or nearly zero (compared to the lighter tire). ...what about on terrain that is "bumpy" thus more resistance? Could the heavy tire pull ahead of the lighter in that situation??

Sophiecentaur- thanks for that too. I never really understood why people in auto racing were always talking about "unsprung weight". I never thought about it or looked it up. Now I know.
 
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  • #9
maximiliano said:
Lsos and Michael C- thanks again...good answer. I see it works the way I thought. The only thing I'll add is that a heavier tire, for my example (slow-speed mountain CLIMBING) won't matter except to a calculator, at least due to the extra mass being lifted via the climb. Yes, it's extra mass fighting the acceleration of gravity...but it's so small a portion of the pie. For example, a "very heavy" tire might weigh 1.3 Kg; while a "cross country" tire may weight .6 Kg. So, I'm looking at an extra .7 Kg by going with the heavy tire (which will not pinch, tear and will grip like crazy on the way down, plus be more stable). That .7 Kg is part of the total mass being accelerated against gravity. Since I weigh 111 Kg. My bike weighs about 13 Kg. My water and tools go maybe 3 Kg. So, the total mass I'm moving is 127Kg. The extra .7 Kg is less than 1/2 of 1% increase in mass, and would only matter to a calculator, stopwatch or over very very long distances.

So, is the bottom line: lighter tires will always require less energy when traveling (all else being equal); BUT, on level ground, the extra energy to maintain a constant speed is either minimal, or nearly zero (compared to the lighter tire). ...what about on terrain that is "bumpy" thus more resistance? Could the heavy tire pull ahead of the lighter in that situation??

Less weight is better, but you obviously need a certain amount in order to get proper strength/ endurance. I don't know how much of a performance penalty .7 kg will make. it doesn't sound like much, although it will be more than simply adding .7 kg on the frame, for example, because of the fact that the tire is rotating (and unsprung, if you have a suspension). And of course there's two tires, so .7 kg all of a sudden becomes 1.4 kg.

However I agree that if a flimsy tire is failing too often then it's clearly not up to the task and probably needs a bit more meat. Although I don't mountain bike so I wouldn't know. Perhaps .7 kg doesn't sound like much, but there's many parts on a bicycle. You reduce .7 here, .1 there, .9 there and over many parts you can end up with significant weight savings. Obviously this all costs money, and there's a point of diminishing returns. It's up to you to decide where that point is.

As for bumpy terrain, the heavier tire will still offer more resistance simply because of more weight and thus friction. Again, it's probably not much.

Only during transitions does a heavier tire have a chance of offering an advantage (in limited instances)...and even then it's dubious.
 
  • #10
I have never used a bike with 'explicit' suspension but is there any noticeable benefit to roadholding after a bump as there definitely is with motor cars? That would / could make a huge difference when madcap rushing down a rutted / stony track. That would spill the shopping out of the basket on the front of my Raleigh Roadster!
 
  • #11
Great thread, here. I've been looking to buy a new mountain bike wheel set, and I've been trying to get to the bottom of how much I should be concerned with dropping weight at the rim (which, naturally is tied to cost). The physics aren't simple, and everyone seems to have an opinion - I'm glad to get some expert input.

I've seen some references that having a lower moment of inertia allow the wheel to spin faster. I assume total (bike and rider) aerodynamics is the dominant resistant force as speeds increase, so having wheels that 'can spin faster' won't really make much difference, since they generally aren't the limiting factor. Is this an accurate assessment?
 
  • #12
This is Physics, guys. If you are concerned with the basic Kinetic Energy and Momentum involved then it is easy to work out (exercise for the student). You've only got to compare the total Momentum Change when accelerating with the change in angular momentum of the wheels. (Or, likewise, the KE involved) to see what proportion of your sweat and tears goes into getting the wheels to rotate. But all terrain bikes can't be made to be all that efficient, can they?
Then you need to decide just how much money you are prepared to spend on the lightest possible wheels that would be strong enough for the particular job.

I guess you could examine other factors like the energy lost when a lightweight wheel flexes as you go over a bump. There would be some surfaces on which this could be very relevant. That sort of information could be difficult to obtain without a test bed but I wouldn't be surprised if the Manufacturers came up with 'convincing' figures that their most expensive wheels were 'the best' for the job. haha. I have this problem in the context of "marine' versions of hardware, compared with camping and home versions.
 
  • #13
SC -

The kinetic energy calcs would be useful, I do think breaking it down to an energy ratio would give me some perspective on where dollars are best spent (besides me losing weight!).

Wheels don't seem to be advertised in the way you reference, likely because they are usually built from several different brands of parts by a third party.

Thanks -

Tim
 
  • #14
It would be easy to find the MI of rim plus tyre. This would tell you whether the rotational energy is significant compared with the whole of the KE.
But, with a bike, there are so many factors which could influence efficiency vs durability vs cost vs flash value. :-)
 

1. What is rotational mass?

Rotational mass, also known as moment of inertia, is a measure of an object's resistance to changes in its rotation. It is similar to mass in linear motion, but for rotational motion.

2. How is rotational mass calculated?

The formula for calculating rotational mass is I = mr^2, where I is the moment of inertia, m is the mass of the object, and r is the distance from the axis of rotation.

3. What is the relationship between rotational mass and energy?

The rotational mass of an object affects how much energy is required to change its rotational motion. The higher the rotational mass, the more energy is required to accelerate or decelerate the object's rotation.

4. How does the distribution of mass affect rotational energy?

The distribution of mass plays a crucial role in determining the rotational energy of an object. Objects with a larger portion of their mass located farther from the axis of rotation have a higher rotational energy.

5. How is rotational energy used in everyday life?

Rotational energy is used in various everyday activities, such as riding a bicycle or driving a car. It is also essential in machines like turbines, which are used to generate electricity. Understanding rotational energy is crucial in designing efficient and effective mechanical systems.

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