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Children Are Source Of Embarrassment

  1. Aug 29, 2005 #1


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    This morning my eight years old daughter asked me: "why is it so difficult to balance on my bicycle when it is not moving?". It was very embarrassing moment, I simply coud not give her a convincing answer.Then I thought of other things which "I" can not explain without using mathematics. So I decided to ask you people for help.
    As well as "wy daughter's bicycle" there is the "spinning top", the issue here "I believe" is how to explain the relation between stability & motion without math?
    The other "more difficult" subject, I thought about, is explaining global & local gauge principles for people with no knowledge of group theory.
    So if you ,like me, can not explain the above, can you list other similar subjects?


    sam :devil: :devil:
  2. jcsd
  3. Aug 29, 2005 #2


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    I think I can explain the bicycle without math, but I don't know if it would be comprehensible to an eight year old. When you lean the bike without the wheels spinning, you have to cause the little bits of rubber on the tire to start moving. This isn't that big a deal because it is just a matter of speeding them up a little. But if the wheels are spinning, then for the bike to tip over would require that you change the direction of the speeding pieces of rubber. The faster they are going, the harder this is to do, just like its harder to turn a speeding car. The result is that the bike is harder to tip over.

    This was an interesting example. It made me think about this stuff, and I think I understand angular momentum better than I did ten minutes ago.
    Last edited: Aug 29, 2005
  4. Aug 29, 2005 #3
    The reason a bike is hard to balance has to do with the center of gravity and the weight distribution. Since the contact area of the tires is pretty small (and the profile of the tires is rounded) the weight distribution from left to right has to be nearly perfect for the bike to stand upright. Imagine a vertical line exactly in the center of the bike (looking at the bike from front to rear). As long as the weight from one side of this line is equal to the other the bike is balanced. When the bike leans to one side gravity pulls harder on that side. If the bike had tires more like a car (flat profile) it would be much more stable and could withstand greater differences in weight from side to side.

    No confusing math, and possibly understandable by a child.
  5. Aug 29, 2005 #4


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    But that doesn't answer the implicit question of why it is not difficult to stay balanced while the bike is moving.
  6. Aug 29, 2005 #5


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    dang kids!!!
  7. Aug 29, 2005 #6
    Ahh kids...they can learn.
    (But will they really?:rolleyes:)

    (<runs and hides away>)
  8. Aug 29, 2005 #7
    The reason the bike does not tip over when moving is because of a gyroscopic property called precession. When the bike is not moving, any lateral force on the bike tire creates a torque about the centre. This torque makes the bike tip. Once the bike is moving, this torque is still there, but since the tire is spinning, the torque applies towarts the whole radius since the tire is rotating. It is NOT because of the threads or centre of mass like stated above.


  9. Aug 29, 2005 #8

    Claude Bile

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    The thing that flips me out is how those Superbikes stay upright when going around corners!

    A spinning wheel is very hard to tip over. This is a neat experiment to do yourself provided you have a spare bike wheel lying around (or are willing to remove one from a bicycle). Spin the wheel then try to tip it over. You will find that when you try to tip it over, the wheel turns about a vertical axis and not a horizontal one. This is why leaning in a particular direction will cause the bike to turn that way (motorbike riders will know what I am talking about :smile: ).

    When I was learning physics in high school, angular momentum was the stuff that spun me out the most, it is not intuitive at all to someone who has not done much physics (unless they ride motorbikes).

  10. Aug 29, 2005 #9

    Well I didn't see that that question was asked, but the reason is a little more complicated. When the bike is moving forward it is said to have momentum in the direction of travel. The faster you move forward on the bike the slower it will fall to one side when it is unbalanced. This slower fall gives us more time to make steering corrections that change the center of gravity of the bike. Many more things come into play like rake, trail, and wheelbase, but thats probably beyond the understanding of most children, and has been discussed here many times in many threads.

    BTW not to be rude, but the post you made read like a bunch of gobbly gook to me. If you got some new understanding from it great, but I don't think it really explained anything.
  11. Aug 29, 2005 #10


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    Thank you for your honesty, I appreaciate it. Let me try to explain better. Look at the wheel as being made up of particles of rubber that are connected together. When the wheels are not moving and the bike is not moving, these particles are not moving. When the bike starts to tip without the wheel spinning, these particles have to start moving. This means they have to speed up, but not by much. If the wheel is spinning, they are already moving. Now if you tip the bike over you will not only have to make the particles tip over with the bike, but also change their direction. This is more difficult to do, just like it requires a lot of force to change the direction of a speeding car. (of course the reason is that the change in momentum is greater, but I am avoiding mathematical concepts). Can you understand what I'm trying to say? I don't take it as a personal offense if you don't, I always appreciate honest criticism.
  12. Aug 29, 2005 #11


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    But you don't explain why. That is basically what you are trying to show.
  13. Aug 29, 2005 #12
    I think the best method of teaching gyroscopic forces is to"

    1. Get one bicycle wheel (Front wheel not the rear, Safer.).
    2. Get two of the screw on foot rests post that attach to bicycle axles.
    3. screw on the foot rests on both sides of the wheels axle.

    The foot rests will act as handles to hold on too.
    Lightly spin the wheel while someone is holding it, Then ask them to move the spinning wheel in any direction to see the differences in Gyroscopic forces.
    When you have somebody holding a spinning wheel and ask them to wobble the wheel side to side they will feel the forces at work.

    Gerald L. Blakley
    Last edited: Aug 30, 2005
  14. Aug 29, 2005 #13


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    I think that is a good demonstration of what exactly gyroscopic forces are and I remember it being helpful in high school. Riding a bike and observing the increased stability is another demonstration. But neither gives an intuitive explaination of why the force should exist.
  15. Aug 29, 2005 #14

    An object in motion tends to stay in motion unless acted on by another force. This is a basic law of physics, and I'm sure I could no more explain why than I could explain why the speed of light is constant.

    Imagine I have 2 balls, and I'm holding them 3 feet from the ground. If I let one go it will fall to the earth rather quickly. Now if I throw the other into the air the same force of gravity will be pulling the ball down but it will take longer for it to hit the earth.
  16. Aug 29, 2005 #15
    Gyroscopic forces have very little to do with a bikes stability as has been demonstrated using counter rotating wheels that cancel out this force. These bikes are perfectly ridable. I believe the gyroscopic force translates into forward momentum, but any other effects are negligible.
  17. Aug 29, 2005 #16


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    Yes, the bikes inertia will tend to slow its tip, but why more so when the bike is moving foward? It has nothing to do with the bike's linear momentum, since all inertial frames are equivalent and it has zero linear momentum in its own frame. The answer has to do with the rotation of the wheels. By the way, did my second explaination get across my point, or was it still incomprehensible? I would like to know because people ask me questions sometimes and I would like to be able give clear explainations.
  18. Aug 29, 2005 #17

    The bike experiences acceleration in the direction of travel so it can not claim to be motionless. And no your post still doesn't make sense to me, sorry. It is in fact much easier to turn the wheel of a car that is moving than one sitting still. Anyone without power steering can attest to that. Given adequate traction a car could make full lock steering turns at maximum velocity.
  19. Aug 29, 2005 #18


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    Once the bike is moving at constant velocity it can claim to be motionless.

    I didn't mean its harder to turn the wheel when the car is moving, I meant it requires a larger external force to turn it. That is a fact. When a particle travels in a circle without changing speed, the magnitude of the force is [itex]m\frac{v^2}{r}[/itex].

    Can you see what I was getting at now? When the speed of the particles of rubber is higher, it requires a greater force to change their direction. Does this make sense?
    Last edited: Aug 29, 2005
  20. Aug 29, 2005 #19
    Constant velocity compared to what? As soon as the bike leans to one side slightly (which is a constant occurence for a bike) it has acceleration in that direction.

    Yes I understand, to reiterate my point; an object in motion tends to stay in motion unless acted on by another force.
    Last edited: Aug 29, 2005
  21. Aug 29, 2005 #20


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    The linear momentum that results from wobbling is quite small. In fact it exists when the bike is at rest with respect to the ground, so it can't answer the question. If the bike moves at 15 mph in a straight line with a wobble superimposed on the path, it is essentialy at rest in a frame moving at 15 mph with respect to the ground in the direction of the bike with no wobble.
  22. Aug 29, 2005 #21


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    Leonhard: I liked your first post best.
    The point is that if you start to fall leftward on a stationary bike, gravity tips the whole thing leftward faster and faster (especially the parts far from the ground). But if the bike tire is moving forward on top, this moves the top of the tire leftward, so the tire turns leftward; the bike moving forward turns left some. Not only does your linear inertia carry you rightward compared to the bike, but it also brings the support leftward - underneath you again.
    This is NOT the case if there's a large counter-rotating inertia attached to the front wheel! If the total angular momentum of the front wheel actually points rightward, starting to fall leftward turns the front wheel to the right - exacerbating the relative leftward lean. Yes, they're "ridable", barely. You turn them by applying shoulder-muscle torque to the handlebars, which pushes the bike into the leftward lean that compensates for your inertia moving relatively rightward (allowing F_centripetal). These are the dangerous machines that are not stable, but having large L_vec they tip over slowly enough that you can "muscle" them back upright. *Much* harder to ride than steering with the back wheel.
  23. Aug 30, 2005 #22
    There was an experiment that was done during WarII that had a cannon fire a round straight out level, a second cannon ball was drop along side the cannon, Both the Cannon ball fired from the cannon and the Cannon ball that was dropped both hit the ground at the same time, This is how the science of dropping bombs in War War II got started.

    Gerald L. Blakley
  24. Aug 30, 2005 #23


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    I know I've seen some long detailed articles on this topic somewhere. The following from Google is representative of what I remember reading, though it's only a secondary source at best.

    http://www.diablocyclists.com/bicyclescience.htm [Broken]

    Another source which provides similar information


    Last edited by a moderator: May 2, 2017
  25. Aug 30, 2005 #24
    Agreed, mine was a bad example. Lets consider what happens when a moving bike falls compared to a stationary one. When we measure the tires profile from the center of the tread to the rim at a 90 degree angle lets say we get 2 inches . When the bike is stationary and falls the contact patch travels 2 inches as we measured. When the tire is moving forward and starts to fall to one side the contact patch does not fall at a 90 degree angle and this increases the distance it has to travel to fall the same amount. This increased distance gives us more time to make corrections and makes the bike fall slower.

    I hope I stated that clearly, it would probably make more sense if I had a diagram.
  26. Aug 31, 2005 #25
    WWII, If a bomber plane was flying 1 mile in elevation It will have to plot a bomb drop 1 mile from the target from 1 mile up, the plot is to fly on course until the trajectory forms a triangle between the bomber plane and its target, Two sides of the triangle must be the same, This is what was discovered in War War II, distance of level flight to target must equal elevation from plane to Target when overhead at any speed. You can also use a perfect sided square as the geometry instead of a triangle if all sides of the square are the same where the plane and target are diagonolly across the points of the square. At level flight, speed or velocity is only used to determine the time it takes for the target to get hit which equals the same length of time the plane will be over head of target if the plane does not change speed or elevation.

    Hope this helps.

    Gerald L. Blakley
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