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There's a physical effect that I can't seem to figure out.

When you are biking (on a bicycle), you have some speed. Then let's say you stop petaling and make a 90º turn to the right. You slow down, but you still have considerable velocity.

However, if you do the math, you had a velocity of say [10,0] (10m/s in the x direction; 0m/s in the y direction). Then you turn, and now you have a velocity of say [0,8]. That vector requires a total ∆V of [-10,8]. But clearly that's not true because you managed to successfully turn without adding energy.

What causes you to be able to turn without adding energy? Is it the elasticity of the tires? The air resistance? Nothing I can think about seems to be able to answer this.

It has to be some effect of contact with the ground or air resistance, because, in space, you do need to impart that ∆V to affect that change.

Thanks in advance.

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  • $\begingroup$ Very roughly, the kinetic energy of you and your bike as a whole (ignoring the motions of "internal" degrees of freedom such as the petals and wheels) drops from $10^2=100$ to $8^2=64$ (there is a factor $\frac{1}{2}m$ in front with $m$ being the total mass). So you don't need to add any energy, in fact you must have lost some energies, probably due to friction. $\endgroup$ – Meng Cheng Aug 9 '15 at 16:52
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In a collision there are two key points to consider, which I think you're getting a bit confused over.

The first is conservation of energy. In an elastic collision, we have that $\sum_{bodies}{\frac{1}{2}m_{body} |\vec{V_{body}}|^2}$ is constant. Now this doesn't impose any conditions on the direction of $\vec {V_{body}}$, so if initially the bike has velocity (10,0), and then afterwards has velocity (0,10), this is completely allowed if we only consider conservation of energy. If you consider a satellite, they continuously change their direction without any input of energy at all. The reason why bodies in general can't arbitrarily change their direction is due to conservation of momentum, the second consideration in a collision.

In free space, the total momentum of a rocket has to be conserved. This means the rocket can only make its crew compartment move forward (with e.g. positive momentum), by giving its exhaust gas backwards velocity, so the positive momentum of the crew capsule cancels out the negative velocity of the exhaust gases. So for a rocket, the concept of $\Delta$V is very important, as the amount of momentum able to be imparted to the important part of the rocket is strictly limited.

This is very different on Earth, as the Earth itself serves as an essentially infinite sink of momentum. If you fire a gun, the gun pushes back on you, you push back on the Earth through your feet and the Earth gains a microscopic backwards velocity. So the concept of $\Delta$V is not nearly as relevant as you can essentially get as much momentum as you want from the Earth.

This is what happens with the bike when turning. From a conservation of energy point of view there is no problem with changing your velocity vector, as long as it doesn't get longer without any energy input. There is a question of where the momentum comes from to change your velocity, but you can clearly see that it comes from the Earth - If you lean over slightly as you're turning then the normal reaction force on the bike will provide a force that changes the direction of your velocity.

Hope that helped clear up any problems!

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  • $\begingroup$ Turning a bike is not collision $\endgroup$ – paparazzo Aug 10 '15 at 15:20
  • $\begingroup$ @ChrisCundy - ok, so I can see that. But now you're riding along and, in the wheel's frame, the ground is moving backwards with some velocity (and the wheel's spinning to keep up). When the wheel turns, the ground is now moving in a different direction, which is converted to its frame as a (slower?!?) velocity in the direction the wheel wants to spin and a normal velocity (which causes slight elastic mutation of the tire, either causing it to break static friction or have a momentum transfer with earth). How does it actually go about the momentum transfer, and why doesn't the tire slow down? $\endgroup$ – iAdjunct Aug 13 '15 at 14:36
  • $\begingroup$ The loss of energy and the elasticity of the tyre don't have anything to do with the problem in its essentials - you could ride a perfectly rigid bike with no loss of speed at it would still be the same problem essentially. The way you transfer momentum between the bike and the Earth is the same way you do when you walk - you lean slightly, then push against the ground, and the frictional forces allow you to push the Earth away, which transfers momentum. In the case of the bike it's the same, except you're leaning to go into a circular path and the friction stops you falling over. $\endgroup$ – Chris Cundy Aug 13 '15 at 14:53
  • $\begingroup$ So, I understand the basis of the momentum transfer, but I still can't see what's actually doing the transferring. Theoretically, the tire itself is, but I don't know the mechanism by which turning the wheel magically turns into transferring momentum into the earth to affect a turn. $\endgroup$ – iAdjunct Aug 19 '15 at 13:00
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When you turn, there is a friction force between your bicycle tires and the road. That friction force is opposite the direction that your turning tire would like to go (it would like to slide straight), and it is this friction force that causes your bicycle to change directions. It shouldn't be surprising that the friction force did some amount of work on your bicycle (and you) to cause your direction to change, so your final velocity is lower than your initial velocity because the friction work came about by converting some of your initial kinetic energy into a change in direction.

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Your inertia is a vector

mass x velocity

If you are turning 10 degrees right your inertia in the new direction is

Inertia X cosine(10) = inertia X 0.9894

The rest of energy is friction of turning

A wheel on a axle is diffent as straight is bacially the circle

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It has to be some effect of contact with the ground or air resistance, because, in space, you do need to impart that ∆V to affect that change.

You do need a $\Delta V$, but you don't have to have energy to do it. In space or on earth, there is an energy change when acceleration has a component along your velocity vector. When acceleration is perpendicular to the velocity, no energy is required (or gained).

In space, objects can remain in a nearly circular orbit almost indefinitely. As the force of gravity is pulling at right angles to the direction of travel, no energy is transferred between the orbiting craft and the gravitational potential energy.

On earth, when you turn (smoothly), the forces from the ground are at right angles to the direction of travel. This causes the velocity vector to rotate, but not change in magnitude. So energy transfer is not required.

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