How do I turn my bicycle? No doubt a very simple question with an easy answer that's been puzzling me:
If I'm riding my bicycle in the $x$ direction with speed $v$ and turn my handlebars I can end up travelling in the $y$ direction with almost the same speed without having to provide any additional energy. I don't have to come to a halt, so there must be some mechanism coupling my $x$ kinetic energy to my $y$ kinetic energy without the need for a force to decrease one and increase the other separately. But what is this mechanism?
I assume it's the friction of my tires when I turn the wheel but I don't quite see it: this force always seems to act away from my intended direction of travel.
 A: Actually unless you lean over you will tip when you turn the wheel. It is the leaning that changes the direction of travel and the handlebars only provide fine tuning of the motion.
The component of gravity perpendicular to the plane of the bike provides for a force that does no work (since its perpendicular to motion also). So the speed is unchanged, but the velocity (vector) rotates until the lean stops.
You can try turning without turning the handle bars by just leaning and it will work just fine.
A: The answer can be very easy or very complex.
Easy answer: when you turn the wheel, there is a lateral force that appears. This does what the centripetal force does all the time - it makes the bicycle turn.
The actual dynamics of bicycles (why you lean, the shape of the fork, why you can ride without hands) is all quite complex... I gave some details in thie earlier answer
A: Your assumption that it is friction that is responsible is correct. Friction acts on the wheels of the bicycle, in the forward direction. It is important that friction works to oppose relative motion, and thus works in the direction that the wheel is aligned to, and not the direction of the bicycle. 
When you turn the handle bar of the bicycle(let us say you want to turn left), the front wheel is not parallel to the back wheel. The friction on the forward wheel is along its direction, that is, to the left. So, the bicycle is experiencing some net force due to friction to the front-left. This force can resolved in the left direction and the front direction, just like the case of uniformly accelerated circular motion. This explains why the bikes momentum changes.
The reason why you don't change speed is that the turning arc is not perfectly circular. Your question about change in kinetic energy is answered by saying that energy is a scalar and it doesn't matter in what direction you are traveling. 
A: 
I assume it's the friction of my tires when I turn the wheel

You are perfectly right, it is (kinetic) friction.
Here is the force on the front wheel of the bike while turning steadily at different angles:



*

*First pic. No kinetic friction works against your motion.

*Second pic. When turning your handle slightly, you introduce a component of force straight to the side. This is perpendicular to the direction.


Remember, velocity is a vector $\vec{v}$. So is force and acceleration, and $\vec{F}=m\vec{a}$. The (resulting) force points in the direction of the acceleration.
Acceleration is change of velocity. The presence of non-zero acceleration $\vec{a}$ will change $\vec{v}$. Since $\vec{a}$ is perpendicular to $\vec{v}$ at this point, the magnetude (your speed) is not changes, only the direction. So you turn. This is a perfect circular motion and $\vec{a}$ is pointing to the center of the circle.


*Third pic. While turning a little (the direction of $\vec{v}$ changes gradually), gradually turning the handle more keeps the force perpendicular to the front wheel and to the motion (direction of $\vec{v}$). 


You are now turning with a smaller circle.


*Fourth pic. Same story.



this force always seems to act away from my intended direction of travel.

The "force" you feel is actually the change in direction (the acceleration). Usually called "centripetal force"; but this is not a force, just a name for that feeling. 


*

*When travelling straight ahead and braking, you feel thrown forward. It might feel like a force going forward. But it isn't; it is you feeling deceleration. Your body wants to continue its motion, but is stopped and all particles of your body have a velocity forward which feels like the push forward.

*When travelling in a circular path, you feel thrown out of the circle of the same reason. That feeling of "force" is always opposite the resulting force, which means opposite the acceleration, which means radially out of the circle.
A: Let's make a few assumptions: let us model the bicycle as a set of two wheels, each of mass $m$, connected together by a rigid, light bar of length $L$.
Let's assume the bicycle is upright, i.e. its wheels both lie in vertical planes. The front wheel is attached to one end of the bar, $F$, and its vertical plane has been rotated by an angle $\theta$ with respect to the bar, to model a turn of the handle bars. The back wheel is attached to the other end, $B$, and its vertical plane is aligned with the bar.

Alright. Now, to analyse what is going on here, we use the method of Instantaneous Centres. The idea is that the bar is rigid, so therefore its rate of change of rotation $\omega$ must be the same value for each point on the bar. This rotation occurs about a particular point in space, called the instantaneous centre, which I'll call point $I$. So, points $F$ and $B$ rotate about point $I$ in that instance, so the velocities of $F$ and $B$ must be perpendicular to the lines connecting those points to $I$. Therefore, we can find $I$ by drawing lines from $F$ and $B$ that are perpendicular to the velocities, and see where those lines cross.
So, say point $F$ moves in the direction of the wheel with velocity $v_F$, and $B$ does similarly with velocity $v_B$. I'm assuming that friction is strong enough so that there is no sliding of the wheels on the ground.

By trigonometry, we can determine the length of the lines of the triangle: length $FI$ is $\frac{L}{\sin\theta}$, and the length $BI$ is $\frac{L}{\tan\theta}$.
A useful equation to use is a definition of angular velocity:
$$\omega = \frac{v}{r}$$
where $v$ is the velocity tangential to the arc of rotation for a particular point, and $r$ is the radial distance between that point and the centre of rotation, $I$.
Therefore,
$$\omega = \frac{v_F \sin\theta}{L} = \frac{v_B \tan\theta}{L}$$
$$\therefore v_B = v_F \cos\theta$$
In order for the rotation to occur, we need the resultant forces at points $F$ and $B$ to be centripetal forces. Because of this, the only considerable possible forces available are frictional forces.
Therefore, there are frictional forces $P_F$ and $P_B$ that cause the turning (in such a way to preserve kinetic energy in an ideal situation) on the bicycle, acting on points $F$ and $B$ respectively.
The centripetal force has a magnitude of $\frac{mv^2}{r}$, therefore:
$$P_F = \frac{mv_F \sin\theta}{L}$$
$$P_B = \frac{mv_F \sin\theta\cos\theta}{L}$$
and the direction of these forces are towards point $I$.
Note the cases of what happens at $\theta = 0°$ and $90°$.
