Does turning sharply on a bicycle conserve more energy than a wide turn? I use a bike to commute, so I spend a lot of time thinking about how to get the most bang out of my momentum.
Aside from the extra distance traveled in a wide turn, does making a sharp turn save you any energy? My guess is no, because these things tend to even out, but it definitely feels like I'm going much faster. I've even considered that maybe taking a turn sharp is worse, because the extra pressure will cause more friction in the bearings and the tires.
Either way it's more fun, though that probably doesn't get a term in the equations.
 A: So this will take a really simplistic look at it, ignoring things like flexibility in the tires/wheels/bike and assuming that you don't go too fast to slide out. 
The work done to turn is the force to turn times the distance of the turn. The force is $mv^2/R$ where $m$ is the mass of the system, $v$ is the speed of the turn, and $R$ is the radius of the turn. So the force gets higher as either the speed increases or the turn gets smaller, as expected. The distance of the turn is the arc-length of a circle (assumption of course) of the radius, which is $d = \theta R$. 
This means the work done is 
$$W = \frac{mv^2}{R} \theta R$$
Or, in other words, if you aren't going to slide out and you can generate the forces you need to make the turn, it doesn't actually matter what radius you choose, the work done is the same either way. 
Now this is where the assumptions become obviously violated. We can't turn instantly ($R = 0$) because there isn't enough friction to keep us upright. And the amount of friction increases as the turn gets sharper. The turning friction dominates the friction along the arc-length (unless your tires are slipping riding in a straight line), so all of this comes together to imply that you save energy by taking a bigger turn faster than a narrower turn slower. 
But you obviously want to take the tightest turn you can at the fastest speed you can without crashing to maximize everything.
A: I don't think your question can be answered without looking in detail at the engineering mechanics of the tyres, bearing friction and losses in any skid. 
The centripetal force for an object is normal to the velocity vector, so does no work. The only work done when there is acceleration along the tangent vector $\hat{\vec{t}}$ to the path:
$${\rm d}_t \vec{v} = {\rm d}_t (v\, \hat{\vec{t}}) = \dot{v} \hat{\vec{t}} + \kappa\,v^2\,\hat{\vec{n}}$$
the second step following from the Frenet–Serret formula (here $\hat{\vec{n}}$ is the unit normal and $\kappa$ path's local curvature), so that instantaneous rate at which energy is input to / lost from the bicycle/rider system is:
$$p = m\, \left<\vec{v},{\rm d}_t \vec{v}\right> = m\, v\,\dot{v} = \frac{m}{2}\,{\rm d}_t(v^2)$$
(whence the kinetic energy formula, btw: you can see it's quite general). To my mind, the only power transfer if the speed is constant is the power dissipated irreversibly deforming the tires and in bearing friction, so this is not a trivial question. Probably the first mechanism will be the main one: to understand this, witness that there must always be either skid or local deformation of a tyre with a nonzero contact area, a proposition whose truth you can grasp by thinking about a nonzero length cylinder rolling on a curved path so that the velocity vector is tangent to the cylinder's axis of rotational symmetry. The relationship $v = \omega\,R$ where $R$ is the radius of curvature can hold only at one plane normal to the cylinder's axis, if the cylinder is rigid (so that $\omega$ is the same everywhere). Put it another way, if no part of the cylinder is skidding, and if $R$ is the radius of curvature of the path of the cylinder's centre of mass, then the angular speed of the plane in the cylinder a distance $\ell$ along the axis would need to fulfill $v(\ell) = \omega(\ell) (R + \ell)$ if each plane cross section stays underformed. This cannot happen indefinitely as the cylinder would undergo torsion and fail from torsional shear. So in practice, what happens in a nonskidding tyre is that the rubber on the inside edge gets deformed so that its angular speed is slower than the rest of the rubber on the same edge, then, when it leaves the ground, it flicks forward to "catch up" with the rest of the rubber. An analogous description holds for each tyre cross-section aside from the one fulfilling $v = \omega\,R$. Moreover, there is a cyclic, oscillatory torsion about the tyre's radius (i.e. orthogonal to the wheel's rotation axis) as the rubber contacts the ground, stays so fleetingly and then lets slip its grip. So the tyre at most of its points is in a constant stretch-shrink deformation cycle locally with period $2\pi/\omega$ and this is what dissipates most energy in turns and also wears out the tyres.
You can experimentally detect signs of this complex deformation by taking your car to one of those indoor carparks where the concrete is very smooth, ideally painted. Open your driver's window and drive as slowly as you can in a tight bend. You'll find that you can hear a loud, squeaky noise from the forward end tyres no matter how slowly you go as the rubber is deformed (and undergoes little skids) whilst it passes through the region in contact with the ground. Maybe do this experiment next time you take your children shopping, so you can witness the hilarious mixture of bewilderedness, embarassment and "what the **** is Papa doing?!" on your chilren's faces as they watch you with you with your head out of one window straining to be sure you can confirm the source of the noise, driving at the same time and with the vegetables and groceries poking out the other open window.
