# Energy conservation on cornering bicycle

## Summary

The decrease in your potential energy is compensated by the increase in your rotational kinetic energy.

## Setup

First let's analyze a similar scenario in another context to build the intuition needed to understand the original problem. Imagine a horizontal rod spinning around a stationary axis and you are hanging at the farther end of the rod. Let's say that now you want to get closer to the axis of rotation and thus you try to pull yourself closer to the axis, however you feel an opposing force, the centrifugal force (a pseudo force encountered in the rotating frame of reference whose magnitude is $$m\omega ^2 r$$). So you need to do some work against this force to get closer to the axis.

## Angular Momentum

But the question arises, where does this work done go? The answer is that this work done contributes to the change in kinetic energy. How? See, when you changed your position, one thing remained constant throughout the change, which is the angular momentum. This is famously stated as the law of conservation of angular momentum. It implies that the final angular momentum is the same as the initial angular momentum, that is

$$\mathrm d (I \omega)=0\quad \Rightarrow \quad I_{\text{initial}}\omega_{\text{initial}}=I_{\text{final}}\omega_{\text{final}}\tag{1}$$

where $$I$$ is the moment of inertia and $$\omega$$ is the angular velocity. In the above case, since you came closer to the axis, the moment of inertia decreased and thus the angular velocity increased while keeping the angular momentum constant

## Energy

From now on, I will denote "final" as "f" and initial as "i".

However, the energy was not conserved during this change, which was expected since you did some non zero positive work to get closer to the axis. So, the change in the energy is

$$\Delta E =\frac 1 2 I_\text{f}\omega^2_\text{f}-\frac 1 2 I_\text{i}\omega^2_\text{i} \tag{2}$$

Since $$I_{\text{i}}\omega_{\text{i}}=I_{\text{f}}\omega_{\text{f}}$$, thus $$(2)$$ simplifies to

$$\Delta E =\frac 1 2 I_\text{i} \omega_\text{i}(\omega_\text{f}-\omega_\text{i})$$

Now since $$\omega_\text{f}>\omega_\text{i}$$, thus $$\Delta E>0$$. This positive energy change, which was manifested as increase in the rotational kinetic energy, is due to the work done by you against the centrifugal force.

## The bicycle turn case

You'd have already observed the analogy and the connection between the above example and the bicycle turning case. In the turning case, the friction is almost always acting along the center of curvature of the turning circle, so we can safely conserve the angular momentum of the cyclist about the center of curvature (in reality, there would be a tangential component of friction which would reduce the angular momentum and the rotational kinetic energy, however it's negligible for a fully inflated tire and we can ignore it). And again, in this case as you lean, you get closer to the axis passing through the center of curvature, and the moment of inertia of the system decreases, so the final angular velocity increases and so does the final rotational kinetic energy. And thus, you speed up while turning around the circle.

## Difference

However, there is a subtle difference in the bicycle case, there is a torque due to gravity, however this torque does not stop us from conserving angular momentum since it is perpendicular to the angular velocity. This case is analogous to a particle moving in uniform circular motion. The gravity's torque only changes the direction of the angular momentum and does not change its magnitude, thus as long as we are only concerned with the magnitude of the angular momentum, which we are, we need not worry about the gravity's torque. However, if you are interested, then you can refer to Wikipedia's page on precession for further reading. So it is precession which makes your bicycle go around a corner/turn after you tilt it and it is the law of angular momentum conservation which makes your bicycle speed up once you turn it. This scenario is very similar to its translational analog, a ball (connected by a thread) moving uniformly in a 2D circle initially and then you reduce (change) its radius

I'm amazed that I have never asked this question myself. The most likely explanation is that you speed up a little when you lean over to turn (assuming you're not using the brakes).

Especially at low speed, I have often felt like leaning over to do a tight turn seems to cause the bike to speed up, but it never occurred to me that this was a real effect. It would be a good intro physics experiment to try to measure this using a speedometer attached to the wheel and some way of measuring the angle of lean.

Except at low speed, the effect would likely be small because the change in potential energy is not large compared to the kinetic energy of a bike.

Judging from the last line of your confusion, I think you have a slight confusion about the nature of energy. Energy is a scalar quantity and not a vector quantity. The total energy has to be conserved, but not in the directions.

A simple demonstration is a block sliding down an inclined plane. The potential energy decreases in the direction of the field (i.e. straight downwards) but the kinetic energy increases in a direction inclined to the field. Still, kinetic energy and potential energies are conserved.

All that really matters is that the sum of $$E_\mathrm k$$ and $$E_\mathrm p$$ should be conserved at the end of the day (or second, or minute). Now if the velocity of the bicycle increases, it will compensate for the decrease in height of the centre of mass. But as @taciteloquence noted, this decrease is probably very less.

For safe turn,

$$\theta=\tan^{-1}\left(\frac{v^2}{Rg}\right)$$ change in height $$\Delta h=h\left(1-\cos\left(\tan^{-1}\left(\frac{v^2}{Rg}\right)\right)\right)$$ change in potential energy $$mg\,\Delta h=mgh\left(1-\cos\left(\tan^{-1}\left(\frac{v^2}{Rg}\right)\right)\right)$$

For realistic values of $$v=5.6\ \mathrm{m\ s^{-1}}$$, $$R=4\ \mathrm m$$ and $$h=1\ \mathrm m$$, you will get at most, an increment of $$0.43\ \mathrm{m\ s^{-1}}$$ which is probably unnoticed. This is further reduced if friction is present to help you.