With rolling resistance and air resistance

More realistically, however, we would expect there to be non-conservative forces acting on the trike, such as rolling resistance and aerodynamic drag. When these forces act upon the trike a simple analytical solution becomes less easy to obtain, at least if the angle of descent is allowed to not remain constant. To parametize the problem we will use an independent variable $s$ that will be distance, in metres, from the summit of the hill or mountain. We will assume this mountain has a road that we are travelling down and $s$ is the distance from the summit along this road.

Constant gravity

We will first assume that the variation in gravitational acceleration between the base and summit of the hill or mountain is negligible and derive the equation of motion based on this assumption.

Formulating the Lagrangian

The first step in deriving the equation of motion is to determine the Lagrangian of the system. The Lagrangian contains both the kinetic and potential energy of the system and hence both will be calculated next.

Kinetic energy

Substituting: $\begin{aligned} v=\dot{s} \end{aligned}$ where $\dot{s}=\dfrac{ds}{dt}$ is the trike's velocity in $\mathrm{m}\cdot \mathrm{s}^{-1}$ into Equation (??) yields: $\begin{aligned} T=\dfrac{m\dot{s}^2}{2}. \end{aligned}$

Potential energy

The potential energy term will only include the conservative force of gravitation. To compute it, we need to be able to formulate the height of the trike above sea level in terms of $s$. If we have inclination/declination angle data for the hill in question, let us define $\theta(s)$ as the declination angle (positive when downhill) at $s$ metres from the summit. If we form an infinitesimal triangle on the hill/mountain, we will see that $dh = -d s \sin{\theta}$ and hence: $\begin{aligned} \dfrac{dh}{ds} &= -\sin{\theta(s)}. \end{aligned}$ Integrating from the summit, we obtain that: $\begin{aligned} h(s) &= h_{\mathrm{summit}} - \int_{0}^s \sin{\theta(\sigma)}d\sigma. \end{aligned}$ Where $h_{\mathrm{summit}}$ is the height of the hill/mountain, or rather its tallest point accessible by road, and we use $\sigma$ as a dummy (integration) variable for distance from the summit.

Hence if gravitational acceleration can be assumed to be constant: $\begin{aligned} V &= mg\left(h_{\mathrm{summit}} - \int_0^s \sin{\theta(\sigma)}d\sigma\right). \end{aligned}$

Final Lagrangian

The Lagrangian is, by definition: $\begin{aligned} L &= T - V. \end{aligned}$

Substituting Equations (??) and (5) into Equation (6). Therefore our Lagrangian is: $\begin{aligned} L &= \dfrac{m\dot{s}^2}{2} - mg\left(h_{\mathrm{summit}} - \int_0^s \sin{\theta(\sigma)}d\sigma\right). \end{aligned}$

Non-conservative forces

It is important to note, however, that nonconservative forces — such as air resistance and rolling resistance — are a major factor in determining the speed of a tricycle going downhill. To account for these effects, we need to calculate generalized forces. Fortunately, since our problem only has a single independent variable — that of distance from the summit — calculating the generalized forces should be straightforward.

Rolling resistance is proportional to the normal force. The normal force could be calculated as: $\begin{aligned} N=mg\cos{\theta(s)}. \end{aligned}$ The rolling resistance force is defined as:

Where $C_{rr}$ is the coefficient of rolling resistance. Substituting Equation (8) into Equation (8) yields: $\begin{aligned} F_r &= mgC_{rr}\cos{\theta(s)}. \end{aligned}$ This force directly opposes motion in the direction of $s$.

As for air resistance (or drag): $\begin{aligned} F_D &= \dfrac{\rho C_D A v^2}{2}. \end{aligned}$ Substituting Equation (1) into Equation (10) yields: $\begin{aligned} F_D &= \dfrac{\rho C_D A \dot{s}^2}{2}. \end{aligned}$ Where $\rho$ is air density in $\mathrm{kg} \cdot \mathrm{m}^{-3}$, $C_D$ is a dimensionless drag coefficient, $A$ is the cross sectional surface area (in $m^2$) that the area acts on. This is also in the opposite direction to motion. Hence the generalized non-conservative forces acting on our tricycle would be:

Euler-Lagrange equation with non-conservative forces

The Euler-Lagrange equation with dissipative forces in the $s$ coordinate is: $\begin{aligned} \dfrac{d}{dt}\dfrac{\partial L}{\partial \dot{s}} - \dfrac{\partial L}{\partial s} &= Q_s. \end{aligned}$ The left-hand side of Equation (12) can also be denoted as $\dfrac{\delta S}{\delta s}$, the functional derivative of $S$ with respect to $s$. $S$ is the action and is defined by:

where $t_0$ is the starting time and $t_f$ is the finishing time. First, we shall calculate $\dfrac{\partial L}{\partial \dot{s}}$, the canonical momentum: Hence: $\begin{aligned} \dfrac{d}{dt}\dfrac{\partial L}{\partial \dot{s}} &= m\ddot{s}. \end{aligned}$ As for the second term in Equation (12), it is: So the functional derivative is therefore: Hence Equation (12) becomes: Rearranging and factorizing yields: Dividing both sides by $m$ yields: $\begin{aligned} \ddot{s} &= g(\sin{\theta(s)}+C_{rr}\cos{\theta(s)})-\dfrac{\rho C_D A\dot{s}^2}{2m} \end{aligned}$ assuming constant gravitational acceleration.

Simplification to first order

Hence Equation (14) becomes, after dividing by $m$: Let $u=\dot{s}^2$, then: Hence by the chain rule: Consequently: Multiplying both sides by 2 yields: Adding the $u$ term to the left-hand side: $\begin{aligned} \dfrac{du}{ds} + \dfrac{\rho C_D A u}{m} &= 2g(\sin{\theta(s)} - C_{rr} \cos{\theta(s)}). \end{aligned}$

Solution

We first seek an integration factor, $I(s)$, so that we can re-write Equation (25) as:

This implies that $I'(s)=I(s)\dfrac{\rho C_D A}{m}$ hence: Multiplying Equation (15) by our integration factor yields: The left-hand side is the derivative of $I(s)u(s)$ with respect to $s$: Integrating both sides with respect to $s$ (using the dummy variable of integration $\sigma$) from 0 to $s$ yields: Since the trike starts at zero velocity $u(0)=0$. Hence we have: $\begin{aligned} e^{\frac{\rho C_D As}{m}}u(s) &= 2g\int_0^s e^{\frac{\rho C_D A\sigma}{m}}(\sin{\theta(\sigma)} - C_{rr} \cos{\theta(\sigma)})d\sigma. \end{aligned}$ Multiplying by $e^{-\frac{\rho C_D As}{m}}$ yields: $\begin{aligned} u(s) = 2ge^{-\frac{\rho C_D As}{m}}\int_0^s e^{\frac{\rho C_D A\sigma}{m}}(\sin{\theta(\sigma)}-C_{rr} \cos{\theta(\sigma)})d \sigma. \end{aligned}$

Square rooting we obtain the velocity: $\begin{aligned} v(s) &= \sqrt{2ge^{-\frac{\rho C_D As}{m}}\int_0^s e^{\frac{\rho C_D A\sigma}{m}}(\sin{\theta(\sigma)}-C_{rr} \cos{\theta(\sigma)})d \sigma}. \end{aligned}$ We neglected the negative possibility because velocity is positive in the downhill direction, which is the only direction the trike will go when under the action of gravity.

Below are the velocities achieved in the above model for various $C_D A$ and $C_{rr}$ values when descending Castle Hill. The simulation did not end at 0 metres above sea level; it ended at about 36 metres above sea level, which is approximately at the high end of Gregory Street. $\theta(s)$ values were calculated from QldDem data on elevation.

Non-constant gravity

Gravitational acceleration depends on the distance from the centre of the source of gravity, so there is some variation in gravitational acceleration between the base and summit of hills and mountains although it is minor. In this section, the equation of motion will be derived while accounting for this variability.

Formulating the Lagrangian

The kinetic energy we use when studying non-constant gravity is the same as before. Consequently, Equation (13) is also true for this Lagrangian.

Potential energy

If we cannot assume constant gravity, then: $\begin{aligned} V &=\dfrac{GMm}{(r_{\mathrm{base}} + h(s))} - \dfrac{GMm}{r_{\mathrm{summit}}}. \end{aligned}$ Factorizing: $\begin{aligned} V &= GMm\left((r_{\mathrm{base}} + h(s))^{-1} - r_{\mathrm{summit}}^{-1}\right). \end{aligned}$ Where $G$ is the gravitational constant and $M$ is the mass of the Earth.

Substituting Equation (4) into Equation (20) yields:

Final Lagrangian

The Lagrangian of our system is therefore:

Non-conservative forces

If gravity cannot be assumed to be constant then:

Substituting Equation (20) into Equation (9) yields: $\begin{aligned} F_r &= \dfrac{GMmC_{rr}\cos{\theta(s)}}{\left(r_{\mathrm{summit}} - \int_0^s \sin{\theta(\sigma)}d\sigma\right)^2} \end{aligned}$

Computing Equation (12)

The conservative force term of the Euler-Lagrange equation is therefore: $\begin{aligned} \dfrac{\partial L}{\partial s} &= \dfrac{GMm\sin{\theta(s)}}{\left(r_{\mathrm{summit}} - \int_0^s \sin{\theta(\sigma)}d\sigma\right)^2}. \end{aligned}$ Hence the functional derivative of $S$ with respect to $s$ is therefore:

The time derivative of generalized momentum is the same as above. As for $Q_s$ it is: $\begin{aligned} Q_s &= -\dfrac{GMmC_{rr}\cos{\theta(s)}}{\left(r_{\mathrm{summit}} - \int_0^s \sin{\theta(\sigma)}d\sigma\right)^2} - \dfrac{\rho C_D A \dot{s}^2}{2}. \end{aligned}$ Substituting Equations (22) and (23) into Equation (12) yields: Rearranging and factorizing: Dividing by $m$: $\begin{aligned} \ddot{s} &= \dfrac{GM}{\left(r_{\mathrm{summit}} - \int_0^s \sin{\theta(\sigma)}d\sigma\right)^2} \left[\sin{\theta(s)}-C_{rr}\cos{\theta(s)}\right] - \dfrac{\rho C_D A\dot{s}^2}{2m}. \end{aligned}$

Simplification to first order

$\ddot{s} = \dot{s} \dfrac{d\dot{s}}{ds}$ hence Equation (24) becomes:

Letting $u=\dot{s}^2$, then $\dfrac{du}{ds} = 2\dot{s}\dfrac{d\dot{s}}{ds}$ and so: Multiplying both sides by 2 yields: Adding the $u$ term to the left-hand side: $\begin{aligned} \dfrac{du}{ds} + \dfrac{\rho C_D A}{m}u &= \dfrac{2GM}{\left(r_{\mathrm{summit}} - \int_0^s \sin{\theta(\sigma)}d\sigma\right)^2} \left[\sin{\theta(s)}-C_{rr}\cos{\theta(s)}\right]. \end{aligned}$

Solution

We first seek an integration factor, $I(s)$, so that we can re-write Equation (25) as:

Hence $I'(s)$ must equal $I(s)\dfrac{\rho C_D A}{m}$, which therefore implies that We can replace the left-hand side with $\dfrac{d}{ds}\left[I(s)u(s)\right]$ hence: Therefore Equation (25) becomes: $\begin{aligned} \dfrac{d}{ds}\left[e^{\frac{\rho C_D As}{m}}u(s)\right] &= 2GMe^{\frac{\rho C_D As}{m}}\dfrac{\sin{\theta(s)}-C_{rr}\cos{\theta(s)}}{\left(r_{\mathrm{summit}} - \int_0^s \sin{\theta(\sigma)}d\sigma\right)^2}. \end{aligned}$ Integrating both sides of Equation (26) from 0 to $s$ with respect to $s$ (calling it a new dummy variable $\sigma$; $\sigma$ in our denominator will be renamed $\xi$) yields: $\begin{aligned} e^{\frac{\rho C_D As}{m}}u(s)-e^0 u(0) &= 2GM\int_0^s e^{\frac{\rho C_D A\sigma}{m}}\dfrac{\sin{\theta(\sigma)}-C_{rr}\cos{\theta(\sigma)}}{\left(r_{\mathrm{summit}} - \int_0^{\sigma} \sin{\theta(\xi)}d\xi\right)^2} d\sigma. \end{aligned}$ The trike's initial velocity is assumed to be zero, hence $u(0)=0$: Multiplying by $e^{-\frac{\rho C_D As}{m}}$ yields: Square rooting $u$ to obtain velocity yields: $\begin{aligned} v(s) &= \sqrt{2GMe^{-\frac{\rho C_D As}{m}}\int_0^s e^{\frac{\rho C_D A\sigma}{m}}\dfrac{\sin{\theta(\sigma)}-C_{rr} \cos{\theta(\sigma)}}{\left(r_{\mathrm{summit}}-\int_0^\sigma \sin{\theta(\xi)} d\xi\right)^2} d \sigma}. \end{aligned}$ Below are the velocities achieved in the above model for various $C_D A$ and $C_{rr}$ values when descending Castle Hill. The simulation did not end at 0 metres above sea level; it ended at about 36 metres above sea level, which is approximately at the high end of Gregory Street.