# Equations of motion of a car with variable mass and drag [closed]

Lately, I've been trying many ways to attain these equations, yet I've not succeeded in any. I'm basically using Newtonian mechanics because it's the most suitable for this case with dissipative forces. Furthermore, I also neglected kinetic friction since ideal rolling is assumed. To sum up, the only dissipative force acting on the car is drag. Also, bear in mind that it is not that simple owing to the fact that the potential energy isn't of the typical form $$mgh$$ since, in this case, the car is travelling on a highway, so the only potential varying is the chemical one proceeding from the combustion of fuel. Having said that, the main problem here is the force due to fuel combustion: I don't know whether to consider it constant or time dependant, and I don't know its expression either. Also, I've never worked with chemical potential, and so, I think we can't make use of $$F = -\frac{d}{dx}U$$. Before providing the main equations/formulae that I'm making use of to try to attain this force, I'll write Newton's 2nd Law in the case of this car. $$m(t)a = F_{fuel} -F_{drag} = F_{fuel}-\kappa v^2$$ Anyways, firstly, I'll present the formula of energy for this car model: $$-\Delta PE = -\eta_{eff} \Delta E_{fuel} = \Delta LKE + n \Delta RKE + E_{drag},$$ where $$\eta_{eff}$$ is the efficiency of the motor, about 25% or $$\frac{1}{4}$$th; $$E_{fuel}$$ is the energy due to the combustion of fuel, and it can be expressed as $$c_f m_f (t)$$, where $$c_f$$ is the specific energy density of the fuel (assume LPG) [J/kg] and $$m_f (t)$$ is the amount of mass of fuel left in the tank (assume a tank with capacity of 30 kg of fuel); $$n$$ is the number of wheels affected by the rotational $$KE$$, $$RKE$$ (I don't know whether to say 2, for the ones affected by traction, or 4 wheels, for the 4 that are rolling); $$LKE$$ is the linear $$KE$$, and $$E_{drag}$$ is the energy due to drag, which can be expressed as $$\int^x_0 \kappa v^2 dx = \int^t_0 \kappa v^3 dt$$.

Thus, the energy formula should reduce to this (assuming wheels are disks): \begin{align*} -\frac{1}{4} c_f \Delta m_f &= \frac{1}{2}m(t)v^2 + \frac{n}{2}I\omega_w^2 + \int^t_0 \kappa v^3 dt\\ -\frac{1}{4} c_f [m_f(t)-m_{of}] &= \frac{1}{2}(m_o + m_f(t))v^2 + \frac{n}{4}m_w r^2\omega_w^2 + \int^t_0 \kappa v^3 dt\\ \frac{1}{4} c_f [m_{of}-m_f(t)] &= \frac{1}{2}(m_o + \frac{n}{2}m_w+ m_f(t))v^2 + \int^t_0 \kappa v^3 dt, \end{align*} where $$m_o$$ is the mass of the car without fuel (1100 kg), $$m_{of}$$ is the initial mass of fuel in the tank (30 kg), and $$m_w$$ is the mass of a wheel.

Moreover, you shall take into account these other expressions too: $$P_m = \frac{dE_m}{dt} = \tau_m \omega_m = \tau_g \omega_g = \tau_d \omega_d = 2\tau_w \omega_w,$$ where the subscript $$m$$ stands for motor, $$g$$ is for gearbox, $$d$$ is the differential, and $$w$$ means wheel. The last torque-ang. velocity relation is different because the torque from the differential is distributed equally (if not taking a curve) on the left and right wheel, so $$\tau_d = \tau_{w, left} + \tau_{w, right}$$, and since $$\tau_{w, left} = \tau_{w, right}$$, we then have $$\tau_d = 2\tau_{w}$$.

Finally, I'll add the initial conditions: $$x_o = 0$$; $$v_o = 0$$; $$a_o =$$ ct..

Having compiled all of these, now I'll attempt to obtain the force:

Knowing that \begin{align*} P_m = \frac{dE_m}{dt} &\approx \frac{d}{dt}\Big[-\frac{1}{4}E_{fuel}\Big]\\ &= -\frac{1}{4}c_f \dot{m_f}\\ &= \frac{d}{dt}\Bigg[\frac{1}{2}(m_o + \frac{n}{2}m_w+ m_f(t))v^2 + \int^t_0 \kappa v^3 dt\Bigg]\\ &=(m_o + \frac{n}{2}m_w + m_f(t))va + \frac{1}{2}\dot{m_f}v^2 + \kappa v^3\\ &= 2\tau_w \omega_w, \end{align*} we get these two equations: $$\begin{cases} 2\tau_w \omega_w = -\frac{1}{4}c_f \dot{m_f}\ ;\\ 2\tau_w \omega_w = (m_o + \frac{n}{2}m_w + m_f(t))va + \frac{1}{2}\dot{m_f}v^2 + \kappa v^3 . \end{cases}$$

Using first equation we get this: \begin{align*} 2\tau_w \omega_w &= 2F_{fuel}r\omega_w = 2F_{fuel}v = -\frac{1}{4}c_f \dot{m_f}\\ F_{fuel} &= -\frac{1}{8}c_f \frac{\dot{m_f}}{v}. \end{align*} The problem is that evaluating the force at $$t=0$$, then $$v(0) = v_o = 0$$, hence, the force at the start is undefined or tends to $$\infty$$.

Now, using the second, we get this other one: \begin{align*} 2F_{fuel}v &= (m_o + \frac{n}{2}m_w + m_f(t))va + \frac{1}{2}\dot{m_f}v^2 + \kappa v^3\\ F_{fuel} &= \frac{1}{v}\Bigg[\frac{1}{2}(m_o + \frac{n}{2}m_w + m_f(t))va + \frac{1}{4}\dot{m_f}v^2 + \frac{1}{2}\kappa v^3 \Bigg]. \end{align*} The problem now is that I can't cancel the velocities because, as I already mentioned, at $$t=0$$, $$v_o = 0$$, therefore, the force would be undefined as in the first case. One possible solution would be to set the initial velocity to a some value close to 0, such as $$v_o = 0.001$$; this way, we would solve both "undefined" forces and we could even cancel velocities. The other way around, would be considering this force constant. I might also need help on finding $$m(t)$$ or $$m_f(t)$$ in terms of $$x$$, $$v$$ or $$a$$ because, if not, I won't be able to reduce the Newton's 2nd law equation to only $$x$$ and its derivatives.

PS: I've actually thought of another way to find $$F_{fuel}$$; it involves torque: \begin{align*} \tau_w = F_{fuel}r &= I\alpha_w\\ F_{fuel} &= I\frac{\alpha_w}{r} = \frac{1}{2}m_w r \alpha_w = \frac{1}{2}m_w a . \end{align*} Is this correct?

• What is the question here? Aug 6 at 18:53
• Sorry, I changed the titled to something shorter and forgot to formulate a question. Aug 6 at 20:17
• You should put the question in the question, not in the title, especially with a long text like that. Aug 6 at 21:38
• I've just done it, please: take a look. Aug 7 at 11:55
• So, if you have a more specific problem then the desire for someone to look at all that and make something useful or of it, then you'll probably get help, but just letting these equations here is a bit rough. What is your specific problem? Aug 7 at 12:08