Conceptual question about rotational and translational kinetic energy

Question

My real life problem is to calculate initial translational and angular velocities of a vehicle in a loss of control to a stop. (The vehicle will translate and rotate about it's center of mass.)

Initial strategy is to use energy-work theorem, therefore:

$K = W$

where $K$ is the initial kinetic energy and $W$ is the work due to friction (F) between the tires and the road surface (assume that's the only external force at play), therefore:

$\frac 1 2 mv^2 + \frac 1 2 I\omega^2 = \int F\,ds + \int \tau\,d\theta$

Assume I know how to calculete the RHS of the equation. The problem with this, is that i have two variables($v$ and $\omega$) with only one equation.

The question is: Is it valid to write two different equations, one for rotation and the othe for translation, as following?

$\frac 1 2 mv^2 = \int F\,ds$

$\frac 1 2 I\omega^2 =\int \tau\,d\theta$

This way i would be able to solve for $v$ and $\omega$, but I am not so sure i can do it without violating some underliying principle..

Answer 1

I would look at it like this

Let's call the friction force of wheel a, Fa

then the contributory effect of this force is two parts one part imparting negative acceleration and the other part which is the negative torque resisting car spin, $T=F_a*da \ $ with "da" being the distance of the tire to the CG of the car.

This will together with the other 3 wheels give a set of four differential equations matrix.

$$F_adt=-(m \alpha+I*da)dt$$

Answer 2

To keep it simple: You can calculate the final kinetic energy from the known final values of translational speed and angular velocity. I guess they are not independent, but have a decidable dependency equation. For ex. the angular velocities of car tyres and other rotating parts of the car are strictly connected to the driving speed if there's no slipping nor sliding.

If a vehicle accelerates there's some slipping and sliding and also air resistance and other energy dissipating frictions in car mechanics. And there can be some climbing upwards along the road elevation profile. They eat a part of the mechanical power, so the work integrals very likely are together more than the kinetic energy calculated from the final translational speed and angular velocities of the rotating parts.