Simple Car Model; Part 2

Last time we ended with these equations:

$$  \dot{\vec{p}} = ( \frac{F(t)}{m } ) * t + C $$



$$ \int \dot{\vec{p}} dt  = ( \frac{F(t)}{m } ) \int ( t + C_{0}) dt $$

$$ {\vec{p}}  = ( \frac{F(t)}{m } )  ( t^{2} +  C_{0}t +  C_{1}) $$



$$ (\dot{x} = \dot{p}\cos (\theta) ,  \dot{y} = \dot{p}\sin (\theta) )$$



This represents a simple 2-wheeled car that is pushed by a force along a 2D plane. Now with this model if we pushed the car with a force it would sail forever along the path of the wheels.

Instead, to make it realistic, we add in friction.  Kinematic friction is really the reason why objects coast to a halt rather than moving forever.

Kinematic friction creates a force - a drag if you will but not drag that is another force that resists the motion of an object moving over another.  When a wheeled vehicle coasts to a stop the friction we are talking about is not the rubber wheels on the road, because that friction must be almost 100% of the normal force (more on this later) to push off the surface and cause motion.

What I am talking about is the friction force created by an object traveling at velocity that resists forward motion.  It is generated in the axles of wheels by the tire rim rotating around the drive axle.  In this case, the kinematic friction is very small.  But that is good and it's a real life force.

I researched online and found some experimental numbers for steel on  greased steel friction that is about what we need.

I got them here http://www.school-for-champions.com/science/friction_coefficient_greased.htm#.U-vXOZUzBBI

Coefficient of Sliding Friction (greased surfaces)
Material 1	Material 2	Static	Kinetic

Steel (mild)

Steel (mild)

-

0.09 - 0.19

Since there are two axle and wheels on our simple car, let's use the high end number once instead of an average number twice.  It would be about the same in the end. $$\mu = 0.19 $$

Let's use Newton's Third Law:
For every action, there is an equal and opposite reaction.

Written another way;  The sum of all forces on an object must be zero.


$$  \sum{F(t)} = 0 $$

Otherwise we could create or destroy energy, we can't so the mass and energy must all be transfered into other ways.

The friction force resists motion so we apply the vector in the opposite direction to the pushing force we apply,  so they subtract.  The nice thing here is that since the push force and friction will always be opposite we can ignore the vector addition and simply subtract them.  That won't happen when we stop applying force, but then the vehicle will coast to a halt.  Which is realistic. That is not to say we shouldn't state the special exception, but that is the reason why in the next equation there should be:

$$  \sum{F(t)} = F_{push} - F_{friction} = 0 = m  \cdot \ddot{\vec{p}}    $$

The equation for friction is  $$ F_{friction} = \mu* F_{normal} $$

which in a 3D plane must be normal towards gravity and in a magnitude that represents the total mass under friction. The wheels push down with the total force of the mass they carry.  So,

$$ F_{friction} = \mu* F_{normal} = \mu* mg $$

Going back to the acceleration equation,

$$  \ddot{\vec{p}}  =  \frac{F(t) - \mu*mg}{m }  $$

$$ \int \ddot{\vec{p}} dt  = \int ( \frac{F(t) - \mu*mg}{m } ) dt  $$ 

Friction is constant as well.

$$ \int \ddot{\vec{p}} dt  =( \frac{F(t) - \mu*mg}{ m } ) \int  dt  $$

$$  \dot{\vec{p}}   =( \frac{F(t) - \mu*mg}{m } )(  t   + C_{0}  )$$

and

$$ \int \dot{\vec{p}} dt   =( \frac{F(t) - \mu*mg}{m } ) \int (  t   + C_{0}  ) dt $$



$$ {\vec{p}} dt   =( \frac{F(t) - \mu*mg}{m } ) (  t^{2}   + C_{0}t + C_{1}  ) $$

We are going to ignore static friction, that is the special force that stops an object from motion just as it lifts off contact.  It is normally higher but for mild steel it was measured in experiment above as negligible.  For steel rims and steel axles this is the friction force added.

We have not solved for any terms, or simplified the equations.  That is the strictly correct way to represent the solutions of an ordinary differential equation. 
Let's make a very generic initial value problem.  Imagine that the simple car always starts off at zero acceleration and zero velocity.  Like real life.

Then the initial value solution is