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I am confused by the nature of the basic physics that occured while watching the MythBusters Knightrider Ramp Experiment. To summarize the experiment, they wanted to see if a car could safely drive up a moving ramp (attached to a semi truck) as performed in the hit 1980's television show. (I didn't think they could fake that stunt in 1980, so I assumed it was possible).

The most common misconception is that the car will experience the "conveyer belt effect", which can be seen when a person speeds up while walking on a moving conveyer, the person's walking speed is added to the speed of the conveyer belt. In this case, there is concern that the speed of the car will be added to the speed of the semi-truck and the car will catapult to the front of the semi-truck trailer. The main problem is that this would invariably damage the whoosh whoosh lights in the front, rendering the vehicle virtually useless, leaving countless poor and usually goodhearted people to be preyed upon by evil chemical treatment plants and ruthless land developers.

However, lucky for KITT, the slow motion cameras show that the wheels on the vehicle slow down almost instantaneously, slowing dow the car and allowing safe entry into the back of the semi-truck. I suppose I could buy that KITT's advanced circuiry allows him to calculate the needed braking in order bring the car to a safe stop, but let's assume that this is not the case. I am confused how and why the tires slow down so dramatically, without some sort of serious new frictional coefficient. What is the basic algebra of this equation? What forces the tires to slow? Please elaborate more beyond was mentioned in the video clip. They mention intertia, but it is not clear which objects are in reference to the others.

Let's take this one step further: what would happen if an actual conveyer belt were attached to the ramp and was moving forward while the car drove up it?

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  • $\begingroup$ Is the car front or rear wheel drive? $\endgroup$
    – user108787
    Oct 9, 2016 at 10:16

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Mythbusters

Let $v_c$ be the speed of the car (and thus the wheels) and $v_r$ the speed of the ramp.

Assuming the wheels roll without slipping, then (where $\omega$ denotes angular velocity):

$$v_c=R\omega_0$$

When the car starts moving up the ramp the relative speed of the car w.r.t. the ramp is:

$$v_{rel}=v_c-v_r$$

Still assuming the wheels aren't slipping as the car moves up the ramp, this causes a friction force $F_f$ to arise and the resulting torque will reduce the angular velocity $\omega_0$ of the wheel:

$$F_f\times R=I\frac{d\omega}{dt}$$ And: $$\omega(t)=\omega_0-\frac{d\omega}{dt}t$$

'CountTo10's' question "Is the car front or rear wheel drive?" is relevant here as with a rear wheel drive, the front wheels will only be free wheeling and their angular deceleration will not significantly affect the car's speed. But it will affect the car's speed once the rear wheels are in contact with the ramp.

The friction force $F_f$ is usually modelled simply as:

$$F_f=\mu F_N$$

Where $\mu$ is a friction coefficient (here the 'rolling resistance') and $F_N$ the Normal force.

As long as the car isn't fully on the ramp yet, $F_N$ is a bit hard to model but is approximately:

$$F_N\approx\frac14mg\sin\alpha$$

Once the car is fully on the ramp it will be subject to gravitational deceleration (regardless of any braking force applied, of course) of:

$$a=-g\sin\alpha$$ So that relative speed (again, without accounting for braking applied by the driver) is given by:

$$v_{rel}(t)=(v_c-v_r)-g(\sin\alpha) t$$

Let's take this one step further: what would happen if an actual conveyor belt were attached to the ramp and was moving forward while the car drove up it?

The reasoning remains much the same. Call the speed of the conveyor belt $v_{belt}$ and assume its vector points in the same direction as $v_c$ and $v_r$, then:

$$v_{rel}=v_c-v_r-v_{belt}$$

In the 'special case' where $v_c=v_r+v_{belt}\implies v_{belt}=v_c-v_r$, no friction force would arise. That would kind of be the 'ideal' way of 'parking' a car inside a truck, as it would be the same as driving uphill for a short distance.

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  • $\begingroup$ Can you add a quick bit about the differences in FWD, RWD and AWD in terms of what will happen before, during and after the moment the vehicle makes contact with the ramp?... just in case I'm wrong in my assumptions. I am guessing AWD will behave the same as FWD. $\endgroup$
    – user58446
    Oct 18, 2016 at 16:38

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