6.3.3 Over-the-hump

Suppose a car of mass m is going over a hump at high speed v. The only two forces acting on it is the downward weight mg and the upward normal contact force N. Is ? Of course not. It would have meant net force is zero and the car travelling along a straight line.

Critical Speed

Seeing the curved path as a circular motion, we realize that a resultant centripetal force in the downward direction is required. If the radius of curvature at the top of the slope is r, we can write

\begin{aligned}({{F}_{{net}}}&=ma)\\mg-N&=m\frac{{{{v}^{2}}}}{r}\\N&=mg-m\frac{{{{v}^{2}}}}{r}\quad \cdots (1)\end{aligned}

From equation (1), it looks like if v is too large, N must become negative. This means that mathematically speaking, the hump must provide a downward normal contact force on the car, if it were to keep the car in circular motion. But this is physically impossible. There is no way the hump can “suck” the car downward.

This suggests that there is a critical speed above which the car will not be able to complete the circular motion. This critical speed vc can be derived by putting N=0 in equation (1). So


Why is there a critical speed?

Basically, vc is the speed at which the required centripetal force is equal to mg. At speed lower than this, the car’s inertia (and gravity) would have carried it along a parabolic path (drawn in red) that cuts inside the circular motion (shown by the dashed line). But the hump is in the way. So it would naturally step in to provide the amount of normal contact force required to keep the car in circular motion.

On the other hand, if the car is traveling at speed higher than vc, its inertia (and gravity) would have carried it along a parabolic path (drawn in blue) that cuts outside the circular path (shown by the dashed line). Since this path does not cause the car to press against the hump, the normal contact force is zero. The hump has no problem with the car becoming air borne.

Video Explanation

Car Driving over Hump

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