In Fig. 6-57, a stuntman drives a car (without negative...

In Fig. $6-57$, a stuntman drives a car (without negative lift) over the top of a hill, the cross section of which can be approximated by a circle of radius $R=250 \mathrm{~m}$. What is Figure 6-57 Problem 82. the greatest speed at which he can drive without the car leaving the road at the top of the hill?

Key Concepts

Centripetal Acceleration

This is the acceleration that points toward the center of the circle during circular motion, and it is given by the formula a = v²/R, where v is the speed of the object and R is the radius of the circle. It is a crucial concept for analyzing forces when an object travels along a curved path, as the net inward force must be equal to the mass times this centripetal acceleration.

Gravitational Force

Gravitational force is the attractive force exerted by the Earth on any object near its surface. In the context of an object moving over a hill, this force acts downward and plays a key role in the net force calculation required for providing the needed centripetal acceleration when the object is at the top of the hill.

Circular Motion

Circular motion involves an object moving along a curved path where the direction of the velocity vector continuously changes. In such motion, even if the speed remains constant, an acceleration known as centripetal acceleration is required to change the direction of the object’s velocity, keeping it moving along the circle.

Normal Force and Critical Speed Condition

At the top of a curved path, such as a hill, the normal force is the force exerted by the surface on the object. The critical condition for the object to remain in contact with the road is when the normal force becomes zero. This condition sets the maximum (or critical) speed by equating the gravitational force to the required centripetal force, leading to the equation v = ?(gR), where g is the acceleration due to gravity and R is the radius of the curvature.

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