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 the greatest speed at which he can drive without the car leaving the road at the top of the hill?

Key Concepts

Centripetal Acceleration

Centripetal acceleration is the acceleration that keeps an object moving in a circular path and always points toward the center of the circle. It arises from the need for a net inward force to change the object's direction continuously and is calculated as v²/R, where v is the speed of the object and R is the radius of the circular path.

Newton’s Second Law in Circular Motion

Newton’s Second Law states that the net force acting on an object equals the mass of the object multiplied by its acceleration. In uniform circular motion, this net force is provided by the centripetal force, meaning that m(v²/R) is the required net inward force to sustain circular motion.

Gravitational Force

Gravitational force is the attractive force exerted by the Earth on any object near its surface, given by the product of the object's mass and the acceleration due to gravity (mg). Near the top of the hill, gravitational force provides a significant portion of, or the entire, force needed for centripetal acceleration.

Normal Force and Contact Condition

The normal force is the support force exerted by a surface to prevent an object from passing through it. At the crest of the hill, if the car's speed increases too much, the required centripetal force exceeds the gravitational force, which could reduce the normal force to zero. The condition where the car just maintains contact with the road (normal force equals zero) is used to determine the maximum safe speed.

Transcript

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