Question

*A car is going over the top of a hill whose curvature approximates a circle of radius 350m. At what velocity will the occupants of the car appear to weigh 10% less than their normal weight

Answer 1

Answer:

v= 18.5 m/s

Explanation:

• When at the top of the hill, the only force that keeps the car in the circular trajectory, is the centripetal force.

• This force is not a new force, is just the net force aiming to the center of the circle.

• In this case, is just the difference between the normal force (always perpendicular to the surface, pointing upward) and the force that gravity exerts on the car (which is known as the weight), pointing downward.

• So, we can write the following expression:

       $F_{cent} = F_{g} - F_{n} (1)$

• It can be showed that the centripetal force is related to the speed by the following expression:

       $F_{cent} = m*\frac{v^{2}}{r} (2)$

• The normal force, it is called the apparent weight, because it would be the weight as measured by a scale.

• Replacing (2) in (1), and solving for Fn, we get:

       $F_{n} = F_{g} - m*\frac{v^{2}}{r} (3)$      

• Now, we need to find the value of v that makes Fn, exactly 10% less than the normal weight (m*g), so we can write the following equation:

        $0.9*F_{g} = F_{g} - m*\frac{v^{2}}{r} (4)$

• Replacing Fg by its value, simplifying, and solving for v, we get:
• $v =\sqrt{0.1*g*r} =\sqrt{0.1*9.8m/s2*350m} = 18.5 m/s (5)$