Question

*A car is going through a dip in the road whose curvature approximates a circle of radius 150m. At what velocity will the occupants of the car appear to weigh 15% more than their normal weight

Answer 1

Answer:

v= 14.85 m/s

Explanation:

• When at the bottom of the dip, 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:

       $F_{cent} = F_{n} - F_{g} (1)$

• 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:

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

• Now, we need to find the value of v that makes Fn, exactly 15% more than the weight m*g, so we can write the following equation:

      $F_{n} = 1.15*F_{g} = m*\frac{v^{2}}{r} +F_{g} (4)$

• Replacing Fg by its value, simplifying, and solving for v, we get:

       $v = \sqrt{0.15*g*r} = \sqrt{9.8 m/s2*0.15*150m} = 14.85 m/s (5)$