Consider a road over a hill where the top is shaped like a circular bump. This circular arc has its center in the hill below. As a 1200 kg car rounds the top of a hill at a speed of 20.0 m/s, it very breifly loses contact with the pavement. What is the radius of this circular bump?
mass is irrelevant
when centripetal acceleration = 1 g, the vehicle leaves the road
Va = v^2/r = g
so r = v^2/g = 400/9.81
around 40 meters
To find the radius of the circular bump, we can use the concept of centripetal force.
When the car loses contact with the pavement, the only force acting on it is gravity. At that moment, the centripetal force provided by gravity is equal to the force required to keep the car moving in a circular path.
The centripetal force (Fc) is given by the equation:
Fc = (mv^2)/r
where m is the mass of the car (1200 kg), v is the speed of the car (20.0 m/s), and r is the radius of the circular bump.
Since the car just loses contact with the pavement, the centripetal force is equal to the gravitational force (mg), where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Setting the centripetal force equal to the gravitational force, we have:
(mv^2)/r = mg
We can now rearrange the equation to solve for r:
r = (mv^2)/(mg)
Plug in the known values:
r = (1200 kg * (20.0 m/s)^2) / (1200 kg * 9.8 m/s^2)
Simplifying the equation:
r = (480,000 kg m^2/s^2) / (11,760 kg m^2/s^2)
r = 40.8 m
Therefore, the radius of the circular bump is approximately 40.8 meters.