A hilly road has a 20 m radius top to one of its hills. What is the minimum speed for a car to just leave the road at the top of this hill?
To find the minimum speed for a car to just leave the road at the top of the hill, we can use the concept of centripetal force.
When a car moves in a circular path, it experiences a centripetal force towards the center of the circle. For the car to just leave the road, the gravitational force acting on the car at the top of the hill must provide the centripetal force required to keep the car moving in a circular path.
The centripetal force (Fc) is given by the equation:
Fc = (m * v^2) / r
Where:
m is the mass of the car
v is the velocity of the car
r is the radius of the circular path
In this case, the centripetal force is provided by the gravitational force (Fg) acting on the car:
Fc = Fg
The gravitational force (Fg) is given by the equation:
Fg = m * g
Where:
m is the mass of the car
g is the acceleration due to gravity (approximately 9.8 m/s^2)
Therefore, we can equate the gravitational force (m * g) to the centripetal force ((m * v^2) / r):
m * g = (m * v^2) / r
We can rearrange this equation to solve for the velocity (v):
v^2 = r * g
v = sqrt(r * g)
By substituting the given values, we can find the minimum speed for the car to just leave the road:
v = sqrt(20 m * 9.8 m/s^2)
v = sqrt(196 m^2/s^2)
v ≈ 14 m/s
So, the minimum speed for the car to just leave the road at the top of the hill is approximately 14 m/s.