A car drives over a hump in the road. The radius of the (circular) hump is 45m. How fast (m/s) would the car need to travel, so that the wheels just barely lose contact with the road?
v^2/R = g
v^2/45 = 9.81
To solve this problem, we need to use the concept of centripetal force. When the car drives over the hump, the centripetal force acting on the car must be equal to the weight of the car in order for the wheels to just barely lose contact with the road.
The centripetal force is given by the formula F = m * v^2 / r, where F is the centripetal force, m is the mass of the car, v is the velocity of the car, and r is the radius of the hump.
The weight of the car, W, is given by W = m * g, where g is the acceleration due to gravity.
Since the car is just about to lose contact with the road, the normal force acting on the car is zero. Therefore, the centripetal force is equal to the weight of the car.
Setting F equal to W, we have:
m * v^2 / r = m * g
Canceling out the mass, we get:
v^2 = r * g
Finally, solving for v, we have:
v = sqrt(r * g)
Substituting the given values of r = 45m and g ≈ 9.8 m/s^2, we can calculate the velocity:
v = sqrt(45 * 9.8) m/s
v ≈ 20.49 m/s
Therefore, the car would need to travel at approximately 20.49 m/s in order for the wheels to just barely lose contact with the road.