What is the maximum speed the car can have as passes the highest pt of the bump before losing contact withroad

Question

A car with mass 938 kg passes over a bump in a road that follows the arc of a circle of radius 50.4 m. The acceleration of gravity is 9.8 m/s^2.

Answer 1

Impossible to calculate.

you didn't count for spring and shock absorber.  They really determine the ability of a car to hug the road.

Good Luck...

Answer 2

The critical point is where the force up due to inertia equals the force down due to gravity. The other answers are incorrect under the assumption that the curve is smooth. A spring and shock combo are only effective on impulse. This isn't really a bump at all, but a hill. A smooth curve will result in them extending slowly until the tires lose contact with the ground.

F=ma is the formula for the force in terms of the product of the mass times the acceleration.

a = v^2/r Is the formula for the acceleration felt through a curve at the square of the velocity divided by the radius.

The force due to gravity is simply:

F = 938 * 9.8

The force due to inertia is simply:

F = 938*v^2/50.4

Set the two forces equal and the mass cancels out

9.8= v^2/50.4

v^2 = 493.92

v = sqrt(493.92)

v = 22 m/s or 50. MPH

Answer 3

You need to know the effect of the springs and shock absorbers to calculate.

Answer 4

centripetal acceleration = v^2/r

(sqrt(9.81*radius)) = velocity