1) A stuntman drives a car (without negative lift) over the top of a hill, the cross section of which can be approximated by a circle of radius R 280 m. What is the greatest speed at which he can drive without the car leaving the road at the top of the hill?
F normal zero at the top so mg - Fn m v^2/r so g v^2/r and v root(gr) work it out.
If my overweight sister-in-law is the driver, I would be surprised to see the hill could stay up, as the car reaches the top.
Use the equation for uniform circular motion. In order to travel in a circle at constant speed, the acceleration needs to be: a v^2 / R Acceleration in this case is provided by gravity. Going over the hill, the net force has to be (mv^2/R) downward. Gravity provides a downward force of mg. If that's greater than mv^2/R, the hill exerts a Normal force upward to reduce the net force. But the reverse is not true - if gravity is not sufficient to make a tight enough circle, the hill cannot exert a downward normal force. The car's momentum lifts it off the descending road. The maximum speed that will allow the car to stay on the road is thus the speed where v^2/R exactly equals g, and the normal force is 0.
F normal zero at the top so mg - Fn m v^2/r so g v^2/r and v root(gr) work it out.
If my overweight sister-in-law is the driver, I would be surprised to see the hill could stay up, as the car reaches the top.
Use the equation for uniform circular motion. In order to travel in a circle at constant speed, the acceleration needs to be: a v^2 / R Acceleration in this case is provided by gravity. Going over the hill, the net force has to be (mv^2/R) downward. Gravity provides a downward force of mg. If that's greater than mv^2/R, the hill exerts a Normal force upward to reduce the net force. But the reverse is not true - if gravity is not sufficient to make a tight enough circle, the hill cannot exert a downward normal force. The car's momentum lifts it off the descending road. The maximum speed that will allow the car to stay on the road is thus the speed where v^2/R exactly equals g, and the normal force is 0.
