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 250 m. What is the greatest speed at which he can drive without the car leaving the road at the top of the hill?The answer is 178 km/h, but I don‘t know how to get this answer.
What a lovely, lovely question!!! The only hint here that you need is find the equation for centripetal force. The car will take off when the force is overcome. Don't forget to convert the answer from m/s to km/h!!!!!! Should take you a maximum of 2 minutes to get the answer from reading this.
The centripetal force should be less than gravitational force(here weight of car) for the car to be on the road. if it exceeds then there will be negative lift and the car may fly. so weight centripetal force mgm(v^2)/r vsqrt(gr) vsqrt(9.81*250)49.5 m/s 178 kmph
m * g m * v^2/r v SQRT(g * r) If g 10 m/s2 and r 250m you get V SQRT(2500) 50 m/s 180 km/h If you take g as 9.81 you will probably get the 178 km/h you are looking for.
m * g m * v^2/r v SQRT(g * r) If g 10 m/s2 and r 250m you get V SQRT(2500) 50 m/s 180 km/h If you take g as 9.81 you will probably get the 178 km/h you are looking for.
What a lovely, lovely question!!! The only hint here that you need is find the equation for centripetal force. The car will take off when the force is overcome. Don't forget to convert the answer from m/s to km/h!!!!!! Should take you a maximum of 2 minutes to get the answer from reading this.
The centripetal force should be less than gravitational force(here weight of car) for the car to be on the road. if it exceeds then there will be negative lift and the car may fly. so weight centripetal force mgm(v^2)/r vsqrt(gr) vsqrt(9.81*250)49.5 m/s 178 kmph
