A racecar with a mass of 1500 kg is drivin around a turn that has a radius of 25m . If the coefficient of friction btw the road and the tires is 0.6, how fast can the car go without skiddin
2007-05-17 11:54:19 · 4 answers · asked by damigurl05 1 in Science & Mathematics ➔ Physics
Any object following a circular path is subject to a centripetal force, this is the force that makes it continuously change direction to follow the circular path. In this case that force is provided by the friction between the tires and the road. the kinematics formula for centripetal force is m*v^2/r where m=mass, v=velocity and r= radius. This force has to be equated to the frictional force which is x*N where x is the friction coefficient (greek letter mu is usally employed) and N the normal force, you dont mention an inclination angle so I assume the road is flat, then that normal force turns out to be the car's weight, that is, m*g where m=mass and g=gravity acceleration.
Centripetal Force=Frictional Force
m*v^2/r=x*m*g solving for v..
v=sqrt(x*g*r) with numbers v=12.13 m/s
2007-05-17 12:28:02 · answer #1 · answered by Jay 2 · 1⤊ 0⤋
Five meters per hour but that sharp a turn would be the same for any vehicle because the turn is so sharp. skidding out would happen at about 15. The Movie Tokyo drift makes you think that in a sliding drift you can go much faster. The truth is the speed is not that much more.
2007-05-17 19:19:31 · answer #2 · answered by Pablo 6 · 0⤊ 0⤋
500 miles per hour
2007-05-17 18:56:11 · answer #3 · answered by oddball_0123 2 · 0⤊ 0⤋
good question
2007-05-17 18:56:54 · answer #4 · answered by Anonymous · 0⤊ 0⤋