Acar goes around a curve on a road that is banked at an angle of 34.5^\circ . even though the road is slick, the car will stay on the road without any friction between its tires and the road when its speed is 18.0 m/s. what is the radius of the curve? a car goes around a curve on a road that is banked at an angle of 34.5^\circ . even though the road is slick, the car will stay on the road without any friction between its tires and the road when its speed is 18.0 m/s. what is the radius of the curve?

A car goes around a curve on a road that is banked at an angle of 30.0 . Even though the road is slick, the car will stay on the road without any friction between its tires and the road when its speed is 21.0 m/s 

What is the radius of the curve? 

Go to the web site below for a diagram of a maximum velocity on banked curve with friction problem!! 

http:// 
hyperphysics.phy-astr.gsu.edu/hbase/mec... 

Force centripetal force = mass * (velocity^2 ÷ radius) 

The centripetal force is a horizontal force directed toward the center of the curve. You need to determine the sum of the forces that are directed toward the center of the curve. 

The road exerts a force on the tires of the car. This force is perpendicular to the inclined surface, and is labeled N. 

For the car to remain at the same vertical position, the sum of the vertical forces must equal 0 N. 

Without friction: 
The horizontal component of the Normal force = N * sin θ 

Eq. #1 
m * (v^2 ÷ r) = N * sin θ 
m * (18^2 ÷ r) = N * sin 34.5° 

The vertical component of N = N * cos θ 
Weight is a negative vertical force = -(m * g) 
For the car to remain at the same vertical position, the sum of the vertical forces must equal 0 N. 

N * cos 34.5°+ -(m * 9.8) = 0 

Eq. #2 
m * 9.8= N * cos 34.5° 
When you divide Eq. #1 by Eq. #2, m on the left side of the equal sign will cancel, and N on the right side of the equal sign will cancel. 

18^2 ÷ (r * 9.8) = sin 34.5° / cos34.5° = tan 34.5° 
Solve for r 

18^2 = r * 9.8 * tan 34.5 
r = 18^2 ÷ (9.8 * tan 34.5) 

r = 48.10 meters