Banked Curves using Radius and speed

In summary, the car can round the curve with zero frictional forces at an angle of 12.1 degrees if the road is banked at that angle.
  • #1
SlainTemplar
12
0
Okay so i have a test and so I'm reviewing for it, and one of the questions i came across was:

1. A curve of radius 180 m is being designed in a new highway to allow cars traveling at 70 km/h to round the curve with zero frictional forces.
a) At what angle must the road be banked?
b) What is the minimum coefficient of friction that will allow a car to successfully go around the curve at 30 km/h?

so i did a, i used Fn = mg/cos theta, and Fn (sin theta) = mv^2/r and substituted for Fn, getting Theta = Tan inverse v^2/gr and i got an angle of 12.1 degrees which is right.

But I'm having a problem getting friction. i tried to assume a mass but that didn't work, and I'm not sure how to start this question. any ideas?
 
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  • #2
Quite complicated! I find it easiest to take the point of view of the car moving in a circle so you have a centrifugal force pushing the car out. And gravitational force down. You have to take the components of each of those forces into the road and along the road. Use the into the road components to get the normal force and the friction force. Then do the forces along the road surface: the inward/downward component will be just equal to the outward/upward components (including friction).
 
  • #3
Would i be safe to assume a mass in order to solve for the gravitational force down?
 
  • #4
Yes. You can do it for each kg of the car mass. Or just leave the m in each of the force expressions and it will cancel out when you calculate the coefficient of friction.
 
  • #5
okay i officially hate this problem. I am playing with a triangle diagram for Fn and Ff, and I am using

Ffx + Fnx = mv^2/r
Ffx = mv^2/r - Fnx

so then i got got a formula for fnx
tan theta = fnx/fny
tan theta = fnx/mg because fny = Fg and Fg = m(9,81)
therefore, Fn = tan theta (mg)

so then i subsituted it into Ffx = mv^2/r - Fnx
giving me Ffx = mv&2/r - tan theta (mg)

im doing all of this because Ff = Ffx/cos theta
so i need ffx to solve for friction, but I am not seeing how mass cancels out. am i doing something wrong?
 
  • #6
normal force Fn = Fc*sin(12.1) + mg*cos(12.1)
Ff = u*Fn
force up = force down
Fc*cos(12.1) + Ff = mg*sin(12.1)
All the forces have m in them so it cancels out at this point when you fill in the details for Fc and Ff.
 
  • #7
Well i ended up completing the question, i ended up just going to my teacher for help. i can see why it was so hard for it to be explained xD there's a lot u have to do and i was on a totally different track.

Thanks for the help though Delphi51 :)
 

Related to Banked Curves using Radius and speed

1. What is a banked curve?

A banked curve is a curved section of a road or track that is designed to allow vehicles to safely travel around it at high speeds. It is sloped inward towards the center of the curve, providing a centripetal force that helps keep the vehicle on the road and prevents it from skidding or sliding off.

2. How does the radius of a curve affect the banking angle?

The radius of a curve is directly proportional to the banking angle. This means that as the radius increases, the banking angle also increases. This is because the larger the radius, the higher the centrifugal force acting on the vehicle, which requires a greater banking angle to counteract it and keep the vehicle on the road.

3. How does speed affect the banking angle?

The speed of a vehicle also affects the banking angle. As the speed increases, the centrifugal force acting on the vehicle also increases. This means that a higher speed requires a higher banking angle to keep the vehicle on the road and prevent it from sliding off the curve.

4. What is the ideal banking angle for a given speed and radius?

The ideal banking angle for a given speed and radius is determined by the equation tanθ = v^2/rg, where θ is the banking angle, v is the speed, r is the radius, and g is the acceleration due to gravity. This equation ensures that the centrifugal force and the centripetal force are equal, allowing the vehicle to safely travel around the curve without skidding or sliding off.

5. What happens if the banking angle is too steep or too shallow?

If the banking angle is too steep, the centripetal force will be greater than the centrifugal force, causing the vehicle to experience a net force towards the center of the curve. This can result in the vehicle losing traction and skidding off the curve. On the other hand, if the banking angle is too shallow, the centrifugal force will be greater than the centripetal force, causing the vehicle to slide outwards and potentially roll over. It is important to calculate the correct banking angle to ensure safe travel around a banked curve.

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