Banked curve angle w/no friction - teacher's work differs

In summary, the angle at which the roadway is banked to allow cars to make the curve at 12 m/s even if friction is 0 should be 17.09 degrees.
  • #1
1MileCrash
1,342
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Homework Statement



At what angle should the roadway on a curve with a 50m radius be banked to allow cars to make the curve at 12 m/s even if friction is 0?

Homework Equations





The Attempt at a Solution



All of the centripetal acceleration comes from normal force from the road on the car.

I choose to represent the road as being along the y-axis and Fn as being along +x axis. Then gravity is directed at angle theta to the left of the -yaxis.

Normal force must equal the component of gravity along -xaxis, which is mgsin(theta),

Since the required centripetal force is mv^2/r, this is 2.88m, mgsin(theta) = 2.88m

gsin(theta) = 2.88

theta = 17.09

My professors work seems to be similar, but at the end she says that:
gtan(theta) = 2.88

theta = 16.37

why?
 
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  • #2
Just to check I understand your axes correctly: so the instantaneous velocity of the car is in the z-direction? And the y-direction is like the direction a pedestrian would walk along to cross the road as quick as possible? And the x-direction is the direction of the normal force?

In this case, gravity will not be directed at angle theta from the y-axis. And the normal force should not equal the component of gravity along the x-axis. I think you should draw the diagram again. I would advise making one of the axis go opposite to the direction of gravity. (Although your method should work too, but I think it makes it a bit more complicated).
 
  • #3
1MileCrash said:

The Attempt at a Solution



All of the centripetal acceleration comes from normal force from the road on the car.

I choose to represent the road as being along the y-axis and Fn as being along +x axis. Then gravity is directed at angle theta to the left of the -yaxis.

Normal force must equal the component of gravity along -xaxis, which is mgsin(theta),

Since the required centripetal force is mv^2/r, this is 2.88m, mgsin(theta) = 2.88m


You calculated N instead of the centripetal force, which is Nsin(phi). The angle gravity encloses with the negative x-axis is not the banking angle. It is phi in the figure.

ehild
 

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Related to Banked curve angle w/no friction - teacher's work differs

1. What is a banked curve angle?

A banked curve angle is the angle at which a curved track or road is tilted or inclined. It is designed to help vehicles navigate the curve more safely and efficiently by reducing the amount of friction between the tires and the road surface.

2. How does banked curve angle affect vehicles with no friction?

Banked curve angles are particularly important for vehicles with no friction, such as roller coasters or cars on an icy road. Without friction, these vehicles would not be able to maintain their speed and would easily slip or slide off the curve. A banked curve angle allows these vehicles to maintain their speed and stay on the track or road.

3. How does the banked curve angle differ for different vehicles?

The banked curve angle differs depending on the speed and weight of the vehicle. Heavier and faster vehicles require a steeper banked angle to counteract the centrifugal force and maintain their speed. Lighter and slower vehicles can navigate the curve with a shallower banked angle.

4. What is the role of friction in banked curves?

Friction plays a crucial role in banked curves, as it helps maintain the stability of the vehicle. When a vehicle turns, the force of friction between the tires and the road surface helps keep the vehicle on the track. Without friction, a vehicle would easily slide off the curve, making banked curves essential for vehicles with no friction.

5. How is the banked curve angle calculated?

The banked curve angle can be calculated using the formula: tan θ = v² / rg, where θ is the banked angle, v is the speed of the vehicle, r is the radius of the curve, and g is the acceleration due to gravity. This formula takes into account the centrifugal force of the vehicle and the gravitational force acting on it.

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