Wize High School Grade 12 Physics Textbook > Circular Motion

Banked Roads

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Car Turning Around a Circular Section of Road


If the road is flat, then the centripetal force is equal to the force of friction on the car.
If there is no friction, there is no force that can supply the centripetal force required to make the car move in a circular path - there is no way that the car can turn!
If the road is banked, the horizontal component of the normal force provides the centripetal force.


Wize Concept
If a car is turning around a curve without friction, then the curve must be banked at some angle θ\thetaθ(in radians) in order for the car to not slide out of the turn.

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Example: Banked Roads
A turn in a highway has a radius of 70 meters and a banking angle of 10o. (Why would it make sense to bank a curve in a road?)

What is the critical speed (the speed for which no friction is required between the car's tires and the surface) for a car on this curve?

Solution:

θ=10°\theta=10\degreeθ=10°

Nsinθ=mv2RNsin\theta=\dfrac{mv^2}{R}Nsinθ=Rmv2​

Ncosθ=mgNcos\theta=mgNcosθ=mg

tanθ=v2grv=grtanθ≅11 m/stan\theta=\dfrac{v^2}{gr}\qquad v=\sqrt{grtan\theta}  \cong 11\ m/stanθ=grv2​v=grtanθ​≅11 m/s

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Example: Car in a Flat Ramp

A car is negotiating a 160160160 m diameter flat ramp and is about to slip off the road with speed of 909090 km/h. What is the coefficient of static friction between the tires and the road?

A) 0.80.80.8                B) 0.60.60.6                              C) 0.40.40.4                     D) 0.150.150.15

First, let's convert the speed:   90 kmhr=90⋅1000 m3600 s=25 ms90 \ \dfrac{km}{hr}=90\cdot\dfrac{1000\ m}{3600\ s}=25\ \dfrac{m}{s}90 hrkm​=90⋅3600 s1000 m​=25 sm​

The only force toward the centre of this circular path is the static friction between tires and the road. 

Since the car is about to slip off, it means that we are talking about the maximum static friction (the car is on the verge of sliding). 

So, the centripetal force is equal to maximum static friction:

Horizontal direction:        ∑F=mac\sum F=ma_c∑F=mac​

                                   Fstatic, max=mv2rF_{static, \ max}=\dfrac{mv^2}{r}Fstatic, max​=rmv2​
                                              μsN=mv2r\mu_sN=\dfrac{mv^2}{r}μs​N=rmv2​

Vertical direction:          ∑F=0\sum F=0∑F=0  since there is no motion along this direction
                                 NB−mg=0N_B-mg=0NB​−mg=0  
                                                N=mgN=mgN=mg

Combining the equations we get:

μsmg=mv2r\mu_s\cancel mg=\dfrac{\cancel mv^2}{r}μs​m​g=rm​v2​

      μs=v2rg\mu_s=\dfrac{v^2}{rg}μs​=rgv2​

            =25280×9.81=0.8=\dfrac{25^2}{80\times9.81}=0.8=80×9.81252​=0.8

A racing car is moving on a circular banked track with radius of 200200200 m at a constant speed. The driver feels a normal force 303030 percent heavier than his true weight. What are the angle of the bank and the speed of the racing car?

v=185 km/h, Θ=39.7 ∘v=185\ km/h,\ \Theta=39.7^{\ ^{\circ}}v=185 km/h, Θ=39.7 ∘

v=145 km/h, Θ=39.7 ∘v=145\ km/h,\ \Theta=39.7^{\ ^{\circ}}v=145 km/h, Θ=39.7 ∘

v=145 km/h, Θ=33.4 ∘v=145\ km/h,\ \Theta=33.4^{\ 
^{\circ}}v=145 km/h, Θ=33.4 ∘               

v=185 km/h, Θ=36.7 ∘v=185\ km/h,\ \Theta=36.7^{\ ^{\circ}}v=185 km/h, Θ=36.7 ∘

I don't know

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Practice: Banked Road and Centripetal Force
A curve with a 120-m radius on a level road is banked at the correct angle for a speed of 20 m/s. 

(a) If an automobile rounds this curve at 30 m/s, what is the minimum coefficient of static friction needed between tires and road to prevent skidding?
(Hint: start by finding the angle of inclination of the curve. For a banked curve with no friction, there is one speed and one angle where it is safe to drive around the curve.)

(b) With this coefficient of static friction and angle of inclination, what would be the slowest possible speed you can safely drive on the curve without slipping? 
(Hint: You will need a slightly modified FBD from above.)

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