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N sin θ = mv 2 r . size 12{N"sin"θ= { { ital "mv" rSup { size 8{2} } } over {r} } } {}

Because the car does not leave the surface of the road, the net vertical force must be zero, meaning that the vertical components of the two external forces must be equal in magnitude and opposite in direction. From the figure, we see that the vertical component of the normal force is N cos θ size 12{N"cos"θ} {} , and the only other vertical force is the car’s weight. These must be equal in magnitude; thus,

N cos θ = mg . size 12{N"cos"θ= ital "mg"} {}

Now we can combine the last two equations to eliminate N size 12{N} {} and get an expression for θ size 12{θ} {} , as desired. Solving the second equation for N = mg / ( cos θ ) size 12{N= ital "mg"/ \( "cos"θ \) } {} , and substituting this into the first yields

mg sin θ cos θ = mv 2 r
mg tan ( θ ) = mv 2 r tan θ = v 2 rg.

Taking the inverse tangent gives

θ = tan 1 v 2 rg (ideally banked curve, no friction). size 12{θ="tan" rSup { size 8{ - 1} } left ( { {v rSup { size 8{2} } } over { ital "rg"} } right )} {}

This expression can be understood by considering how θ size 12{θ} {} depends on v size 12{v} {} and r size 12{r} {} . A large θ size 12{θ} {} will be obtained for a large v size 12{v} {} and a small r size 12{r} {} . That is, roads must be steeply banked for high speeds and sharp curves. Friction helps, because it allows you to take the curve at greater or lower speed than if the curve is frictionless. Note that θ size 12{θ} {} does not depend on the mass of the vehicle.

In this figure, a car from the backside is shown, turning to the left, on a slope angling downward to the left. A point in the middle of the back of the car is shown which shows one downward vector depicting weight, w, and an upward arrow depicting force N, which is a linear line along the car and is at an angle theta with the straight up arrow. The slope is at an angle theta with the horizontal surface below the slope. The force values, N multipliy sine theta equals to centripetal force, the net force on the car and N cosine theta equal to w are given below the car.
The car on this banked curve is moving away and turning to the left.

What is the ideal speed to take a steeply banked tight curve?

Curves on some test tracks and race courses, such as the Daytona International Speedway in Florida, are very steeply banked. This banking, with the aid of tire friction and very stable car configurations, allows the curves to be taken at very high speed. To illustrate, calculate the speed at which a 100 m radius curve banked at 65.0° should be driven if the road is frictionless.


We first note that all terms in the expression for the ideal angle of a banked curve except for speed are known; thus, we need only rearrange it so that speed appears on the left-hand side and then substitute known quantities.


Starting with

tan θ = v 2 rg size 12{"tan"θ= { {v rSup { size 8{2} } } over { ital "rg"} } } {}

we get

v = ( rg tan θ ) 1 / 2 . size 12{v= \( ital "rg""tan"θ \) rSup { size 8{1/2} } } {}

Noting that tan 65.0º = 2.14, we obtain

v = ( 100 m ) ( 9.80 m /s 2 ) ( 2 . 14 ) 1 / 2 = 45.8 m/s.


This is just about 165 km/h, consistent with a very steeply banked and rather sharp curve. Tire friction enables a vehicle to take the curve at significantly higher speeds.

Calculations similar to those in the preceding examples can be performed for a host of interesting situations in which centripetal force is involved—a number of these are presented in this chapter’s Problems and Exercises.

Got questions? Get instant answers now!

Take-home experiment

Ask a friend or relative to swing a golf club or a tennis racquet. Take appropriate measurements to estimate the centripetal acceleration of the end of the club or racquet. You may choose to do this in slow motion.

Phet explorations: gravity and orbits

Move the sun, earth, moon and space station to see how it affects their gravitational forces and orbital paths. Visualize the sizes and distances between different heavenly bodies, and turn off gravity to see what would happen without it!

Gravity and Orbits

Section summary

  • Centripetal force F c size 12{F rSub { size 8{c} } } {} is any force causing uniform circular motion. It is a “center-seeking” force that always points toward the center of rotation. It is perpendicular to linear velocity v size 12{v} {} and has magnitude
    F c = ma c ,

    which can also be expressed as

    F c = m v 2 r or F c = mr ω 2 ,

Questions & Answers

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Mahmud Reply
The Critical Angle Derivation So the critical angle is defined as the angle of incidence that provides an angle of refraction of 90-degrees. Make particular note that the critical angle is an angle of incidence value. For the water-air boundary, the critical angle is 48.6-degrees.
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the ratio of the density of a substance to the density of a standard, usually water for a liquid or solid, and air for a gas.
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it is the product of mass ×velocity of an object
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Instantaneous velocity is defined as the rate of change of position for a time interval which is almost equal to zero
The potential in a region between x= 0 and x = 6.00 m lis V= a+ bx, where a = 10.0 V and b = -7.00 V/m. Determine (a) the potential atx=0, 3.00 m, and 6.00 m and (b) the magnitude and direction of the electric ficld at x =0, 3.00 m, and 6.00 m.
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algum profe sabe .. Progressivo ou Retrógrado e Acelerado ou Retardado   V= +23 m/s        V= +5 m/s        0__>              0__> __________________________>        T= 0               T=6s
Practice Key Terms 5

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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