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Banked Curves. Banked Curves. Consider a car rounding a circular banked curve at a constant speed. Banked Curves. Consider a car rounding a circular banked curve at a constant speed Road friction is negligible. Banked Curves.

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PowerPoint Slideshow about ' Banked Curves' - felicia-korbin


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Banked curves1
Banked Curves

  • Consider a car rounding a circular banked curve at a constant speed


Banked curves2
Banked Curves

  • Consider a car rounding a circular banked curve at a constant speed

  • Road friction is negligible


Banked curves3
Banked Curves

  • Consider a car rounding a circular banked curve at a constant speed

  • Road friction is negligible


Banked curves4
Banked Curves

  • Consider a car rounding a circular banked curve at a constant speed

  • Road friction is negligible

  • Ɵ is banking angle

  • Consider a car rounding a circular banked curve at a constant speed

  • Road friction is negligible

Ɵ


Banked curves5
Banked Curves

  • Consider a car rounding a circular banked curve at a constant speed

  • Road friction is negligible

  • Ɵ is banking angle

  • Notethe direction of the

    centre of the circular

    curve is along the

    horizontal plane

  • Consider a car rounding a circular banked curve at a constant speed

  • Road friction is negligible

To centre of curve

Ɵ


Banked curves6
Banked Curves

  • What are the real forces acting on the car?

To centre of curve

Ɵ


Banked curves7
Banked Curves

  • Only the normal force and the earth's gravitational force act on the car

FN

To centre of curve

Fg

Ɵ


Banked curves8
Banked Curves

  • Only the normal force and the earth's gravitational force act on the car

  • Let's get rid of the car so we can

    see the forces more clearly!

FN

To centre of curve

Fg

Ɵ


Banked curves9
Banked Curves

  • What is the direction of the car's acceleration?

FN

To centre of curve

Fg

Ɵ


Banked curves10
Banked Curves

  • Since the car is moving in UCM, the acceleration is centripetal, or towards the centre of the circle.

FN

a

To centre of curve

Fg

Ɵ


Banked curves11
Banked Curves

  • Since the car is moving in UCM, the acceleration is centripetal, or towards the centre of the circle.

  • How do we set up th XY plane?

FN

a

To centre of curve

Fg

Ɵ


Banked curves12
Banked Curves

  • Since the car is moving in UCM, the acceleration is centripetal, or towards the centre of the circle.

  • The acceleration is lined up with

    the positive x-axis.

Y

FN

X

a

To centre of curve

Fg

Ɵ


Banked curves13
Banked Curves

  • Since the car is moving in UCM, the acceleration is centripetal, or towards the centre of the circle.

  • The acceleration is lined up with

    the positive x-axis.

  • What real force is not

    lined up with the

    x and y axis?

Y

FN

X

a

To centre of curve

Fg

Ɵ


Banked curves14
Banked Curves

  • Since the car is moving in UCM, the acceleration is centripetal, or towards the centre of the circle.

  • The acceleration is lined up with

    the positive x-axis.

  • The normal force is not

    lined up with the

    x and y axis.

Y

FN

X

a

To centre of curve

Fg

Ɵ


Banked curves15
Banked Curves

  • Since the car is moving in UCM, the acceleration is centripetal, or towards the centre of the circle.

  • The acceleration is lined up with

    the positive x-axis.

  • The normal force is not

    lined up with the

    x and y axis.

  • How do we solve the

    problem?

Y

FN

X

a

To centre of curve

Fg

Ɵ


Banked curves16
Banked Curves

  • Since the car is moving in UCM, the acceleration is centripetal, or towards the centre of the circle.

  • The acceleration is lined up with

    the positive x-axis.

  • The normal force is not

    lined up with the

    x and y axis.

  • Resolve FN into

    components

FNx

Y

FN

X

FNy

a

To centre of curve

Fg

Ɵ


Banked curves17
Banked Curves

  • Why is Ɵ also found in the component diagram?

FNx

Y

FN

X

FNy

a

Ɵ

To centre of curve

Fg

Ɵ


Banked curves18
Banked Curves

  • Why is Ɵ also found in the component diagram?

  • Watch carefully!

FNx

Y

FN

X

FNy

a

Ɵ

To centre of curve

Ɵ

Fg


Banked curves19
Banked Curves

  • Why is Ɵ also found in the component diagram?

  • Watch carefully!

FNx

Y

FN

X

FNy

a

Ɵ

To centre of curve

90 -θ

Ɵ

Fg


Banked curves20
Banked Curves

  • Why is Ɵ also found in the component diagram?

  • Do you see why now?

FNx

Y

FN

X

FNy

a

Ɵ

To centre of curve

θ

90 -θ

Ɵ

Fg


Banked curves21
Banked Curves

  • Why is Ɵ also found in the component diagram?

  • Do you see why now?

  • Let's call | FN | = FN

FNx

Y

FN

X

FNy

a

Ɵ

To centre of curve

θ

90 -θ

Ɵ

Fg


Banked curves22
Banked Curves

  • Why is Ɵ also found in the component diagram?

  • Do you see why now?

  • Let's call | FN | = FN

  • In terms of FN , what is

    FN X = ?

FNx

Y

FN

X

FNy

a

Ɵ

To centre of curve

θ

90 -θ

Ɵ

Fg


Banked curves23
Banked Curves

  • Why is Ɵ also found in the component diagram?

  • Do you see why now?

  • Let's call | FN | = FN

  • In terms of FN , what is

    FN X = + FN sinθ

FNx

Y

FN

X

FNy

a

Ɵ

To centre of curve

θ

90 -θ

Ɵ

Fg


Banked curves24
Banked Curves

  • Why is Ɵ also found in the component diagram?

  • Do you see why now?

  • Let's call | FN | = FN

  • In terms of FN , what is

    FN X = + FN sinθ

    FN y = ?

FNx

Y

FN

X

FNy

a

Ɵ

To centre of curve

θ

90 -θ

Ɵ

Fg


Banked curves25
Banked Curves

  • Why is Ɵ also found in the component diagram?

  • Do you see why now?

  • Let's call | FN | = FN

  • In terms of FN , what is

    FN X = + FN sinθ

    FN y = + FN cosθ

FNx

Y

FN

X

FNy

a

Ɵ

To centre of curve

θ

90 -θ

Ɵ

Fg


Banked curves26
Banked Curves

  • Why is Ɵ also found in the component diagram?

  • Do you see why now?

  • Let's call | FN | = FN

  • In terms of FN , what is

    FN X = + FN sinθ

    FN y = + FN cosθ

  • Label a in terms of

    a positive scalar.

FNx

Y

FN

X

FNy

a

Ɵ

To centre of curve

θ

90 -θ

Ɵ

Fg


Banked curves27
Banked Curves

  • Why is Ɵ also found in the component diagram?

  • Do you see why now?

  • Let's call | FN | = FN

  • In terms of FN , what is

    FN X = + FN sinθ

    FN y = + FN cosθ

  • Label a in terms of

    a positive scalar.

FNx

Y

FN

X

FNy

a = +a

Ɵ

To centre of curve

θ

90 -θ

Ɵ

Fg


Banked curves28
Banked Curves

  • FN X = + FN sinθ FN y = + FN cosθ

  • Let's deal with the x-forces

FNx

Y

FN

X

FNy

a = +a

Ɵ

To centre of curve

θ

90 -θ

Ɵ

Fg


Banked curves29
Banked Curves

  • FN X = + FN sinθ FN y = + FN cosθ

  • Let's deal with the x-forces

    Fnet x = max

FNx

Y

FN

X

FNy

a = +a

Ɵ

To centre of curve

θ

90 -θ

Ɵ

Fg


Banked curves30
Banked Curves

  • FN X = + FN sinθ FN y = + FN cosθ

  • Let's deal with the x-forces

    Fnet x = max

    FN X = max

FNx

Y

FN

X

FNy

a = +a

Ɵ

To centre of curve

θ

90 -θ

Ɵ

Fg


Banked curves31
Banked Curves

  • FN X = + FN sinθ FN y = + FN cosθ

  • Let's deal with the x-forces

    Fnet x = max

    FN X = max

    + FN sinθ = max

FNx

Y

FN

X

FNy

a = +a

Ɵ

To centre of curve

θ

90 -θ

Ɵ

Fg


Banked curves32
Banked Curves

  • FN X = + FN sinθ FN y = + FN cosθ

  • Let's deal with the x-forces

    Fnet x = max

    FN X = max

    + FN sinθ = max

  • For UCM, what is ax

    in terms of v and r ?

FNx

Y

FN

X

FNy

a = +a

Ɵ

To centre of curve

θ

90 -θ

Ɵ

Fg


Banked curves33
Banked Curves

  • FN X = + FN sinθ FN y = + FN cosθ

  • Let's deal with the x-forces

    Fnet x = max

    FN X = max

    + FN sinθ = m(v2/ r)

FNx

Y

FN

X

FNy

a = +a

Ɵ

To centre of curve

θ

90 -θ

Ɵ

Fg


Banked curves34
Banked Curves

  • FN X = + FN sinθ FN y = + FN cosθ

  • Let's deal with the x-forces

    Fnet x = max

    FN X = max

    + FN sinθ = m(v2/ r)

    equation #1

FNx

Y

FN

X

FNy

a = +a

Ɵ

To centre of curve

θ

90 -θ

Ɵ

Fg


Banked curves35
Banked Curves

  • FN X = + FN sinθ FN y = + FN cosθ #1FN sinθ = mv2/r

FNx

Y

FN

X

FNy

a = +a

Ɵ

To centre of curve

θ

90 -θ

Ɵ

Fg


Banked curves36
Banked Curves

  • FN X = + FN sinθ FN y = + FN cosθ #1FN sinθ = mv2/r

  • Let's deal with the y-forces

FNx

Y

FN

X

FNy

a = +a

Ɵ

To centre of curve

θ

90 -θ

Ɵ

Fg


Banked curves37
Banked Curves

  • FN X = + FN sinθ FN y = + FN cosθ #1FN sinθ = mv2/r

  • Let's deal with the y-forces

    Fnet y = ?

FNx

Y

FN

X

FNy

a = +a

Ɵ

To centre of curve

θ

90 -θ

Ɵ

Fg


Banked curves38
Banked Curves

  • FN X = + FN sinθ FN y = + FN cosθ #1FN sinθ = mv2/r

  • Let's deal with the y-forces

    Fnet y = 0 b/c ay = 0

FNx

Y

FN

X

FNy

a = +a

Ɵ

To centre of curve

θ

90 -θ

Ɵ

Fg


Banked curves39
Banked Curves

  • FN X = + FN sinθ FN y = + FN cosθ #1FN sinθ = mv2/r

  • Let's deal with the y-forces

    Fnet y = 0

    FN Y + Fg = 0

FNx

Y

FN

X

FNy

a = +a

Ɵ

To centre of curve

θ

90 -θ

Ɵ

Fg


Banked curves40
Banked Curves

  • FN X = + FN sinθ FN y = + FN cosθ #1FN sinθ = mv2/r

  • Let's deal with the y-forces

    Fnet y = 0

    FN Y = 0

    + FN cosθ – mg = 0

FNx

Y

FN

X

FNy

a = +a

Ɵ

To centre of curve

θ

90 -θ

Ɵ

Fg


Banked curves41
Banked Curves

  • FN X = + FN sinθ FN y = + FN cosθ #1FN sinθ = mv2/r

  • Let's deal with the y-forces

    Fnet y = 0

    FN Y = 0

    + FN cosθ – mg = 0

    + FN cosθ = ?

FNx

Y

FN

X

FNy

a = +a

Ɵ

To centre of curve

θ

90 -θ

Ɵ

Fg


Banked curves42
Banked Curves

  • FN X = + FN sinθ FN y = + FN cosθ #1FN sinθ = mv2/r

  • Let's deal with the y-forces

    Fnet y = 0

    FN Y = 0

    + FN cosθ – mg = 0

    + FN cosθ = mg

FNx

Y

FN

X

FNy

a = +a

Ɵ

To centre of curve

θ

90 -θ

Ɵ

Fg


Banked curves43
Banked Curves

  • FN X = + FN sinθ FN y = + FN cosθ #1FN sinθ = mv2/r

  • Let's deal with the y-forces

    Fnet y = 0

    FN Y = 0

    + FN cosθ – mg = 0

    + FN cosθ = mg

    equation #2

FNx

Y

FN

X

FNy

a = +a

Ɵ

To centre of curve

θ

90 -θ

Ɵ

Fg


Banked curves44
Banked Curves

#2FN cosθ = mg #1FN sinθ = mv2/r

  • How can we eliminate FN and m?

FNx

Y

FN

X

FNy

a = +a

Ɵ

To centre of curve

θ

90 -θ

Ɵ

Fg


Banked curves45
Banked Curves

#2FN cosθ = mg #1FN sinθ = mv2/r

  • Divide #1 by #2

FNx

Y

FN

X

FNy

a = +a

Ɵ

To centre of curve

θ

90 -θ

Ɵ

Fg


Banked curves46
Banked Curves

#2FN cosθ = mg #1FN sinθ = mv2/r

  • Divide #1 by #2

  • FN sinθ = mv2/r

    FN cosθ = mg

FNx

Y

FN

X

FNy

a = +a

Ɵ

To centre of curve

θ

90 -θ

Ɵ

Fg


Banked curves47
Banked Curves

#2FN cosθ = mg #1FN sinθ = mv2/r

  • Divide #1 by #2

  • FN sinθ = mv2/r

    FN cosθ = mg

  • sinθ/ cosθ = ?

FNx

Y

FN

X

FNy

a = +a

Ɵ

To centre of curve

θ

90 -θ

Ɵ

Fg


Banked curves48
Banked Curves

#2FN cosθ = mg #1FN sinθ = mv2/r

  • Divide #1 by #2

  • FN sinθ = mv2/r

    FN cosθ = mg

  • sinθ/ cosθ = v2/(r g)

FNx

Y

FN

X

FNy

a = +a

Ɵ

To centre of curve

θ

90 -θ

Ɵ

Fg


Banked curves49
Banked Curves

#2FN cosθ = mg #1FN sinθ = mv2/r

  • Divide #1 by #2

  • FN sinθ = mv2/r

    FN cosθ = mg

  • sinθ/ cosθ = v2/(r g)

  • ? = v2/(r g)

FNx

Y

FN

X

FNy

a = +a

Ɵ

To centre of curve

θ

90 -θ

Ɵ

Fg


Banked curves50
Banked Curves

#2FN cosθ = mg #1FN sinθ = mv2/r

  • Divide #1 by #2

  • FN sinθ = mv2/r

    FN cosθ = mg

  • sinθ/ cosθ = v2/(r g)

  • tan θ = v2/(r g)

FNx

Y

FN

X

FNy

a = +a

Ɵ

To centre of curve

θ

90 -θ

Ɵ

Fg


Banked curves51
Banked Curves

#2FN cosθ = mg #1FN sinθ = mv2/r

  • Divide #1 by #2

  • FN sinθ = mv2/r

    FN cosθ = mg

  • sinθ/ cosθ = v2/(r g)

  • tan θ = v2/(r g)

  • Scalar Banked curve

    equation Memorize !

FNx

Y

FN

X

FNy

a = +a

Ɵ

To centre of curve

θ

90 -θ

Ɵ

Fg


Example: Assuming negligible friction, with what constant speed in km/h can a 2500.0 kg car go around a 15.0° banked circular curve of radius 575 m? What is the magnitude of the normal force on the car?


Example: Assuming negligible friction, with what constant speed in km/h can a 2500.0 kg car go around a 15.0° banked circular curve of radius 575 m? What is the magnitude of the normal force on the car?

  • Given:Θ = 15.0° r = 575 m m = 2500.0 kg

  • Unknowns:v = ? FN= ?

  • Formulas:tan θ = v2/(r g) or v = ( r g tan θ )1/2

    FN cosθ = mg or FN = mg/cosθ

  • Sub:v=( 575 X10.0 tan 15.0°)1/2 =39.3 m/s=141 km/h

    FN = mg/cosθ = 2500X10 / cos 15.0° = 25800 N


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