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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

slide53

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?

slide54

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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