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## PowerPoint Slideshow about ' Banked Curves' - felicia-korbin

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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

- Consider a car rounding a circular banked curve at a constant speed
- Road friction is negligible

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 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 Curves

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

FN

To centre of curve

Fg

Ɵ

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 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 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 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 Curves

- 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 Curves

- 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 Curves

- 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 Curves

- 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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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