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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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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 • What are the real forces acting on the car? 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 • What is the direction of the car's acceleration? 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 • 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 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 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 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 Curves • Why is Ɵ also found in the component diagram? FNx Y FN X FNy a Ɵ 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