1. IB Physics 12 Mr. Jean September 25 th, 2013

2. The plan: Video clip of the day Banked Turns without Friction Banked Turns with Friction

3. Video Clip of the day: http://youtu.be/uENITui5_jU https://www.youtube.com/watch?v=3Xc_e h-OlvUhttps://www.youtube.com/watch?v=3Xc_e h-OlvU

4. Banked turns:

5. Frictionless Banked Turns: F net = F c Thus F n sinθ = (mv 2 ) / r F n cos θ mg F n sinθ = (mv 2 ) / r Too messy! Clean up! F n cos θ mg

6. Frictionless Banked Turns: F net = F c Thus F n sinθ = (mv 2 ) / r F n cos θ mg F n sinθ = (mv 2 ) / r F n cos θ mg

7. With some simplification we can see that this reduces to something decent. tan θ = v 2 rg This is the maximum speed an object can move around a banked turn without friction.

8. Example #1: A curve has a radius of 50 meters and a banking angle of 15. What is the critical, speed (the speed for which no friction is required between the car's tires and the surface) for a car on this curve?

11. Frictionless Banked Turns: Mr. Jean invents a stupid sport for this next year’s Olympics. It is called Rollerblading on ice. The track is circular with a radius of 6 meters. The track has a slope of 25 degrees and the coefficient of friction between the rollerblades and the ice is 0. What is the maximum speed which Olympic competitors will be able to reach (assuming they somehow get to this speed)? What would be the world record lap time?

15. Lap times What happens if a lap time is slower than _____ seconds? What happens if a lap time is quicker than _____ seconds?

16. Frictionless Banked Turns: Rollerblading on ice. The track is circular with a radius of 6 meters. The track has a slope of 90 degrees and the coefficient of friction between the rollerblades and the ice is 0.

17. θ = 90 degrees  AAAAAAAA!!!!!!!! θ = 89.9 degrees  183.64m/s θ = 89 degrees  58.1m/s θ = 75 degrees  14.1m/s θ = 60 degrees  10.1m/s θ = 45 degrees  7.67m/s θ = 30 degrees  5.83m/s θ = 1 degrees  1.01m/s

18. Conical Pendulum Motion:

21. T = Tension in Newton's T cos θ is balanced by the object's weight, mg. –Thus T * cos(θ) = mg

22. Conical Pendulum Motion: T sin θ that is the unbalanced central force that is supplying the centripetal force necessary to keep the block moving in its circular path: –Thus T sin θ = F c = ma c. –Thus T sin θ = F c = (mv 2 ) r

23. How long does it take for the object to complete one complete circle? HINT: v = 2 π r f

25. Banked Turns with Friction:

26. Important assumptions for Banked turns with Friction: F net = F c = F f + F g Let’s look at the frictional force first: 1.F f = μ * F n 2.F f = μ * F g * cos (10) F f = μmg cos(Θ)

27. Let’s look at the gravitational force: 1.F g = F g * sin(10) F f = μmg sin(10) F net = F c = F f + F g

28. Banked Curves (with friction) The Problem: A car with the mass of 1500kg is traveling in uniform circular motion along a circular curve with radius of 50 meters on a road that is banked at 10 degrees. The coefficient of friction is 0.4. What is the maximum velocity in which this car can take the curve?

30. Finding the sum of all center seeking forces. (Use previous diagram to highlight forces)

31. Frictional Force:

32. Gravitational Force:

33. Centripetal Force: