Presentation is loading. Please wait.

Presentation is loading. Please wait.

IB Physics 12 Mr. Jean September 15 th, 2015. The plan: Video clip of the day –https://www.youtube.com/watch?v=QcdlGjAP 0xshttps://www.youtube.com/watch?v=QcdlGjAP.

Similar presentations


Presentation on theme: "IB Physics 12 Mr. Jean September 15 th, 2015. The plan: Video clip of the day –https://www.youtube.com/watch?v=QcdlGjAP 0xshttps://www.youtube.com/watch?v=QcdlGjAP."— Presentation transcript:

1 IB Physics 12 Mr. Jean September 15 th, 2015

2 The plan: Video clip of the day –https://www.youtube.com/watch?v=QcdlGjAP 0xshttps://www.youtube.com/watch?v=QcdlGjAP 0xs Conical Pendulums Investigation Banked turns with Friction

3 New IA Criteria:

4 Conical Pendulum: http://www.youtube.com/watch?v=5C4RJl FABichttp://www.youtube.com/watch?v=5C4RJl FABic http://www.youtube.com/watch?v=h- aStiXBaushttp://www.youtube.com/watch?v=h- aStiXBaus

5 Conical Pendulum Motion:

6

7

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

9 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

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

11

12

13 Banked Turns with Friction:

14 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)

15 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

16 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?

17

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

19 Frictional Force:

20 Gravitational Force:

21 Centripetal Force:

22 Chapter #5 If you are wondering where we are: Giancoli –P. 117 to 122  Gravitational Constant –P. 122 to 127  Newton & Kepler’s Synthesis

23

24 Newton’s Law of Universal Gravitation: F g = force of gravity in newtons (N) m 1 = first mass in kilograms (Kg) m 2 = second mass in kilograms (Kg) r = distance between centers of mass in meters G = Universal Gravity Constant (***next slide for units***)

25 Universal Gravitational Constant:

26 How to masses act on each other:

27 In the case of the Earth-Moon system, the moon is accelerating towards the Earth. The moon has a tangential component to its velocity. So it keeps moving in a circle around the Earth.

28 Kepler’s Empirical Equations: Johannes Kepler’s (1571 – 1630) was the famous German astronomer who laid the framework for understanding planetary motion.

29 Kepler’s First Law: Planets move in elliptical orbits around the sun. As an average these orbits are nearly circular.

30 Kepler’s Second Law: An imaginary line between the Sun and a planet sweeps out equal areas in equal time intervals.

31 K = is a CONSTANT for all planet’s as they travel around the sun. r = average distance from sun (m) T = period of planet’s revolution around the sun (seconds) Kepler’s Third Law:

32 The squared product of the period for a planet’s revolution around the Sun and the cube of the average distance from the Sun is a constant and the same for all planets. r = average distance from sun (m) T = period of planet’s revolution around the sun (s)

33

34 Satellites in space: A satellite in space moves around a heavy body. To keep the satellite from smashing back into the Earth (or planet it is orbiting), scientists set the force of gravity equal to the centripetal force. F g = F c

35 Example Question: You find yourself in space. In fact you are walking on a large asteroid. Your mass is 70kg, the asteroid has a mass of 8.0 x 10 5 kg and the radius between the two centers of mass is 80 meters. What is your weight on the asteroid?

36 m 1 = 70 (kg) m 2 = 8.0 x 10 5 (kg) r = 80 meters (m)

37 This weekend: Please read Giancoli Chapter #5 –P. 117 to 122  Gravitational Constant –P. 122 to 127  Newton & Kepler’s Synthesis


Download ppt "IB Physics 12 Mr. Jean September 15 th, 2015. The plan: Video clip of the day –https://www.youtube.com/watch?v=QcdlGjAP 0xshttps://www.youtube.com/watch?v=QcdlGjAP."

Similar presentations


Ads by Google