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Circular Motion Monday Dec. 3.

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Presentation on theme: "Circular Motion Monday Dec. 3."— Presentation transcript:

1 Circular Motion Monday Dec. 3

2 Forces in more complex situations
First we addressed constant forces that act along only one axis. Then we addressed constant forces along two dimensions. Most forces are not constant; they can change in both magnitude and direction. Now we deal with the simplest case of continually changing forces: circular motion. © 2014 Pearson Education, Inc.

3 The qualitative velocity change method for circular motion
Instantaneous velocity is tangent to the displacement for any instant. In circular motion, the system object travels in a circle and the velocity is tangent to the circle at every instant. Even if an object is moving at constant speed around a circle, its velocity changes direction. A change in velocity means there is acceleration! Lets explore that…. © 2014 Pearson Education, Inc.

4 Conceptual Exercise 4.1: Direction of a racecar's acceleration
Determine the direction of the racecar's acceleration at points A, B, and C in the figure as the racecar travels at constant speed on the circular path. This is left for the instructor to solve for students © 2014 Pearson Education, Inc.

5 Newton's second law and circular motion
The sum of the forces exerted on an object moving at constant speed along a circular path at any moment points toward the center of that circle in the same direction as the object's acceleration. © 2014 Pearson Education, Inc.

6 Let’s draw some force diagrams!!
A rollercoaster enters the track in an upright position and is at the bottom of the loop That same rollercoaster is now at the top of the loop The rollercoaster is now at the top of a hill on the track

7 Determining the factors that might affect acceleration
Imagine our experience in a car following the circular curve of a highway. The faster the car moves along a highway curve, the greater the risk that the car will skid off the road. The tighter the turn radius (r), the greater the risk that the car will skid. We guess the acceleration depends on v and r. © 2014 Pearson Education, Inc.

8 Radial acceleration: An object at constant speed v on a circular path of radius r
Relationships between acceleration with velocity and the radius: We can define the radial acceleration as: This is usually chosen as the radial or x direction when working problems The acceleration points toward the center. This is a linear quantity! The SI units are m/s2.

9 Example 4.2: Blackout When a fighter pilot pulls out from the bottom of a power dive, his body moves at high speed along a segment of an upward-bending, approximately circular path. While his body moves up, his blood tends to move straight ahead (tangent to the circle) and begins to fill the easily expandable veins in his legs. This can deprive his brain of blood and cause a blackout if the radial acceleration is 4 g or more and lasts several seconds (memorize). During a dive, an airplane moves at a modest speed of v = 80 m/s through a circular arc of radius r = 150 m. Is the pilot likely to black out? I'm leaving this for the instructor to solve © 2014 Pearson Education, Inc.

10 Fighter Pilot Chapter 4 example problem
During a dive, an airplane moves at a modest speed of v = 80 m/s through a circular arc of radius r = 150 m. Is the pilot likely to black out? © 2014 Pearson Education, Inc.

11 Period Period is the time interval it takes an object to travel around an entire circular path one time. Period has units of time, so the SI unit is s. Here is the equation: Do not confuse the symbol T for period with the symbol (t) for the time!! = C = X V V © 2014 Pearson Education, Inc.

12 Finding time period when given number of revolutions…
T = time/revolutions Convert time to seconds before hand. © 2014 Pearson Education, Inc.

13 Circular motion component form of Newton's second law
© 2014 Pearson Education, Inc.

14 Circular Motion: Mumbo Jumbo Translated
FNetx or FNetr FNety © 2014 Pearson Education, Inc.

15 Remember these?! Let’s write Fnet equations!
A rollercoaster enters the track in an upright position and is at the bottom of the loop That same rollercoaster is now at the top of the loop The rollercoaster is now at the top of a hill on the track

16 Problem-solving strategy: Processes involving constant-speed circular motion
Sketch and translate Sketch the situation described in the problem statement. Label it with all relevant information. Choose a system object and a specific position to analyze its motion. © 2014 Pearson Education, Inc.

17 Problem-solving strategy: Processes involving constant-speed circular motion (Cont'd)
© 2014 Pearson Education, Inc.

18 Problem-solving strategy: Processes involving constant-circular motion
Convert the force diagram into the radial r-component form of Newton's second law. For objects moving in a horizontal circle (unlike this example), you may also need to apply a vertical y-component form of Newton's second law.

19 Problem-solving strategy: Processes involving constant-speed circular motion
Solve and evaluate Solve the equations formulated in the previous two steps. Evaluate the results to see if they are reasonable (e.g., the magnitude of the answer, its units, limiting cases). © 2014 Pearson Education, Inc.

20 Critical Velocity Critical velocity is either the minimum or maximum speed needed to stay on the “circular” course. At the top of a loop the critical velocity is the minimum speed needed to stay on the loop At the top of a hill critical velocity is the maximum speed you can go and still be on the hill Both of these scenarios involve having the Normal force from the loop or the hill act on the object. So the critical velocity is the speed when FN=0!! We will derive the critical velocity equation on the next slide.

21 Deriving Critical Velocity
© 2014 Pearson Education, Inc.

22 Tip for circular motion
There is no special force that causes the radial acceleration of an object moving at constant speed along a circular path. This acceleration is caused by all of the forces (Fnet! Sometimes called Centripetal Force) exerted on the system object by other objects. Add the radial components of these regular forces. This sum is what causes the radial acceleration of the system object. © 2014 Pearson Education, Inc.

23 Conceptual difficulties with circular motion
When sitting in a car that makes a sharp turn, you feel thrown outward, inconsistent with the idea that the net force points toward the center of the circle (inward). Centrifugal force is the feeling that you have… it is not an actual force!! This is caused by inertia!! © 2014 Pearson Education, Inc.

24 A made up example… A bug has landed on your mothers antique record player. Because you are evil and are curious about physics you turn on the player to watch the bug get really dizzy. You time how long it takes the bug to go around the record player 3 times. The clock reading comes out to be 4.5 seconds. You also happen to know that bug is standing approximately .35 meters from the center of the player. What is the static friction force involved is the bugs mass is estimated to be .05 kg? What is the coefficient of friction? Are these estimated numbers correct? How do you know? © 2014 Pearson Education, Inc.

25 LONG PROBLEM SHORT… MEMORIZE
© 2014 Pearson Education, Inc.

26 AS HARD AS IT IS GONNA GET! Bookwork #27… sorta (Banked Curve)
A car moves around a 25 m radius highway curve that is banked at 8 relative to the horizontal and has a coefficient of friction of 0. At what max speed should the car travel so that it makes the turn without slipping. © 2014 Pearson Education, Inc.

27 For the love of all things good….
Just memorize this for problems with one other force besides Fg, that you are splitting into x and y components. Banked Curves and tension circular motion © 2014 Pearson Education, Inc.

28 Answer box for circular homework worksheet
266.67 9.65 6.83 5.81 x 103 3.85 2.64 x 107 2.64 x 104 2.78

29 Newton’s Law of Gravitation
© 2014 Pearson Education, Inc.

30 Some background about Newton and his Law of Gravitation
Newton was among the first to hypothesize that the Moon moves in a circular orbit around Earth because Earth pulls on it, continuously changing the direction of the Moon's velocity. He wondered if the force exerted by Earth on the Moon was the same type of force that Earth exerted on falling objects, such as an apple falling from a tree. © 2014 Pearson Education, Inc.

31 Newton background continued…
Newton compared the acceleration of the Moon to see if it could be modeled as a point particle near Earth's surface to the acceleration of the moon observed in its orbit © 2014 Pearson Education, Inc.

32 Dependence of gravitational force on mass
Newton deduced by recognizing that acceleration would equal g only if the gravitational force was directly proportional to the system object's mass. Newton deduced by applying the third law of motion. © 2014 Pearson Education, Inc.

33 Here’s how Newton’s Law of Gravitation is different than the basic Force of Gravity we have already discussed… You might wonder why if Earth pulls on the Moon, the Moon does not come closer to Earth in the same way that an apple falls from a tree. The difference in these two cases is the speed of the objects. The apple is at rest with respect to Earth before it leaves the tree, and the Moon is moving tangentially. If Earth stopped pulling on the Moon, it would fly away along a straight line(similar to your whirly stoppers!). © 2014 Pearson Education, Inc.

34 The law of universal gravitation
Law of Universal Gravitation Equation: G is the universal gravitational constant, indicating that this law works anywhere in the universe for any two masses. © 2014 Pearson Education, Inc.

35 Making sense of the gravitational force that everyday objects exert on Earth
It might seem counterintuitive that objects exert a gravitational force on Earth; Earth does not seem to react every time someone drops something. Acceleration is force divided by mass, and the mass of Earth is very large, so Earth's acceleration is very small. For example, the acceleration caused by the gravitational force a tennis ball exerts on Earth is 9.2 x 10–26 m/s2. © 2014 Pearson Education, Inc.

36 Yes… we still have assumptions
The law of universal gravitation in the form of applies only to objects: With spherical symmetry (i.e., in the shape of a sphere and with density that varies only with distance from the object's center) and Whose separation distance is much bigger than their size © 2014 Pearson Education, Inc.

37 For objects on earth: The numbers in the blue box equal 9. 8
For objects on earth: The numbers in the blue box equal 9.8! CRAZY RIGHT?! © 2014 Pearson Education, Inc.

38 Example Problem How would the weight of an object be effected if the object was located on a planet that was 3X as massive as Earth and had a radius that was half the size? (density remains the same) An unknown planet is ½ as dense as Earth with 4 time the volume. What is the force of gravity at the surface of this unknown planet?

39 Satellites and astronauts: Putting it all together
Geostationary satellites stay at the same location in the sky. This is why a satellite dish always points in the same direction. These satellites must be placed at a specific altitude that allows the satellite to travel once around Earth in exactly 24 hours while remaining above the equator. An array of such satellites can provide communications to all parts of Earth. © 2014 Pearson Education, Inc.

40 Astronauts are NOT weightless in the International Space Station; they are Normal Forceless!
Earth exerts a gravitational force on them! This force causes the astronaut and space station to fall toward Earth at the same rate while they fly forward, staying on the same circular path. The astronaut is in free fall (as is the station). If the astronaut stood on a scale in the space station, the scale would read zero even though the gravitational force is nonzero. Weight is a way of referring to the gravitational force, not the reading of a scale (normal force). © 2014 Pearson Education, Inc.

41 Combining Gravitational and circular ideas!
For planets: The gravitational force (Fg) is the centripetal Force (Fc) © 2014 Pearson Education, Inc.

42 Example Problem A satellite wishes to orbit the earth at a height of 100 km above the surface of the earth. Determine the speed, acceleration and orbital period of the satellite. (Given: Mearth = 5.98 x 1024 kg, Rearth = 6.37 x 106 m)

43 These were around before Newton!


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