1 / 20

Physics 203 College Physics I Fall 2012

Physics 203 College Physics I Fall 2012. S. A. Yost. Chapter 5. Circular Motion, Universal Gravitation. Announcements. Chapter 4’s homework, HW04, was due today. Chapter 5’s homework, HW05, is due next Tuesday.

winter-marsh
Télécharger la présentation

Physics 203 College Physics I Fall 2012

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Physics 203College Physics IFall 2012 S. A. Yost Chapter 5 Circular Motion, Universal Gravitation

  2. Announcements • Chapter 4’s homework, HW04, was due today. • Chapter 5’s homework, HW05, is due next Tuesday. • Read Chapter 5, except the last section “Nonuniform Circular Motion” for Thursday. • Today we will do an inclined plane problem in class, then look at circular motion.

  3. Quiz: Question 1 • Which of the following is the frictional force on a sliding object always proportional to? • (a) the contact surface area • (b) its weight • (c) the normal force • (d)the contact pressure

  4. Quiz: Question 2 • For a given object on a surface, which is larger? • (a) the coefficient of kinetic friction • (b) the coefficient of static friction • (c) Neither, they are equal. • (d) Either could be larger, depending on the object.

  5. Quiz: Question 3 a A rocket of mass m is near earth, accelerating into space, as its engines produce a thrust of Fth. The rocket is propelled upward by (a) the force of the rocket on the earth. (b) the force of the earth on the rocket. (c) the force of the rocket on its exhaust. (d) the force of the exhaust on the rocket. (e) the force of the rocket exhaust on empty space.

  6. Quiz: Question 4 a A rocket of mass m is near earth, accelerating into space, as its engines produce a thrust of Fth. The force of the rocket on the earth at this time is (a) Fth downward (b) Fth upward (c) mg downward (d) mg upward (e) zero

  7. Inclined Plane • A block slides down the inclined plane shown, starting at the top. How long does it take to reach the bottom? 5 m 3 m 4 m

  8. Inclined Plane • Fnet= ma • Fnet = mg sin q = 3/5 mg • a = 3/5 g = 5.88 m/s2 • Time to reach bottom… • x = ½ a t2 • 5 m = (2.94 m/s2) t2 • t = 1.3 s. y N q Fnet x mg q sin q = 3/5 a = g/5 = 1.96 m/s2

  9. Uniform Circular Motion In uniform circular motion, and object goes around the circumference at constant speed. Which way does the velocity vector point? It is always tangential to the path. → → R v

  10. Uniform Circular Motion The circle has radius R. If the period is T, what is the magnitude of the velocity vector? It is the length the radius vector traces out divided by the period: v = 2pR/T. → → R v

  11. Uniform Circular Motion What is the direction of the acceleration vector? The speed is constant, so the component of acceleration in the direction of motion must be zero. a is perpendicular to v. a points inward, since v turns that way. → → → a R v → → → → → a is called the centripetal acceleration.

  12. Uniform Circular Motion What is the magnitude of the acceleration? Each revolution, the velocity vector traces out a circle of radius 2pv. Then a = 2pv/T. Recall that T = 2pR/v. Then a = v2/R. → → → → a R v v

  13. Force in Uniform Circular Motion • From the fact that there is acceleration directed toward the center of a circle in uniform circular motion, we can infer that the net force on the object is F = ma ,also directed toward the center of the circle. • Note! This is not a new force due to the circular motion. It is the result of all of the forces acting on the object together. → →

  14. Circular Motion If you twirl an object around your head on a string and then let go, which way does it travel? (a) (b) (c) view from above

  15. Roller Coaster v A roller coaster car goes over a hill with radius of curvature R at speed v. How fast must it go for the riders to feel weightless at the top? R • Note: it is ok that the motion isn’t uniform – it’s the instantaneous speed and radius that matter.

  16. Roller Coaster Feeling “weightless” means there is no contact force between the rider and car: FN= 0 • Gravity must be responsible for the entire centripetal acceleration: mg = mv2/R g = v2/R. v = √ g/R → FN → mg

  17. Conical Pendulum • A 120 g mass hanging from a string of length 1.50 m swings in a circle of radius 24 cm. • a) What is the direction of the net force on the mass? • Toward the center. • b) Find the tension in the string and the period of revolution. L = 1.50 m → Fnet R = 0.24 m

  18. Conical Pendulum • The net force is F = mv2/R. • What forces are responsible for the net force? • Fnet = T + mg. • Use F = ma, separated into horizontal and vertical components. L = 1.50 m → FT → → → → → → Fnet R = 0.24 m → mg

  19. Vertical Circular Motion • A mass is rotated at constant speed in a vertical circle on the end of the rod. Which vector(s) could correctly show the force of the rod on the mass? A B C D

  20. Fg Fc Vertical Circular Problem • Suppose the mass is 1.5 kg and the rod is 35 cm long. If the rod is tilted at q = 30o with respect to the horizontal, and it is spinning around at 200 rpm, what is the force of the rod at this instant? Fr q

More Related