1 / 25

Welcome back to Physics 215

Welcome back to Physics 215. Today’s agenda: Velocity and acceleration in two-dimensional motion Motion under gravity -- projectile motion Acceleration on curved path. Current homework assignment. HW2: Ch.2 (Knight textbook): 48, 78, 82 Ch.3 (Knight textbook): 28, 32, 42

saoirse-plunkett
Download Presentation

Welcome back to Physics 215

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. Welcome back to Physics 215 Today’s agenda: Velocity and acceleration in two-dimensional motion Motion under gravity -- projectile motion Acceleration on curved path

  2. Current homework assignment HW2: • Ch.2 (Knight textbook): 48, 78, 82 • Ch.3 (Knight textbook): 28, 32, 42 • Ch.4 (Knight textbook): 40 • due Wednesday, Sept 15th in recitation Reminder about course website: http://www.phy.syr.edu/courses/PHY215.10Fall/index.html

  3. Exam 1: next Thursday (9/23/10) • In room 208 (here!) at the usual lecture time • Material covered: • Textbook chapters 1 - 4 • Lectures up through 9/21 (slides online) • Wed/Fri Workshop activities • Homework assignments • Work through practice exam problems (posted on website) • Work on more practice exam problems next Wednesday in recitation workshop

  4. Displacement in 2D Motion y Ds sI sF s – vector position Displacement s = sF - sI, also a vector! O x

  5. 2D Motion in components • x and y motions decouple • vx = dx/dt vy = dy/dt • ax = dvx/dt ay = dvy/dt • If acceleration is only non-zero in 1 direction, can choose coordinates so that 1 component of acceleration is zero • e.g., motion under gravity

  6. Motion under gravity ax = 0 vx = v0x x= x0 + v0xt ay = -g vy = v0y - gt y= y0 + v0yt - (1/2)gt2 y v0y = v0sin(q) v0x = v0cos(q) v0 q x Projectile motion...

  7. A ball is ejected vertically upward from a cart at rest. The ball goes up, reaches its highest point and returns to the cart. In a second experiment, the cart is moving at constant velocity and the ball is ejected in the same way, where will the ball land? • In front of the cart. • B. Behind the cart. • C. Inside the cart. • D. The outcome depends on the speed of the cart.

  8. Projectile question • A ball is thrown at 45o to vertical with a speed of 7 m/s. Assuming g=10 m/s2, how far away does the ball land?

  9. A battleship simultaneously fires two shells at enemy ships. If the shells follow the parabolic trajectories shown, which ship will be hit first? A. A B. Both at the same time C. B D. need more information

  10. Projectile motion y x R : when is y=0 ? t[vy1-(1/2)gt] = 0 i.e., T = (2v)sinq/g  R (x-eqn.)  hmax (y-eqn.)

  11. Maximum height and range

  12. Motion on a curved pathat constant speed Is the acceleration of the object equal to zero?

  13. Velocity is tangent to path Ds sI sF O v = Ds/Dt lies along dotted line. As Dt  0 direction of v is tangent to path

  14. Motion on a curved pathat constant speed

  15. Subtracting vectors Recall that vF+ (-vI) = Dv Dv vF vI same as -vI vF Dv

  16. For an object moving at constant speed along a curved path, the acceleration is not zero.

  17. For which of the following motions of a car does the change in velocity vector have the greatest magnitude? (All motions occur at the same constant speed.) A. A 90° right turn at constant speed B. A U-turn at constant speed C. A 270° turn on a highway on-ramp D. The change in velocity is zero for all three motions.

  18. A car moves along the path shown. Velocity vectors at two different points are sketched. Which of the arrows below most closely represents the direction of the average acceleration? A. B. C. D.

  19. A child is riding a bicycle on a level street. The velocity and acceleration vectors of the child at a given time are shown. Which of the following velocity vectors may represent the velocity at a later time? a A. B. C. D.

  20. A biker is riding at constant speed clockwise on the oval track shown below. Which vector correctly describes the acceleration at the point indicated?

  21. Biker moving around oval at constant speed As point D is moved closer to C, angle approaches 90°.

  22. Summary • For motion at constant speed, instantaneous acceleration vector is perpendicular to velocity vector • Points ``inward’’ • What is the magnitude of the acceleration vector?

  23. Acceleration vectors for ball swung in a horizontal circle at constant speed v v1 v2 R q v1 q v2 What is the magnitude of the acceleration?a = v2/R

  24. DEMO: shooting the bear … • Bear released at same time ball projected from end of tube • What happens to the bear and the ball? • Does outcome depend on angle relative to floor?

  25. Reading assignment • Circular motion • 4.5 – 4.7 in textbook

More Related