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Announcements

Announcements. First quiz on Friday in class (no more than 15 minutes) Kinematics with constant acceleration Solutions to group problems are on the Schedule page First 15 minutes of class: don’t be late!!!

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Announcements

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  1. Announcements • First quiz on Friday in class (no more than 15 minutes) • Kinematics with constant acceleration • Solutions to group problems are on the Schedule page • First 15 minutes of class: don’t be late!!! • Unless you have received prior approval from Prof. Gerton, YOU MUST COME TO YOUR OWN SECTION FOR THE QUIZ: BOTH SECTIONS ARE FULL!!! • Use the help lab (JFB rotunda) for homework help! • Schedule on course website • Prof. Gerton’s hours: M, F 3:00 – 4:00 PM; Tuesday 8:30 – 10:00 AM • smartPhysics homework • Must be within 1% of “true” answer to get it correct • Don’t round off your intermediate values during computation!!!

  2. Classical Mechanics Lecture 3: Relative and Circular Motion Today's Concepts: a) Relative Motion b) Centripetal Acceleration

  3. Relative Motion These are VECTORS!!! More generally: Object “a” with respect to (or relative to) object “b” Also:

  4. Prelecture 3, Question 1 A B C D

  5. A girl stands on a moving sidewalk that moves to the right at 2 m/s relative to the ground. A dog runs toward the girl in the opposite direction along the sidewalk at a speed of 8 m/s relative to the sidewalk. What is the speed of the dog relative to the ground? vdog,belt= 8 m/s vbelt,ground= 2 m/s CheckPoint: Moving Walkway A) 6 m/sB) 8 m/sC) 10 m/s

  6. vdog,belt= 8 m/s +x vbelt,ground= 2 m/s vdog, ground = vdog, belt + vbelt, ground = (-8 m/s) + (2 m/s) = -6 m/s CheckPoint Result: Moving Walkway What is the speed of the dog relative to the ground? About 90% of you got this right – lets try it again. A) 6 m/s B) 8 m/s C) 10 m/s

  7. vdog,belt= 8 m/s vbelt,ground= 2 m/s ACT: Moving Walkway, part II A girl stands on a moving sidewalk that moves to the right at 2 m/s relative to the ground. A dog runs toward the girl in the opposite direction along the sidewalk at a speed of 8 m/s relative to the sidewalk. What is the speed of the dog relative to the girl? A) 6 m/sB) 8 m/sC) 10 m/s

  8. vdog,belt= 8 m/s vbelt,ground= 2 m/s What is the speed of the dog relative to the girl? A) 6 m/s B) 8 m/s C) 10 m/s A) The dog is traveling along the belt at 8 m/s but the girl is traveling towards the dog at 2 m/s so in the girl's perspective the dog is moving 6 m/s. B) The girl is traveling in reference to the belt just like the dog, and the girl's velocity with respect to the belt is 0 m/s. The dog is traveling with a velocity of 8 m/s with respect to the belt which in turn means it is also traveling at 8 m/s with respect to the girl. C) The girl is moving 2m/s towards the dog, and the dog is moving 8m/s towards the girl. Therefore you add the two velocities 8+2=10.

  9. vdog,belt= 8 m/s vbelt,ground= 2 m/s What is the speed of the dog relative to the girl? A) 6 m/sB) 8 m/sC) 10 m/s B) The girl is traveling in reference to the belt just like the dog, and the girl's velocity with respect to the belt is 0m/s. The dog is traveling with a velocity of 8 m/s with respect to the belt which in turn means it is also traveling at 8 m/s with respect to the girl. Use the velocity addition formula: vdog, girl = vdog, belt + vbelt, girl = -8 m/s + 0 m/s = -8 m/s

  10. DEMO: Toy Truck on a Moving Surface

  11. Which direction should I point the toy bulldozer to get it across the cardboard fastest? A) To the left B) Straight across C) To the right ACT: Tractor on a Moving Surface ABC Cardboard moves to left.

  12. A man starts to walk along the dotted line painted on a moving sidewalk toward a fire hydrant that is directly across from him. The width of the walkway is 4 m, and it is moving at 2 m/s relative to the fire-hydrant. If his walking speed is 1 m/s, how far away will he be from the hydrant when he reaches the other side? ACT: 2D Relative Motion 1 m/s • 2 m • 4 m • 6 m • 8 m Moving sidewalk 2 m/s 4 m

  13. 1 m/s Time to get across: Δt = distance / speed = (4 m) / (1 m/s) = 4 s 4 m If the sidewalk wasn’t moving:

  14. Just the sidewalk: 2 m/s 4 m

  15. 1 m/s 4 m Combining motions: Student response: The concept in regards to the boat crossing the stream. The equation says to add the two velocities, but then the solution involves rather finding the length of the hypotenuse. 2 m/s

  16. How far away will he be from the hydrant when he reaches the other side? 1 m/s • 2 m • 4 m • 6 m • 8 m D 2 m/s 4 m D = (speed of sidewalk) ∙ (time to get across) = (2 m/s) ∙ (4 s) = 8 m

  17. Three swimmers can swim equally fast relative to the water. They have a race to see who can swim across a river in the least time. Relative to the water, Bethswims perpendicular to the flow, Annswims upstream at 30 degrees, and Carlyswims downstream at 30 degrees. Who gets across the river first? A) Ann B) Beth C) Carly y x ACT: Racing Across a River Beth Ann Carly

  18. y x Look at just water & swimmers Time to get across = D / Vy B A C D 30o 30o Beth: Vy = Vo Ann: Vy = Vo cos(30o) Carly: Vy = Vo cos(30o)

  19. y x ACT: 3 Swimmers, part II Three swimmers can swim equally fast relative to the water. They have a race to see who can swim across a river in the least time. Relative to the water, Bethswims perpendicular to the flow, Annswims upstream at 30 degrees, and Carlyswims downstream at 30 degrees. Who gets across the river second? A) Ann C) Carly D) Both same Ann Carly Ann Carly

  20. Announcements • First quiz on Friday in class (no more than 15 minutes) • Kinematics with constant acceleration (Units 1 and 2) • Solutions to group assignments are on the Schedule page • No notes allowed: just a calculator and pen/pencil • First 15 minutes of class: don’t be late!!! • Unless you have received prior approval from Prof. Gerton, YOU MUST COME TO YOUR OWN SECTION FOR THE QUIZ: BOTH SECTIONS ARE FULL!!! • Use the help lab (JFB rotunda) for homework help! • Schedule on course website • Prof. Gerton’s hours: M, F 3:00 – 4:00 PM; Tuesday 8:30 – 10:00 AM

  21. The rock will fly tangent to the circle when the string is broken. At this point the velocity vector will point tangent to the circle. Once the string is cut, the rock no longer experiences centripetal, and will fly straight out from this point with the velocity it had at the moment the string is cut. CheckPoint: Rock on String A girl twirls a rock on the end of a string around in a horizontal circle above her head as shown from above in the diagram. If the string brakes at the instant shown, which of the arrows best represents the resulting path of the rock? A B C D Top view looking down

  22. Show Prelecture

  23. Centripetal Acceleration

  24. Why is acceleration towards the center? Student comment: I don't really understand Centripetal Acceleration. How is the car accelerating toward the center of the circle?

  25. Why is acceleration towards the center? Student comment: I don't really understand Centripetal Acceleration. How is the car accelerating toward the center of the circle? a must be in same direction as dv

  26. Speed and Angular Velocity ωis the rate at which the angleθchanges: in radians!!! angular velocity (angular frequency) θ Once around: rotational period

  27. Another way to see it… dθ R

  28. ACT Prof. Gerton is standing on the equator. What is the direction of his acceleration? • Eastward • Westward • Inward, toward the earth’s center • Outward, away from the earth’s center

  29. ACT Prof. Gerton is standing on the equator. What is the magnitude of his acceleration? • 34 m/s2 • 3.4 m/s2 • 0.34 m/s2 • 0.034 m/s2 Hints:

  30. Centripetal acceleration due to Earth’s rotation is small

  31. ACT Prof. Gerton now moves from the equator to Salt Lake City. How does the magnitude of his acceleration change? • It increases. • It remains the same. • It decreases. • I’m not sure. North Pole South Pole

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