Lect03 - Physics 211 Lecture 3 Today's Concepts a Relative...

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Unformatted text preview: Physics 211 Lecture 3 Today's Concepts: a) Relative Motion Relative b) Centripetal acceleration Physics 211 Lecture 3, Slide 1 Physics How familiar are you with the concepts of relative How motion and centripetal acceleration from your high school course. A) I already know this stuff B) It seems familiar, but I need a review B) C) We didn't learn this in high school C) Stuff you asked about: Stuff I would like to talk more about centripetal motion. The concept was little confusing. would could you explain the what the correct answer is for the speed of the dog relative to girl and how you were able to get that answer. the omega thing.. I couldn't understand it by just watching prelecture once... I think I have to watch it again. What in the world was the relativistic speed stuff as v approaches the speed of light? I know the general properties, but it confused the heck out of me in the middle of the prelecture. If it doesn't matter and we aren't going to calculate it at any time in the class, then my recommendation is just to leave it out I find that the notation for the relative velocities is confusing. If we could briefly discuss the relationship between angular and linear velocity, that linear would be a good refresher. I think understood all of it well. The slide used calculus to find the centripetal acceleration. Will we need to use calculus too in this class? I like pudding Relative Motion vac = vab + vbc Physics 211 Lecture 3, Slide 4 Physics Preflight A girl stands on a moving sidewalk that is moves to the right at 2 m/s m/s relative to the ground. A dog runs toward the girl in the opposite direction relative along the sidewalk at a speed of 8 m/s relative to the sidewalk. relative What is the speed of the dog relative to the ground? Vdog,belt = 8m/s 8m/s Vbelt,ground = 2m/s 2m/s A) 6 m/s m/s B) 8m/s B) 8m/s C) 10m/s C) 10m/s Physics 211 Lecture 3, Slide 5 Physics What is the speed of the dog relative to the ground? Vdog,belt = 8m/s 8m/s Vbelt,ground = 2m/s 2m/s A) 6 m/s m/s B) 8m/s B) 8m/s C) 10m/s C) 10m/s V(dog, ground) = V(dog, belt) + V(belt, ground) -->V dog,ground V(dog = -8m/s + 2m/s = -6m/s Physics 211 Lecture 3, Slide 6 Physics Preflight/Act A girl stands on a moving sidewalk that is moves to the right at 2 m/s m/s relative to the ground. A dog runs toward the girl in the opposite direction relative along the sidewalk at a speed of 8 m/s relative to the sidewalk. relative What is the speed of the dog relative to the girl? Vdog,belt = 8m/s 8m/s Vbelt,ground = 2m/s 2m/s A) 6 m/s m/s B) 8m/s B) 8m/s C) 10m/s C) 10m/s Physics 211 Lecture 3, Slide 7 Physics What is the speed of the dog relative to the girl? Vdog,belt = 8m/s 8m/s Vbelt,ground = 2m/s 2m/s A) 6 m/s m/s B) 8m/s B) 8m/s C) 10m/s C) 10m/s A) The dog moves left at 8m/s relative to the sidewalk, the girl moves right at 2m/s The also relative to the sidewalk. 8 - 2 = 6m/s 6m/s B) The girl is stationary on the sidewalk, so this is simply the dog's speed relative The to the sidewalk. C) The answer is 10m/s because the dog is running toward the girl at a speed of The 8m/s and the girl is moving toward the dog at 2m/s. Physics 211 Lecture 3, Slide 8 Physics Tractor Demo (moving cardboard) 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 ABC Physics 211 Lecture 3, Slide 9 Physics 37 Act 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 relative walking speed is 1 m/s, how far away will he be from the hydrant when he reaches the other side? 1 m/s A) B) C) D) 2 m 4 m 6 m 8 m Moving sidewalk 2 m/s 4m Physics 211 Lecture 3, Slide 10 Physics 1 m/s If the sidewalk wasn’t moving: Time to get across: Time Δt = distance / speed distance = 4m / 1m/s =4s 4m vman , sidewalk 1 m/s Just the sidewalk: 2 m/s 4m vsidewalk ,hydrant 1 m/s Combination of motions: 2 m/s 4m vman ,hydrant = vman , sidewalk + vsidewalk ,hydrant 1 m/s D 2 m/s 4m D = (speed of sidewalk) x (time to get across) (time = (2 m/s) x (4 s) = 8 m y ACT ACT 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, Beth (B) swims perpendicular to the flow, swims Ann (A) swims upstream at 30 degrees, and Carly (C) swims (C) swims downstream at 30 degrees. x Who gets across the river first? A) Ann B) Beth C) Carly ABC 42 Look at just water & swimmers y x Time to get across = Distance / Vy B A C 30o 30o Vy,Beth = V0 Vy,Ann = V0cos(30) cos(30) Vy,Carly = V0cos(30) cos(30) ACT ACT y 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, Beth (B) swims perpendicular to the flow, swims Ann (A) swims upstream at 30 degrees, and Carly (C) swims swims (C) swims downstream at 30 degrees. x Who gets across the river second? A) Ann B) Same C) Carly ABC 42 Preflight/Act 4 Demo! 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. A B If the string brakes at the instant shown, which of the arrows best represents the resulting path of the rock? C D since the rock is no longer attached to the string there is no centripetal acceleration so the rock simply travels in a straight path Top view looking down Physics 211 Lecture 3, Slide 19 Physics Show Prelecture Physics 211 Lecture 3, Slide 20 Physics Physics 211 Lecture 3, Slide 21 Physics v = ωR ω is the rate at which the angle θ changes: is changes: dθ ω= dt θ Once around: v = Δx/Δt = 2πR/T ω = Δθ/Δt = 2π/Τ Δθ Physics 211 Lecture 3, Slide 22 Physics v = ωR Another way to see it: dθ R vdt =Rdθ dθ v=R dt v = Rω Physics 211 Lecture 3, Slide 23 Physics We can ignore this acceleration due to Earth's rotation since its small Physics 211 Lecture 3, Slide 24 Physics ...
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