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Relative and Circular Motion a) Relative motion b) Centripetal acceleration Mechanics Lecture 3, Slide 1 Mechanics Lecture 3, Slide 2 What is the speed of Mike relative to the station? A. B. C. D. -1 m/s 30 m/s 29 m/s 31 m/s Mechanics Lecture 3, Slide 3 Relative Motion in 1 dimension Mechanics Lecture 3, Slide 4 Relative Position and Reference Frames Position of Mike in the ground frame is the vector sum of the position vector of Mike in the train reference frame and the position vector of the train in the ground reference frame. Mechanics Lecture 3, Slide 5 Relative Motion and Reference Frames Differentiate the position vectors to obtain the velocity vectors Mechanics Lecture 3, Slide 6 Relative Motion and Reference Frames Mechanics Lecture 3, Slide 7 Relative Motion and Reference Frames Mechanics Lecture 3, Slide 8 Prelecture 3, Questions 1 again A. B. C. 0% 0% 0% 0% D. Mechanics Lecture 3, Slide 9 CheckPoint 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 A) 6 m/s B) 8 m/s C) 10 m/s Mechanics Lecture 3, Slide 10 What is the speed of the dog relative to the ground? vdog,belt = 8 m/s vbelt,ground = 2 m/s A) 6 m/s B) 8 m/s C) 10 m/s +x vdog, ground = vdog, belt + vbelt, ground = (-8 m/s) + (2 m/s) = -6 m/s Mechanics Lecture 3, Slide 11 CheckPoint 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? vdog,belt = 8 m/s vbelt,ground = 2 m/s A) 6 m/s B) 8 m/s C) 10 m/s About 55% of you got this right – lets try it again. Mechanics Lecture 3, Slide 12 What is the speed of the dog relative to the girl? A. B. C. vdog,belt = 8 m/s 0% 0% 0% vbelt,ground = 2 m/s A) 6 m/s B) 8 m/s C) 10 m/s A) Because the girl is actually moving and the two vectors are opposite, so together they make 6 m/s B) Because the girl is not moving relative to the belt, and the dog is going 8 m/s relative to the belt, the dog is also moving 8 m/s relative to the girl.. C) The dog and girl are running towards each other so when you add the two velocities together it would be 8+2. Mechanics Lecture 3, Slide 13 What is the speed of the dog relative to the girl? vdog,belt = 8 m/s vbelt,ground = 2 m/s A) 6 m/s B) 8 m/s C) 10 m/s B) Because the girl is not moving relative to the belt, and the dog is going 8 m/s relative to the belt, the dog is also moving 8 m/s relative to the girl. Using the velocity formula: vdog, girl = vdog, belt + vbelt, girl = -8 m/s + 0 m/s = -8 m/s Mechanics Lecture 3, Slide 14 Relative Motion in 2 Dimensions Speed relative to shore Mechanics Lecture 3, Slide 15 Relative Motion in 2 Dimensions Direction w.r.t shoreline Mechanics Lecture 3, Slide 16 Relative Motion in 2 Dimensions Caveat !!!! The simple addition of velocities as shown only works for speeds much less than the speed of light…need special relativity at v~c. Mechanics Lecture 3, Slide 17 Moving Sidewalk Question A. B. C. D. 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? A) 2 m B) 4 m C) 6 m D) 8 m 0% 0% 0% 0% Mechanics Lecture 3, Slide 18 If the sidewalk wasn’t moving: Time to get across: Dt = distance / speed = 4m / 1m/s =4s vman ,sidewalk Mechanics Lecture 3, Slide 19 Just the sidewalk: vsidewalk ,hydrant Mechanics Lecture 3, Slide 20 Combination of motions: vsidewalk ,hydrant vman ,sidwalk vsidewalk ,hydrant Mechanics Lecture 3, Slide 21 D D = (speed of sidewalk) ∙ (time to get across) = (2 m/s) ∙ (4 s) = 8m Mechanics Lecture 3, Slide 22 Swim Race A. B. C. 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 swims perpendicular to the flow, Ann swims upstream at 30 degrees, and Carly swims downstream at 30 degrees. Who gets across the river first? A) Ann B) Beth C) Carly Ann Beth Carly y x 0% 0% 0% 0% Mechanics Lecture 3, Slide 23 Look at just water & swimmers Time to get across = D / Vy B D A C 30o 30o Vy,Beth = Vo y Vy,Ann = Vocos(30o) x Vy,Carly = Vocos(30o) Mechanics Lecture 3, Slide 24 Clicker Question 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 swims perpendicular to the flow, Ann swims upstream at 30 degrees, and Carly swims downstream at 30 degrees. Who gets across the river second? A) Ann B) Carly C) Both same Ann Carly y x Mechanics Lecture 3, Slide 25 Accelerating (Non-Inertial) Frames of Reference Accelerating Frame of Reference Confusing due to the fact that the acceleration can result in what appears to be a “push or pull”. Mechanics Lecture 3, Slide 26 Accelerated Frames of Reference Accelerating Frame of Reference Accelerometer can detect change in velocity Mechanics Lecture 3, Slide 27 Inertial Frames of Reference Inertial Frames of Reference Non-accelerating frames of reference in a state of constant, rectilinear motion with respect to one another. An accelerometer moving with any of them would detect zero acceleration. Mechanics Lecture 3, Slide 28 CheckPoint 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 breaks at the instant shown, which of the arrows best represents the resulting path of the rock? A B C D After the string breaks, the rock will have no force acting on it, so it cannot accelerate. Therefore, it will maintain its velocity at the time of the break in the string, which is directed tangent to the circle. Top view looking down Mechanics Lecture 3, Slide 29 Show Prelecture https://www.smartphysics.com/Course/PlaySlideshow?unitItemID=192821 Mechanics Lecture 3, Slide 30 Rotating Reference Frames Speed is constant. Direction Changing Acceleration Mechanics Lecture 3, Slide 31 Rotating Reference Frames Mechanics Lecture 3, Slide 32 Rotating Reference Frames Mechanics Lecture 3, Slide 33 Rotating Reference Frames Mechanics Lecture 3, Slide 34 Rotating Reference Frames Mechanics Lecture 3, Slide 35 Centripetal Acceleration Constant speed in circular path Acceleration directed toward center of circle What is the magnitude of acceleration? Proportional to: 1. Speed 1. time rate of change of angle or angular velocity Mechanics Lecture 3, Slide 36 Centripetal Acceleration Mechanics Lecture 3, Slide 37 Centripetal Acceleration Mechanics Lecture 3, Slide 38 Centripetal Acceleration Mechanics Lecture 3, Slide 39 v = wR w is the rate at which the angle q changes: w dq dt q Once around: v Dx / Dt = 2pR / T w Dq / Dt = 2p / T Mechanics Lecture 3, Slide 40 v = wR Another way to see it: dq R v dt =R dq dq vR dt v = Rw Mechanics Lecture 3, Slide 41 Centripetal Acceleration-Example Mechanics Lecture 3, Slide 42 Centripetal Acceleration due to Earth’s rotation Mechanics Lecture 3, Slide 43 Mechanics Lecture 3, Slide 44