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Relative and Circular Motion. a ) Relative motion b) Centripetal acceleration. What is the speed of Mike relative to the station?. -1 m/s 30 m/s 29 m/s 31 m/s. Relative Motion in 1 dimension. Relative Position and R eference F rames.
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Relative and Circular Motion a) Relative motion b) Centripetal acceleration
What is the speed of Mike relative to the station? • -1 m/s • 30 m/s • 29 m/s • 31 m/s
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.
Relative Motion and Reference Frames Differentiate the position vectors to obtain the velocity vectors
CheckPoint A girl stands on a moving sidewalk that moves to the right at 2 m/srelative 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/sB)8 m/sC)10 m/s
What is the speed of the dog relative to the ground? 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 A)6 m/sB)8 m/sC)10 m/s
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/sB)8 m/sC)10 m/s About 55% of you got this right – lets try it again.
What is the speed of the dog relative to the girl? vdog,belt= 8 m/s vbelt,ground= 2 m/s A)6 m/sB)8 m/sC)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.
What is the speed of the dog relative to the girl? vdog,belt= 8 m/s vbelt,ground= 2 m/s A)6 m/sB)8 m/sC)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
Relative Motion in 2 Dimensions Speedrelative to shore
Relative Motion in 2 Dimensions Direction w.r.t shoreline
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.
Moving Sidewalk Question • 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? • 2 m • 4 m • 6 m • 8 m 1 m/s Moving sidewalk 2 m/s 4 m
If the sidewalk wasn’t moving: 1 m/s Time to get across: Dt = distance / speed = 4m / 1m/s = 4 s 4 m
Just the sidewalk: 2 m/s 4 m
Combination of motions: 1 m/s 4 m 2 m/s
1 m/s D 2 m/s 4 m D = (speed of sidewalk) ∙ (time to get across) = (2 m/s) ∙ (4 s) = 8 m
Swim Race y x • 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) AnnB) BethC) Carly Beth Ann Carly
Look at just water & swimmers y x Time to get across=D / Vy B A C D 30o 30o Vy,Beth = Vo Vy,Ann = Vocos(30o) Vy,Carly = Vocos(30o)
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) AnnB) Carly C) Both same y x Ann Carly
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”.
Accelerated Frames of Reference Accelerating Frame of Reference Accelerometer can detect change in velocity
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.
CheckPoint 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. 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 Top view looking down
Show Prelecture https://www.smartphysics.com/Course/PlaySlideshow?unitItemID=192821
Rotating Reference Frames Speed is constant. Direction Changing Acceleration
Centripetal Acceleration • Constant speed in circular path • Acceleration directed toward center of circle • What is the magnitude of acceleration? • Proportional to: • Speed • time rate of change of angle or angular velocity
Once around: v = Dx /Dt = 2pR / T w = Dq /Dt = 2p / T v = wR wis the rate at which the angleqchanges: q
v = wR Another way to see it: v dt =R dq dq R v = Rw
