Transcript Relative Motion and Reference Frames

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.
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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
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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
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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
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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
vR
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