relative and circular motion - astronomyrelative and circular motion a) relative motion b)...
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Relative and Circular Motiona) Relative motionb) Centripetal acceleration
Mechanics Lecture 3, Slide 1
Mechanics Lecture 3, Slide 2
Time on Video Prelecture
Looks like mostly everyone here has viewed entire prelecture…GOOD!Thank you to those of you who did!
Mechanics Lecture 3, Slide 3
Lecture Thoughts
Mechanics Lecture 3, Slide 4
What is the speed of Mike relative to the station?
A. -1 m/sB. 30 m/sC. 29 m/sD. 31 m/s
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Mechanics Lecture 3, Slide 5
Relative Motion in 1 dimension
Mechanics Lecture 3, Slide 6
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 7
Relative Motion and Reference Frames
Differentiate the position vectors to obtain the velocity vectors
Mechanics Lecture 3, Slide 8
Relative Motion and Reference Frames
Mechanics Lecture 3, Slide 9
Relative Motion and Reference Frames
Mechanics Lecture 3, Slide 10
Prelecture 3, Questions 1 again
A.
B.
C.
D.
0% 0%0%0%-12m
Mechanics Lecture 3, Slide 11
vbelt,ground = 2 m/s
vdog,belt = 8 m/s
A) 6 m/s B) 8 m/s C) 10 m/s
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?
Mechanics Lecture 3, Slide 12
vdog, ground = vdog, belt + vbelt, ground= (−8 m/s) + (2 m/s) = −6 m/s
+x
vbelt,ground = 2 m/s
vdog,belt = 8 m/s
What is the speed of the dog relative to the ground?
A) 6 m/s B) 8 m/s C) 10 m/s
Mechanics Lecture 3, Slide 13
42% of you got this right –let’s do this again
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?
vbelt,ground = 2 m/s
vdog,belt = 8 m/s
A) 6 m/s B) 8 m/s C) 10 m/s
Mechanics Lecture 3, Slide 14
What is the speed of the dog relative to the girl? A.
B.
C.
0% 0%0%
vbelt,ground = 2 m/s
vdog,belt = 8 m/s
A) 6 m/s B) 8 m/s C) 10 m/s
C) The dog and girl are running towards each other so when you add the two velocities together it would be 8+2.
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..
Mechanics Lecture 3, Slide 15
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
What is the speed of the dog relative to the girl?
vbelt,ground = 2 m/s
vdog,belt = 8 m/s
A) 6 m/s B) 8 m/s C) 10 m/s
Mechanics Lecture 3, Slide 16
Relative Motion in 2 Dimensions
Speed relative to shore
Mechanics Lecture 3, Slide 17
Relative Motion in 2 Dimensions
Direction w.r.t shoreline
Mechanics Lecture 3, Slide 18
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 19
Moving Sidewalk Question A.
B.
C.
D.
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0%10%
A man starts to walk along the dotted line painted on a moving sidewalktoward a fire hydrant that is directly across from him. The width of thewalkway 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
Mechanics Lecture 3, Slide 20
Time to get across:
∆t = distance / speed= 4m / 1m/s = 4 s
If the sidewalk wasn’t moving:
sidewalkmanv ,
Mechanics Lecture 3, Slide 21
Just the sidewalk:
hydrantsidewalkv ,
Mechanics Lecture 3, Slide 22
D = (speed of sidewalk) ∙ (time to get across)= (2 m/s) ∙ (4 s) = 8 m
D
Mechanics Lecture 3, Slide 23
Combination of motions: Vector solution
jiv
jiv
jiv
vvv
hydrantman
hydrantsidewalk
sidewalkman
hydrantsidewalksidewalkmanhydrantman
ˆ2ˆ1
ˆ2ˆ0
ˆ0ˆ1
,
,
,
,,,
+−=
+=
+−=
+=
( )( )
myjyjt
stiit
jyitjijitr
tvrtr
jir
tvrtr
cross
crosscross
crosscrosshydrantman
crosshydrantmanhydrantmancrosshydrantman
hydrantman
hydrantmanhydrantmanhydrantman
8ˆˆ2
4ˆ0ˆ4
ˆˆ0ˆ2ˆ1ˆ0ˆ4)(
)(
ˆ0ˆ4
)(
,
,0,,
0,
,0,,
=⇒=
=⇒=−
+=×+−++=
×+=
+=
×+=
Mechanics Lecture 3, Slide 24
Mechanics Lecture 3, Slide 25
Swim Race A.
B.
C.
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0%
60%
20%
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
x
y
AnnBeth
Carly
Mechanics Lecture 3, Slide 26
AB
C
Vy,Beth = Vo
30o 30o
Vy,Ann = Vocos(30o)
Vy,Carly = Vocos(30o)
Time to get across = D / Vy
D
Look at just water & swimmers
x
y
Mechanics Lecture 3, Slide 27
x
y
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 Carlyswims downstream at 30 degrees.
Who gets across the river second?
A) Ann B) Carly C) Both same
Ann Carly
Mechanics Lecture 3, Slide 28
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 29
Accelerated Frames of Reference
Accelerating Frame of Reference
Accelerometer can detect change in velocity
Mechanics Lecture 3, Slide 30
Inertial Frames of Reference
Inertial Frames of Reference
Non-accelerating frames of reference in a state of constant, rectilinear motionwith respect to one another. An accelerometer moving with any of them woulddetect zero acceleration.
Mechanics Lecture 3, Slide 31
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
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.
CheckPoint
Mechanics Lecture 3, Slide 32
Survey
Mechanics Lecture 3, Slide 33
Rotating Reference Frames
Mechanics Lecture 3, Slide 34
Rotating Reference Frames
Direction ChangingSpeed is constant.
Acceleration
Mechanics Lecture 3, Slide 35
Rotating Reference Frames
Mechanics Lecture 3, Slide 36
Rotating Reference Frames
Mechanics Lecture 3, Slide 37
Rotating Reference Frames
Mechanics Lecture 3, Slide 38
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 39
Centripetal Acceleration
Mechanics Lecture 3, Slide 40
Centripetal Acceleration
Mechanics Lecture 3, Slide 41
Centripetal Acceleration
Mechanics Lecture 3, Slide 42
v = ωR
Once around:
ω = ∆θ / ∆t = 2π / Τv = ∆x / ∆t = 2πR / T
ω is the rate at which the angle θ changes:
θ
dtdθω =
Mechanics Lecture 3, Slide 43
dθ v dt
R
=R dθ
Another way to see it:
v = ωR
v = RωdtdRv θ
=
Mechanics Lecture 3, Slide 44
Centripetal Acceleration-Example
Mechanics Lecture 3, Slide 45
Centripetal Acceleration due to Earth’s rotation
Mechanics Lecture 3, Slide 46
Mechanics Lecture 3, Slide 47
River Rescue
N
E
dockriverv ,
riverboatv ,
dockboatv ,
Dock Frame
θ
Mechanics Lecture 3, Slide 48
River Rescue-Dock Frame
jir
tvrtrjivv
jiv
jivv
dockchild
dockchilddockchilddockchild
dockriverdockchild
riverboat
childriverriverchild
ˆ6.0ˆ5.2
)(
ˆ)0(ˆ)1.3(
ˆ)sin*8.24(ˆ)cos*8.24(
ˆ0ˆ0
0
0
,
,,,
,,
,
,,
+−=
×+=
+==
+=
+==
θθ N
E
dockchildv ,
ij
boatchildv ,
Dock Frame0,dockchildr -2.5i,0.6j )(, meetdockchild ttr =
( )
( )jittr
tjijittr
tvrtvrtr
tjijitvrttr
jivvtv
meetboatchild
meetmeetboatchild
boatchilddockchildboatchildboatchildboatchild
meetmeetdockchilddockchildmeetdockchild
riverboatboatriverboatchild
ˆ0ˆ0)(
ˆ)sin*8.24(ˆ)cos*8.24(ˆ6.0ˆ5.2)(
)(
ˆ)0(ˆ)1.3(ˆ6.0ˆ5.2)(
ˆ)sin*8.24(ˆ)cos*8.24()(
,
,
,,,,,
,,,
,,,
00
0
+==
×−+−++−==
×+=×+=
×+++−=×+==
−+−=−==
θθ
θθ
Mechanics Lecture 3, Slide 49
River Rescue-Dock Frame
( ) ( ) hrttt
tt
tt
meetmeetmeet
meetmeet
meetmeet
1037.06.05.28.24
118.24
6.08.24
5.2
8.246.0sin0sin*8.246.0
8.245.2cos0cos*8.245.2
2222
=+−=⇒=
×
+
×
−
×=⇒=×−
×−
=⇒=×−−
θθ
θθ
N
E
dockchildv ,
ij
boatchildv ,
Dock Frame0,dockchildr -2.5i,0.6j )(, meetdockchild ttr =
Mechanics Lecture 3, Slide 50
River Rescue-Dock Frame
( )( )
( ) ( )
( ) ( ) kmd
tttrd
jitttr
tjijittr
tvrttrhrt
meetmeetdockchild
meetmeetdockchild
meetmeetdockchild
meetdockchilddockchildmeetdockchild
meet
26.26.01037.01.35.2
6.01.35.2)(
ˆ6.0ˆ1.35.2)(
ˆ)0(ˆ)1.3(ˆ6.0ˆ5.2)(
)(1037.0
22
22,
,
,
,,, 0
=+×+−=
+×+−===
+×+−==
×+++−==
×+===
N
E
dockchildv ,
ij
boatchildv ,
Dock Frame0,dockchildr -2.5i,0.6j )(, meetdockchild ttr =
)(, meetdockchild ttrd ==
deg4.166
2333.01037.08.24
6.08.24
6.0sin
9721.01037.08.245.2
8.245.2cos
=
=×
=×
=
−=×−
=×
−=
θ
θ
θ
meet
meet
t
t
jiv riverboatˆ)786.5(ˆ108.24, +−=
Mechanics Lecture 3, Slide 51
River Rescue-child frame
jir
tvrtrjivtv
jivv
jivv
jiv
jivv
childdock
childdockchilddockchilddock
riverboatboatchild
dockchildchilddock
dockriverdockchild
riverboat
childriverriverchild
ˆ6.0ˆ5.2
)(
ˆ)sin*8.24(ˆ)cos*8.24()(
ˆ)0(ˆ)1.3(
ˆ)0(ˆ)1.3(
ˆ)sin*8.24(ˆ)cos*8.24(
ˆ0ˆ0
0
0
,
,,,
,,
,,
,,
,
,,
−=
×+=
−+−=−=
+−=−=
+==
+=
+==
θθ
θθ
N
Echilddockv , i
j
childboatv ,
Child Frame
0,childdockr 2.5i,-0.6j
)(, meetchilddock ttr =
θ
Mechanics Lecture 3, Slide 52
River Rescue-child frame( )
( )
( ) ( ) hrttt
tt
tt
jittr
tjijittr
tvrtvrtr
tjijitvrttr
meetmeetmeet
meetmeet
meetmeet
meetboatchild
meetmeetchildboat
riverboatchilddockchildboatchildboatchildboat
meetmeetchilddockchilddockmeetchilddock
1037.06.05.28.24
118.24
6.08.24
5.2
8.246.0sin0sin*8.246.0
8.245.2cos0cos*8.245.2
ˆ0ˆ0)(
ˆ)sin*8.24(ˆ)cos*8.24(ˆ6.0ˆ5.2)(
)(
ˆ)0(ˆ)1.3(ˆ6.0ˆ5.2)(
2222
,
,
,,,,,
,,,
00
0
=+−=⇒=
×
+
×
−
×=⇒=×+−
×−
=⇒=×+
+==
×++−==
×+=×+=
×+−+−=×+==
θθ
θθ
θθ
N
Echilddockv , i
j
childboatv ,
Child Frame
0,childdockr 2.5i,-0.6j
)(, meetchilddock ttr =
θ
Mechanics Lecture 3, Slide 53
River Rescue( )
( )
( ) ( )
( ) ( ) kmd
tttrd
jitttr
hrt
tjijitvrttr
meetmeetchilddock
meetmeetchilddock
meet
meetmeetchilddockchilddockmeetchilddock
26.26.01037.01.35.2
6.01.35.2)(
ˆ6.0ˆ1.35.2)(
1037.0
ˆ)0(ˆ)1.3(ˆ6.0ˆ5.2)(
22
22,
,
,,, 0
=−+×−=
−+×−===
−×−==
=
×+−+−=×+==
deg4.166
2333.01037.08.24
6.08.24
6.0sin
9721.01037.08.245.2
8.245.2cos
=
=×
=×
=
−=×−
=×
−=
θ
θ
θ
meet
meet
t
t
Mechanics Lecture 3, Slide 54