Relative and Circular Motion - AstronomyRelative and Circular Motion a) Relative motion b)...

Relative and Circular Motiona) Relative motionb) Centripetal acceleration

Mechanics Lecture 3, Slide 1

Time on Video Prelecture

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Lecture Thoughts


What is the speed of Mike relative to the station?

A. -1 m/sB. 30 m/sC. 29 m/sD. 31 m/s

0%

30%

70%

0%


Relative Motion in 1 dimension


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


Prelecture 3, Questions 1 again

A.

B.

C.

D.

0% 0%0%0%-12m


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?


vdog, ground = vdog, belt + vbelt, ground= (−8 m/s) + (2 m/s) = −6 m/s

+x


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


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?


vdog,belt = 8 m/s

A) 6 m/s B) 8 m/s C) 10 m/s


What is the speed of the dog relative to the girl? A.

B.

C.

0% 0%0%


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..


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?


vdog,belt = 8 m/s

A) 6 m/s B) 8 m/s C) 10 m/s


Relative Motion in 2 Dimensions

Speed relative to shore



Direction w.r.t shoreline



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.

B.

C.

D.

0%

90%

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


Time to get across:

∆t = distance / speed= 4m / 1m/s = 4 s

If the sidewalk wasn’t moving:

sidewalkmanv ,


Just the sidewalk:

hydrantsidewalkv ,


D = (speed of sidewalk) ∙ (time to get across)= (2 m/s) ∙ (4 s) = 8 m

D


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,,

=⇒=

=⇒=−

+=×+−++=

×+=

+=

×+=


Swim Race A.

B.

C.

20%

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


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


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


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 motionwith respect to one another. An accelerometer moving with any of them woulddetect zero acceleration.


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


Survey


Rotating Reference Frames



Direction ChangingSpeed is constant.

Acceleration


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


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 =


dθ v dt

R

=R dθ

Another way to see it:

v = ωR

v = RωdtdRv θ

=


Centripetal Acceleration-Example


Centripetal Acceleration due to Earth’s rotation


River Rescue

N

E

dockriverv ,

riverboatv ,

dockboatv ,

Dock Frame

θ


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



( ) ( ) 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 ,




( )( )

( ) ( )

( ) ( ) 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 ,


)(, 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, +−=


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


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


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


Relative and Circular Motion - AstronomyRelative and Circular Motion a) Relative motion b)...

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