Download - 3c Free Fall Kinematics 3D
-
8/19/2019 3c Free Fall Kinematics 3D
1/28
Physics 111: Lecture 2, Pg 1
Physics: Lecture 2Physics: Lecture 2
Today’s AgendaToday’s Agenda
Recap of 1-D motion with constant acceleration
1-D free fall example
3-D Kinematics
hoot the mon!ey
"ase#all $n%epen%ence of x an% y components
-
8/19/2019 3c Free Fall Kinematics 3D
2/28
Physics 111: Lecture 2, Pg 2
Review:Review:
&or constant acceleration we foun%:
at v v 0 +=
2 0 0 at 2
1t v x x ++=
const a =
x
a
v t
t
t
v)(v 2
1v
) x 2a(x v v
0 av
0 2 0
2
+=
−=−
&rom which we %eri'e%:
-
8/19/2019 3c Free Fall Kinematics 3D
3/28
Physics 111: Lecture 2, Pg 3
Recall what you saw:Recall what you saw:
12
2 2
3 2
(2
2 0 0
at 2
1t v x x ++=
-
8/19/2019 3c Free Fall Kinematics 3D
4/28
Physics 111: Lecture 2, Pg (
1-D Free-Fall1-D Free-Fall
)his is a nice example of constant acceleration *gra'ity+: $n this case, acceleration is cause% #y the force of gra'ity:
sually pic! y-axis upwar%.
/cceleration of gra'ity is %own.:
y
ay 0 − g
gt v v y
0 y −=
2 y 0 0
t g 2
1t v y y −+=
y
a
v
t
t
t
g ay −=
-
8/19/2019 3c Free Fall Kinematics 3D
5/28
Physics 111: Lecture 2, Pg
Gravity acts:Gravity acts:
g %oes not %epen% on the nature of the material
alileo *14(-14(2+ figure% this out without fancy cloc!s5 rulers
%emo - feather 5 penny in 'acuum
6ominally, g 0 7891 ms2
/t the e;uator g 0 78
-
8/19/2019 3c Free Fall Kinematics 3D
6/28
Physics 111: Lecture 2, Pg 4
Pro!le":Pro!le":
)he pilot of a ho'ering helicopter%rops a lea% #ric! from a heightof 1>>> m8 ?ow long %oes it ta!eto reach the groun% an% how fastis it mo'ing when it gets there@*neglect air resistance+
1000 m
-
8/19/2019 3c Free Fall Kinematics 3D
7/28Physics 111: Lecture 2, Pg <
Pro!le":Pro!le":
&irst choose coor%inate system8
Arigin an% y -%irection8
6ext write %own position e;uation:
RealiBe that v 0y = 0 8
2 0y 0 gt
2 1t v y y +=
2 0 gt
2
1y y −=
1000 m
y = 0
y
-
8/19/2019 3c Free Fall Kinematics 3D
8/28Physics 111: Lecture 2, Pg 9
Pro!le":Pro!le":
ol'e for time t when y = 0 gi'en that y 0 =1000 m.
Recall:
ol'e for v y :
y 0 = 1000 m
y
s314sm819
m1000 2
g
y 2 t
2 0 8
8=
×==
2 0
gt 2
1y y -=
y = 0
+* 0 2
y 0 2 y y y a2 v v -- =
sm140
gy 2 v 0 y
−=
±=
-
8/19/2019 3c Free Fall Kinematics 3D
9/28Physics 111: Lecture 2, Pg 7
Lecture 2#Lecture 2# Act 1 Act 1
1D ree all1D ree all Alice and $ill are standing at the to% o a cli o heightAlice and $ill are standing at the to% o a cli o height H H & $oth throw a !all with initial s%eed& $oth throw a !all with initial s%eed v v 0 0 # Alice straight# Alice straight downdown and $ill straightand $ill straight u%u%& The& The
s%eed o the !alls when they hit the ground ares%eed o the !alls when they hit the ground are v v A A andand v v BB res%ect ively&res%ectively& 'hich o the ollowing is true:'hich o the ollowing is true:
(a)(a) v v A A ** v v BB (!)(!) v v A A ++ v v BB (c)(c) v v A A ,, v v BB
v v 0 0
v v 0 0
$ill$illAliceAlice
H H
v v A A v v BB
-
8/19/2019 3c Free Fall Kinematics 3D
10/28Physics 111: Lecture 2, Pg 1>
Lecture 2#Lecture 2# Act 1 Act 11D Free all1D Free all
ince the motion up an% #ac! %own is symmetric, intuition shoul% tell you that v = v 0
Ce can pro'e that your intuition is correct:
v v 0 0
$ill$ill
H H
v v = v = v 0 0
( ) 0 H H g 2 v v 2 0 2
=−−=− +*.uation:.uation:
This loo/s 0ust li/e $ill threwThis loo/s 0ust li/e $ill threwthe !all down with s%eedthe !all down with s%eed v v 0 0 # so# so
the s%eed at the !otto" shouldthe s%eed at the !otto" should!e the sa"e as Alice’s !all&!e the sa"e as Alice’s !all&
y = 0 y = 0
-
8/19/2019 3c Free Fall Kinematics 3D
11/28Physics 111: Lecture 2, Pg 11
Lecture 2# Act 1Lecture 2# Act 11D Free all1D Free all
'e can also 0ust use the e.uation directly:'e can also 0ust use the e.uation directly:
( )H 0 g 2 v v 2 0 2
−−=− +*Alice:Alice:
v v 0 0
v v 0 0
AliceAlice $ill$ill
y = 0 y = 0
( )H 0 g 2 v v 2 0 2 −−=− +*$ill:$ill:sa"e sa"e
-
8/19/2019 3c Free Fall Kinematics 3D
12/28Physics 111: Lecture 2, Pg 12
-D 3ine"atics-D 3ine"atics
)he position, 'elocity, an% acceleration of a particle in 3%imensions can #e expresse% as:
r r = x i i + y j j + z k k
v v = v x i i + v y j j + v z k k *i i , j j , k k unit vectos )
aa = a x i i + ay j j + az k k
Ce ha'e alrea%y seen the 1-D !inematics e;uations:
a !v
!t
! x
!t = =
2
2 v
!x
!t = x x(t = )
-
8/19/2019 3c Free Fall Kinematics 3D
13/28Physics 111: Lecture 2, Pg 13
-D 3ine"atics-D 3ine"atics
&or 3-D, we simply apply the 1-D e;uations to each of thecomponent e;uations8
Chich can #e com#ine% into the 'ector e;uations:
r r = r r (t) v v = ! r r " !t aa = ! 2 r r " !t 2
a ! x
!t x =
2
2
v !x
!t x =
x x(t = )
a ! y
!t y =
2
2
v !y
!t y =
y y t = ( )
a ! z
!t z =
2
2
v !z
!t z =
z z t = ( )
-
8/19/2019 3c Free Fall Kinematics 3D
14/28Physics 111: Lecture 2, Pg 1(
-D 3ine"atics-D 3ine"atics
o for constant acceleration we can integrate to get:
aa = const
v v = v v 0 + aa t
r r = r r 0 + v v 0 t + 1 " 2 aa t 2
*where aa, v v , v v >, r r , r r >, are all 'ectors+
-
8/19/2019 3c Free Fall Kinematics 3D
15/28Physics 111: Lecture 2, Pg 1
2-D 3ine"atics2-D 3ine"atics
=ost 3-D pro#lems can #e re%uce% to 2-D pro#lems whenacceleration is constant:
hoose y axis to #e along %irection of acceleration
hoose x axis to #e along the other. %irection of motion
4a"%le4a"%le: )hrowing a #ase#all *neglecting air resistance+
/cceleration is constant *gra'ity+
hoose y axis up: ay = #g
hoose x axis along the groun% in the %irection of thethrow
lost mar#les
-
8/19/2019 3c Free Fall Kinematics 3D
16/28Physics 111: Lecture 2, Pg 14
5546 and 5y6 co"%onents o "otion are46 and 5y6 co"%onents o "otion areinde%endent&inde%endent&
/ man on a train tosses a #all straight up in the air8
Eiew this from two reference frames:
Reference frameon the mo'ing train8
Reference frame
on the groun%8
art
-
8/19/2019 3c Free Fall Kinematics 3D
17/28Physics 111: Lecture 2, Pg 1<
Pro!le":Pro!le":
=ar! =cwire clo##ers a fast#all towar% center-fiel%8 )he#all is hit 1 m *y o + a#o'e the plate, an% its initial 'elocity is348 ms *v + at an angle of 3>o * + a#o'e horiBontal8 )hecenter-fiel% wall is 113 m *D+ from the plate an% is 3 m *h+high8
Chat time %oes the #all reach the fence@Chat time %oes the #all reach the fence@
Does =ar! get a home run@Does =ar! get a home run@
θ
v v
h
D
y 0
-
8/19/2019 3c Free Fall Kinematics 3D
18/28Physics 111: Lecture 2, Pg 19
Pro!le"&&&Pro!le"&&&
hoose y axis up8
hoose x axis along the groun% in the %irection of the hit8
hoose the origin (0,0) to #e at the plate8
ay that the #all is hit at t = 0 , x = x 0 = 0
F;uations of motion are:
v x = v 0x v y = v 0y # gt
x = v x t y = y 0 + v 0y t # 1 " 2 gt 2
-
8/19/2019 3c Free Fall Kinematics 3D
19/28Physics 111: Lecture 2, Pg 17
Pro!le"&&&Pro!le"&&&
se geometry to figure out v 0x an% v 0y :
y
x
g
θ
v v
v 0x
v 0y
&in% v 0x = $v $ cos θ8an% v 0y = $v $ sin θ8
y 0
-
8/19/2019 3c Free Fall Kinematics 3D
20/28Physics 111: Lecture 2, Pg 2>
Pro!le"&&&Pro!le"&&&
)he time to reach the wall is: t = % " v x *easy+
Ce ha'e an e;uation that tell us y(t) = y 0 + v 0y t + a t 2 " 2
o, weGre %one8888now we Hust plug in the num#ers:
&in%:
v x 0 348 cos*3>+ ms 0 3184 ms
v y 0 348 sin*3>+ ms 0 1982 ms
t 0 *113 m+ *3184 ms+ 0 389 s
y(t) 0 *18> m+ I *1982 ms+*389 s+ -
*>8+*789 ms2+*389 s+2
0 *18> I 483 - 4289+ m 0 &7 ""
ince the wall is " high, =ar! gets the homer
-
8/19/2019 3c Free Fall Kinematics 3D
21/28Physics 111: Lecture 2, Pg 21
Lecture 2#Lecture 2# Act 3 Act 38otion in 2D8otion in 2D
)wo foot#alls are thrown from the same point on a f lat fiel%8 "oth are thrown at an angle of 30 o a#o'e the horiBontal8 &a'' 2 has twice the initialspee% of a'' 18 $f a'' 1 is caught a %istance %1 from the thrower, how far away from the thrower %2 will the recei'er of a'' 2 #e when he catches it@
*a+ %2 = 2%1 *#+ %2 = 4%1 *c+ %2 = 8%1
-
8/19/2019 3c Free Fall Kinematics 3D
22/28Physics 111: Lecture 2, Pg 22
Lecture 2#Lecture 2# Act 3 Act 39olution9olution
)he %istance a #all will go is simply x = (oizonta' s*ee!) x (time in ai) = v 0x t
2 y 0 0
t g
2
1t v y y −+=
)o figure out time in air., consi%er the
e;uation for the height of the #all:
0 t g 2
1t v 2
y 0 =− Chen the #all is caught, y = y 0
0 t g 2
1v t
y 0 =
− g
v 2 t y 0 =
t = 0 *time of throw+
*time of catch+
twosolutions
-
8/19/2019 3c Free Fall Kinematics 3D
23/28Physics 111: Lecture 2, Pg 23
Lecture 2#Lecture 2# Act 3 Act 39olution9olution
g
v 2 t y 0 = o the time spent in the air is proportional to v 0y :
ince the angles are the same, #oth v 0y an% v 0x for a'' 2
are twice those of a'' 18
a'' 1
a'' 2
v 0y ,1
v 0x ,1
v 0y ,2
v 0x ,2
v 0,1
v 0,2
"all 2 is in the air tice as long as #all 1, #ut it also has tice the horiBontal spee%, so it will go times as far
x = v 0x t
-
8/19/2019 3c Free Fall Kinematics 3D
24/28
Physics 111: Lecture 2, Pg 2(
9hooting the 8on/ey9hooting the 8on/ey(tran.uili;er gun)(tran.uili;er gun)
Chere %oes the Boo!eeperaim if he wants to hit the mon!ey@
* ?e !nows the mon!ey willlet go as soon as he shoots +
-
8/19/2019 3c Free Fall Kinematics 3D
25/28
Physics 111: Lecture 2, Pg 2
9hooting the 8on/ey&&&9hooting the 8on/ey&&&
$f there were no gra'ity, simply aim
at the mon!ey
r = r 0
r =v 0 t
-
8/19/2019 3c Free Fall Kinematics 3D
26/28
Physics 111: Lecture 2, Pg 24
9hooting the 8on/ey&&&9hooting the 8on/ey&&&
r r = v v 0 t #1 " 2 g g t
2
Cith gra'ity, still aim at the mon!ey r r = r 0 # 1 " 2 g g t 2
Dart hits the
mon!ey
-
8/19/2019 3c Free Fall Kinematics 3D
27/28
Physics 111: Lecture 2, Pg 2<
Reca%:Reca%:9hooting the "on/ey&&&9hooting the "on/ey&&&
x x == x x 0 0
y y = #1 " 2 g g t 2
)his may #e easier to thin! a#out8
$tGs exactly the same i%ea
x x == v v 00 t t
y y = #1 " 2 g g t 2
-
8/19/2019 3c Free Fall Kinematics 3D
28/28
Reca% o Lecture 2Reca% o Lecture 2
Recap of 1-D motion with constant acceleration8 *)ext: 2-3+
1-D &ree-&all *)ext: 2-3+
example
3-D Kinematics *)ext: 3-3 5 3-(+
hoot the mon!ey *Fx8 3-11+
"ase#all pro#lem
$n%epen%ence of x an% y components