Download Slide lecture for chapter 4

Chapter 4: Motion in Two Dimensions
Chapter 4 Goals:
• To define and understand the ideas of vector position,,
displacemen, velocity and acceleration
• To understand what is meant by the path (trajectory)
of an object as it moves in space
• To work out the implications of motion with a constant
acceleration: projectile motion
• To work out the implications of motion on a circle:
circular motion
• To semi-quantitatively understand ‘curvy’ motion
• To discuss how motion can be described from a
moving frame of reference
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Kinematics in ‘higher dimensions’ I:
position and displacement as vector
• position is now a vector:
r : x ˆi  y ˆj  z kˆ
• the three components of the position: r:= <x, y, z>
• if the arena is 2d, drop the z-stuff
• displacement is change in position: Dr = rB – rA or
Dr = rf – ri or, best of all, Dr = r(t+Dt) – r(t)
• it is a vector with tail at r(t) and tip at r(t+Dt)
• displacement vector not usually drawn in standard
position but may be, especially if you are adding a
second displacement to the first and you don’t really
care about ‘initial’ position, or you take that to be zero
• the displacement vector is origin-independent!
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Kinematics in ‘higher dimensions’II
• how do we get velocity from this?
• start with average velocity: of course, vavg := Dr/Dt
Dr(t)
y
r(t)
path of
object
r(t+Dt)
x
• vavg is a vector: it’s a vector (Dr) times a number (1/Dt)
• magnitude: |vavg | = = |Dr /Dt| = |Dr |/|Dt|
• direction: same as direction of Dr
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Kinematics in ‘higher dimensions’ III:
the (instantaneous) velocity vector
• velocity is now a vector:
d
 Dr 
v(t )  r (t )  : lim   [drop ' t ' argument]
Dt 0 Dt 
dt
dx ˆ
dˆi dy ˆ
dˆj dz ˆ
dkˆ
 i  x  j y  k  z
dt
dt dt
dt dt
dt
dx ˆ
dy ˆ
dz ˆ
 i  0  j 0  k  0
dt
dt
dt
• here we used the fact that the unit vectors do not
vary with time as the body moves around
• the three components of the velocity v:= <vx, vy , vz >
where vx(t)=dx(t)/dt and similarly for y and z
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Kinematics in ‘higher dimensions’ IV
• we have let Dt get really small (Dt  0) and called it dt
• as that shrunk, so did Dr shrink : call it dr
dr(t) Dr(t)
y
r(t)
path of
object
r(t+dt)
r(t+Dt)
x
• magnitude: |v| = |dr /dt| = speed
v  v x2  v 2y  v z2
• direction: tangent to the path in space!!!
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Kinematics in ‘higher dimensions’ V:
the acceleration vector
• acceleration is now a vector:


d
 Dv 
a(t )  v(t )  : lim a avg : lim 

Dt 0
Dt 0 Dt 
dt
dvx ˆ
dˆi dv y ˆ
dˆj dvz ˆ
dkˆ

i  vx 
j  vy 
k  vz
dt
dt dt
dt dt
dt
: a ˆi  0  a ˆj  0  a kˆ  0
x
y
z
• three components of the acceleration a:= <ax, ay, az >
where ax(t)= dvx(t)/dt2 = d2x(t)/dt2 and similarly for y
and z
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Kinematics in ‘higher dimensions’ VI
• precisely the same procedure gives us the acceleration
• however, the direction of the acceleration is not
tangent to the path: it will have components both tangent
to (at), and normal to (ar for ‘radial’), the path
a(t)
v(t)
y
path
of
object
Dv(t)
v(t+Dt)
v(t+dt)
dv(t)
x
a is parallel
to dv and
rescaled by
1/dt
dv
2
2
2
2
2
a
:

a
:


a

a

a

a

a
x
y
z
t
r
• magnitude:
dt
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What kind of motion can occur near the Earth?
• gravity is down: a = – g j [g = 9.81 m/s2 ≈ 10 m/s2]
• object is projected with an initial velocity vi
• g and vi define a plane in space: the xy plane
• vertical component vyi horizontal component vxi
• let projection angle be qi so vi = {vi ,qi}
• vyi = vi sin qi and vxi = vi cos qi so vi = < vxi , vyi >
• assume projectile starts at the origin: <xi, yi> = <0,0>
•now allow x and y motion to unfold by components
v y t   v yi  gt
v x t   v xi (constant!!)
1
y(t )  v yit  2 gt 2
xt   v xit (steady!!)
• We still don’t know the ‘shape’ of the path, but we
see that the motion remains in the xy plane forever
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Some questions about projectile motion
• what is the time to maximum height? (call it T)
v y T   0  v0 y
v0 sin q 0
 gT  T 

g
g
voy
• clearly, the time to return to starting height is 2T
• what is the maximum height ? (call it ym)
 v0 sin q 0  1  v0 sin q 0 
  g 

ym  y (T )  v0 yT  gT  v0 sin q 0 
2
g
g




1
2
2
v02 sin 2 q 0 v02 sin 2 q 0 v02 sin 2 q 0



g
2g
2g
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2
One more question about projectile motion, on
level ground (ending y is same as yi = 0)
• what is the range – the maximum distance along the
ground? (call it R) . . . This is the x value at t = 2T
 2v0 sin q 0  2vo2 cosq 0 sin q 0
 
R : x(2T )  v0 x 2T  v0 cosq 0 
g


this can be rewritten using double - angle identity :
sin 2   2 sin  cos
v02 sin 2q 0 
R
g
• note that R = 0 for qi = 0° and 90°
• note that the range for qi is the same as the
range for 90° qi
• the maximum range is at qi = 45°
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
g
What is the actual Trajectory in Projectile
Motion? What is y(x)?
• solve the x(t) equation for t and insert into y(t) equation
x
1
t
into y (t )  vi sin q i t  2 gt 2
vi cosq i

x
 y ( x)  vi sin q 0 
 vi cosq i
 x tan q i 
 1 
x
  g 
2

 vi cosq i



2
gx2
2vi2 cos 2 q i
• this is a parabolic path in space!!
• Gravity’s Rainbow… {show Active figure 04_07}
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Side view of the parabolic trajectory, as seen in the
plane of g and vi
r
rm y
m
r
• note that the position vector is pretty complicated
• confusingly, we use q for the angle of the velocity
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The velocity vector is less complicated
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Example of a problem involving projectile motion
You have locked yourself out of the dorm and your roommate
throws your keys to you from a window that is 12 m above the
ground, with a launch speed of 4 m/s, at a launch angle of 50°.
a) how much higher do the keys rise?
b) where should you stand to catch the keys?

vi2 sin 2 qi 4 m/s 2 sin 50
a) ym 

2g
2 10 m/s 2


b) trajectory equation : y ( x)  x tan q i 

2
 .47 m
gx2
2vi2 cos 2 q i
We know y   12.0 m and so putting all # s in we get
 12.0  1.19 x  .76 x 2 or .76 x 2  1.19 x  12.0  0
x

1.19 
1.19 2  4.76 (12.0)
2.76 

1.19  1.42  36.48
1.52
1.19  6.16
 4.84 m out from the base of the wall
1.52
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Another example
A projectile is launched at a certain angle with a certain speed.
a) suppose qi corresponds to the maximum range (call it
Rm). What is ym in terms of Rm?
b) suppose the angle corresponds to the situation where
the range is the same as the maximum height ym. What is qi?


vi2 sin 2  45 vi2
a) The launch angle is qi  45 so Rm 

g
g

 
1
vi2 sin 2 45 vi2 2 Rm
 ym 


2g
g 2
4
vi2 sin 2q i  vi2 sin 2 q i
b) equate expressions for R and ym :

g
2g
2vi2 sin q i cosq i vi2 sin 2 q i
use sin/cos form for R :

g
2g
 tan q i  4  q i  Arctan4   76 
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Kinematics of motion on a curved path I
• the car moves on a flat (2d) curve
at speed v, of radius r
• for now assume a circle: constant
speed v = V; constant radius r = R…
• more complicated: varying r and
varying v…
• goal is to understand the
acceleration
• assume a time Dt elapses
• for now v and r change in
direction only, both by angle Dq
• note the triangle made by ri, rf
and Dr, with Dq in there
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Kinematics of motion on a curved path II
• the velocity vectors form a
precisely similar triangle, using
vi, vf and Dv, with Dq in there
Dr
Dv

Dr : R :: Dv : V or

but dr dt : V and dv dt : a so we have V

2
V
result : ac 
R

V
Dr Dt Dv Dt
 divide both sides by Dt :

R
V
dr dt dv dt
 take limit as Dt  0 :

R
V
R
for the magnitude of a c
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R
a
V
Kinematics of motion on a curved path III
• Now take that same Dt 0 limit
• Dv gets shorter and shorter…
• in the limit, dv points exactly
along a, and becomes exactly
perpendicular to v itself
• result: direction of a is
perpendicular to the path (and recall
that v was tangent to the path)
• this all assumed constant speed V
and constant radius R
• Conclusion: the centripetal acceleration!!
• direction is toward the circle’s center
• magnitude is ac = V2/R = w2R since V = w R
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Kinematics of motion on a curved path IV
• what if v and r are changing?
• new unit vectors rhatand that which are path-dependent
• rhat points radially out from the center and is ┴ to path
• that points tangentially to the path and is ║to path
that
rhat
rhat
that
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Kinematics of motion on a curved path V
• then we can say ac = – rhatV2/R = – rhatw2R
• fact #1: if radius r varies but speed V does not, then
there is only centripetal acceleration, as before (R  r)
• so, the acceleration is purely radial (centripetal)
• BUT if the speed v is changing TOO, then the
acceleration has a tangential component as well
v
 v tˆ ( true generally) so a : dv
dt
 v dtˆ
dt
 tˆ dv
dt
ˆ
 it can be shown that dt
 define
at  dv
dt
2
v
v
 rˆ
so first term is just a c  rˆ
dt
r
r
so second term is just a t  tˆat
ˆa c  tˆ dv
 thus we have a :  tˆa t  r

with v  wr and   dw
dt
dt
2
v
 rˆ
r
we can get a :  tˆr  rˆw 2 r
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Kinematics of motion on a curved path IV
The concept of the centripetal force
• if a body moves on a curve, there must be a centripetal
force (toward the instantaneous center) acting on it
• if the speed is a constant, then Fc is the net force
• it is not a ‘new’ force, it is a new name: it might be N,
or W, or T, or some combination
• We can always say Fc = mac so Fc = mv2/r
• OK, so what is the centrifugal force,huh??
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