Chapter 4 – 2D and 3D Motion I. Definitions II. Projectile motion III. Uniform circular motion IV. Non-uniform circular motion V. Relative motion
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Chapter 4 – 2D and 3D Motion
I. Definitions
II. Projectile motion
III. Uniform circular motion
IV. Non-uniform circular motion
V. Relative motion
Position vector: extends from the origin of a coordinate system to the particle.
)1.4(ˆˆˆ kzjyixr
)2.4(ˆ)(ˆ)(ˆ)( 12121212 kzzjyyixxrrr
I. Definitions
Average velocity:
)3.4(ˆˆˆ ktzj
tyi
tx
trvavg
Displacement vector: represents a particle’s position change during a certaintime interval.
Instantaneous velocity:
)4.4(ˆˆˆˆˆˆ kdtdzj
dtdyi
dtdx
dtrdkvjvivv zyx
-The direction of the instantaneous velocity of a particle is always tangent to the particle’s path at the particle’s position
Instantaneous acceleration:
Average acceleration:)5.4(12
tv
tvvaavg
)6.4(ˆˆˆˆˆˆ kdtdvj
dtdv
idtdv
dtvdkajaiaa zyx
zyx
II. Projectile motion
Motion of a particle launched with initial velocity, v0 and free fall accelerationg.
- Horizontal motion: ax=0 vx=v0x= constant
- Vertical motion: ay= -g = constant
Range (R): horizontal distance traveled by a projectile before returning to launch height.
)7.4()cos( 0000 tvtvxx x
)8.4(21)sin(
21 2
002
00 gttvgttvyy y
)9.4(sin 00 gtvvy )10.4()(2)sin( 02
002 yygvvy
The horizontal and vertical motions are independent from each other.
- Trajectory: projectile’s path.
)11.4()cos(2
)(tan
cos21
cossin
cos)8.4()7.4(
200
2
0
2
000000
00
vgxxy
vxg
vxvy
vxt
000 yx
- Horizontal range: R = x-x0; y-y0=0.
)12.4(2sincossin2
cos21tan
cos21
cos)sin(
21)sin(0
cos)cos(
0
202
000
022
0
2
0
2
000000
200
0000
gvv
gR
vRgR
vRg
vRvgttv
vRttvR
(Maximum for a launch angle of 45º )
Overall assumption: the air through which the projectile moves has no effect on its motion friction neglected.
122: A third baseman wishes to throw to first base, 127 feet distant. His best throwing speed is 85 mi/h. (a) Ifhe throws the ball horizontally 3 ft above the ground, how far from first base will it hit the ground? (b) Fromthe same initial height, at what upward angle must the third baseman throw the ball if the first baseman is tocatch it 3 ft above the ground? (c) What will be the time of flight in that case?
mfoot
mfeet
smmi
ms
hhmi
91.01305.03
/3811609
3600185
x
yv0
h=3ft
B3 B1xmax0 xB1=38.7m
Vertical movementHorizontal movement
sttm
gttvyy y
43.09.491.0021
2
200
34.16)43.0)(/38(380max
00max
Bfrommssmtxtvxx x
The ball will hit ground at 22.3 m from B1
x
v0
h=3ft
B3 B1
y
38.7m6.72sin5.013.0
cossin144463.1899.4sin38
cos387.38
1cos387.38cos7.38
9.4sin38sin9.4
210
00
002
00
stvt
mv
tvtvgttvyy
x
yy
θ
N7: In Galileo’s Two New Sciences, the author states that “for elevations (angles of projection) which exceedor fall short of 45º by equal amounts, the ranges are equal…” Prove this statement.
45
45
45
2
1
290sin452sin'
290sin452sin
20
20
20
20
gv
gvR
gv
gvR
x
v0
x=R=R’?
y
θ=45º
02sin: max0
20 hatdgvRRange
babababababa
sincoscossin)sin(sincoscossin)sin(
)2cos()2sin(90cos)2cos(90sin'
)2cos()2sin(90cos)2cos(90sin
20
20
20
20
gv
gvR
gv
gvR
III. Uniform circular motionMotion around a circle at constant speed.
)13.4(2
rva
)14.4(2
vr
T
tancossintan
sincos
ˆsinˆcosˆˆˆˆ
ˆˆˆ)cos(ˆ)sin(ˆˆ
222
222
22
x
y
yx
xypp
ppyx
aa
radiusalongdirecteda
rv
rvaaa
jrvi
rvjv
rviv
rvj
dtdx
rvi
dtdy
rv
dtvda
jrxv
ir
yvjvivjvivv
- Period of revolution:
- Acceleration: centripetal
vy
vx
-Velocity: tangent to circle in the direction of motion.
Magnitude of velocity and acceleration constant. Direction varies continuously.
54. A cat rides a merry-go-round while turning with uniform circular motion. At time t1= 2s, the cat’s velocityis: v1= (3m/s)i+(4m/s)j, measured on an horizontal xy coordinate system. At time t2=5s its velocity is:v2= (-3m/s)i+(-4m/s)j. What are (a) the magnitude of the cat’s centripetal acceleration and (b) the cat’saverage acceleration during the time interval t2-t1?
v1
v2
x
y
In 3s the velocity is reversed the cat reaches the oppositeside of the circle
mrsm
rsvrT
smv
77.4/5
32/543 22
2222
/23.577.4/25 smmsm
rvac
2
2212
/33.3
ˆ)/67.2(ˆ)/2(3
ˆ)/8(ˆ)/6(
sma
jsmisms
jsmismtvva
avg
avg
IV. Non-Uniform circular motion
22tr aaa
- A particle moves with varying speed in a circular path.
- The acceleration has two components: - Radial ar = -ac= -v2/r
- Tangential at = dv/dt
- at causes the change in the speed of the particle.
- In uniform circular motion, v = constant at = 0 a = ar
rrv
dtvd
aaa rt ˆˆ2
V. Relative motion
1D
Particle’s velocity depends on reference frame
)15.4(BAPBPA vvv
Frame moves at constant velocity
)16.4()()()( PBPABAPBPA aavdtdv
dtdv
dtd
0
Observers on different frames of reference measure the same accelerationfor a moving particle if their relative velocity is constant.
20
0
5.00
)(
flightflightylaunchland
flightsxbglaunchland
gttvyy
tvvxxx
jvivvv
jvivv
ivv
ysxg
yxrel
ss
ˆˆ)(
ˆˆ
ˆ
000
00,0
75. A sled moves in the negative x direction at speed vs while a ball of ice is shot from the sled with a velocityv0= v0xi+ v0yj relative to the sled. When the ball lands, its horizontal displacement ∆xbg relative to the ground(from its launch position to its landing position) is measured. The figure gives ∆xbg as a function of vs. Assumeit lands at approximately its launch height. What are the values of (a) v0x and (b) v0y?The ball’s displacement ∆xbs relative to the sled can also be measured. Assume that the sled’s velocity is notchanged when the ball is shot. What is ∆xbs when vs is (c) 5m/s and (d) 15m/s?
Launch velocity relativeto ground
Displacements relative to ground
syyxy
sxbg
yflightflighty
vgv
gvv
gv
vvx
gv
tgttv
00000
020
222)(
25.00
a b
∆xbg= a+ b vs
smvg
smgvsmvx
smvsm
mgv
xx
sbg
yy
/10)/10(220/100
/6.19/20
802
00
00
smvsgv
t yflight /226.19104
2 220
0
Displacements relative to the sled Relative to the sled, the displacement does not dependon the sled’s speedAnswer (c)= Answer (d)mssmtvx flightxbs 404)/10(0
(iii) A dog wishes to cross a river to a point directly opposite as shown. It can swim at2m/s in still water and the river is flowing at 1m/s. At what angle with respect to theline joining the starting and finishing points should it start swimming?
θ
start
finish
1m/s
rv
ismvrˆ)/1(
x
y
jiv ˆcos2ˆsin20
smvv
smjvsmjiivvv
ff
frf
/3cos2
3001sin2
/)ˆ(/)ˆcos2ˆsin2ˆ(0
0vfv
(ii) A particle moves with constant speed around the circle below. When it is atpoint A its coordinates are x=0, y=3m and its velocity is (5m/s)i. What are itsvelocity and acceleration at point B? Express your answer in terms of unitvectors.
ismim
smirvaB
ˆ)/3.8(ˆ3/25ˆ 2222
x
A
B
yismvAˆ)/5(
jsmvBˆ)/5(
ismaBˆ)/3.8( 2
120. A hang glider is 7.5 m above ground level with a velocity of 8m/s at an angle of 30º below thehorizontal and a constant acceleration of 1m/s2, up. (a) Assume t=0 at the instant just described and writean equation for the elevation y of the hang glider as a function of t, with y=0 at ground level. (b) Use theequation to determine the value of t when y=0. (c) Explain why there are two solutions to part B. Which onerepresents the time it takes the hang glider to reach ground level? (d) how far does the hang glider travelhorizontally during the interval between t=0 and the time it reaches the ground? For the same initial positionand velocity, what constant acceleration will cause the hang glider to reach ground level with zero velocity?Express your answer in terms of unit vectors.
x
y
v0= 8m/sh=7.5m
0
30º22
00 5.045.721 ttyattvyy y
jiv
ja
oˆ30sin8ˆ30cos8
ˆ1
ststtty 3,55.045.700 212
t(s)
y(m)
0
If the ground was not solid, the glider would swoop down,passing through the surface, then back up again, with the twotimes of passing being t=3s, t=5s.
1 2 3 4 5
7.5
jsmisma
smaatavv
movementHorizontal
xxxxx
ˆ)/1.1(ˆ)/85.1(
/85.175.393.60
:
22
20
0ˆˆ0
jvivvy fyfxf
mssmtvd x 78.20)3()/93.6(0max
stttavv
smaayyavv
movementVertical
yyy
yyyyy
75.31.1400
/1.1154)(2
:
0
220
20
2
40. A ball is to be shot from level ground with certain speed. The figure below shows the range R it willhave versus the launch angle θ0 at which it can be launched. The choice of θ0 determines the flight time; lettmax represent the maximum flight time. What is the least speed the ball will have during its flight if θ0 ischosen such as that the flight time is 0.5tmax?
305.0sinsin25.0sin2 10
000max
00 gv
gvt
gvt flight
gvt
gvt
ttvgttvyy y
0max0
00
200
200
21sinsin2
9.4sin5.00
R(m)
θ0
100
200
smsmvv
smvg
vg
vmg
vR
formRGraph
x /4230cos)/5.48(
/5.4890sin2402sin
45240
min
0
20
20
0
20
0max
max00min 5.030cos tforvvtrajectoryhalfatvv xx
The lowest speed occurs at the top of the trajectory (half of total time offlight), when the velocity has simply an x-component.
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