Motion in 2 Dimensions Definitions
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1 Motion in 2 Dimensions Demonstrations: “Monkey gun” Dropping cube/shooting cube Why do the cubes hit simultaneously? What is Δx? What is v f ? Definitions Position: j y i x r ˆ ˆ + = Displacement: Average Velocity: Instantaneous Velocity: i f r r r - = Δ t r v Δ Δ ≡ Same direction as Δr. dt r d t r v t = Δ Δ ≡ → Δ 0 lim Tangential to r vs t curve. Speed | | v v = Transp More definitions: Average Acceleration: Instantaneous Acceleration: t v a Δ Δ ≡ 2 2 0 lim dt r d dt v d t v a t = = Δ Δ ≡ → Δ Can be non-zero if: Speed changes Direction changes Both speed and direction change Q1 Special case: j a i a a y x ˆ ˆ constant + = = t a v t v i + = ) ( t a v t v x ix x + = ) ( Don’t change in time. Integrate: Integrate: 2 2 1 ) ( t a t v r t r i i + + = t a v t v y iy y + = ) ( 2 2 1 t a t v x x x ix i + + = 2 2 1 t a t v y y y iy i + + = Two “independent” constant motions!
Transcript of Motion in 2 Dimensions Definitions
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Motion in 2 Dimensions
�Demonstrations: “Monkey gun”Dropping cube/shooting cube
� Why do the cubes hit simultaneously?� What is ∆x?� What is vf?
Definitions
Position: jyixr ˆˆ +=�
Displacement:
Average Velocity:
Instantaneous Velocity:
if rrr ��� −=∆
tr
v∆∆≡�
� Same directionas ∆r.
dtrd
tr
vt
��� =
∆∆≡
→∆ 0lim Tangential to
r vs t curve.
Speed || vv �=
Transp
More definitions:Average Acceleration:
Instantaneous Acceleration:tv
a∆∆≡�
�
2
2
0lim
dtrd
dtvd
tv
at
���� ==
∆∆≡
→∆
Can be non-zero if:�Speed changes�Direction changes�Both speed and direction change
Q1
Special case: jaiaa yxˆˆconstant +==�
tavtv i��� +=)(
tavtv xixx +=)(
Don’t change in time.Integrate:
Integrate:2
21
)( tatvrtr ii���� ++=
tavtv yiyy +=)(
2
21
tatvxx xixi ++= 2
21
tatvyy yiyi ++=
Two “independent” constant motions!
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Concept Questions:Throw one ball straight up with speed vyi.Throw a second with the same vyi, but also nonzero vxi.
vyi
12
Q2. Which ballgoes highest?
1. Ball 12. Ball 23. Same
Concept Questions:Throw one ball straight up with speed vyi.Throw a second with the same vyi, but also nonzero vxi.
vyi
12
Q3. Which ball hits ground first?1. Ball 12. Ball 23. Same
Concept Questions:Throw one ball straight up with speed vyi.Throw a second with the same vyi, but also nonzero vxi.
vyi
12
Q4. Which ballhits groundmoving fastest?
1. Ball 12. Ball 23. Same
Application: Projectile Motion!
ax = 0 and ay = -g
x motion:
y motion:
vx(t) = vxi
x(t) = xi + vxit
y(t) = yi + vyit - ½gt2
vy(t) = vyi - gt
Note:
jvivjvivv iiiiiyixiˆsinˆcosˆˆ θθ +=+=�
x
y
θθθθi
Transp
vi
BUT ONLY INITIALLY!!!!
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Projectile Motion - Example
A ball rolls off of a shelf 2.20 m high with an initial velocity of 0.75 m/s.
2.20 m
Vix=0.75 m/s
Find how far from the shelf the ball first hits the ground.
Example - Continued
Knowns Unknowns
Vix=0.75 m/s
General strategy: Use TIME as the common denominator between the two dimensions.
X Y X Y
ax=0 m/s
ay =-9.8 m/s2
-2.20 m=∆y
Viy=0 m/s
?=∆x
HERE: Use known Y’s to find time ball takes to fall...find X at that time
Example - Continued
�+=∆ 2
21 attvy i -2.20=0(t) + .5(-9.8)t2
t = +/- 1.49 s--- Choose positive
WHY two solutions?
All falls are parabolic.
We started at top (vx=0)
Two solutions, but only one really happened!
+t-t
Found time for ball to fall.
Example - Continued
Knowns Unknowns
Vix=0.75 m/sTime to FALL is same as time to MOVE in X direction.
So…
X Y X Y
ax=0 m/s
ay =-9.8 m/s2
-2.20 m=∆y
Viy=0 m/s
?=∆x
=+=∆ 2
21
tatvx xix 0.75(1.49)+1/2(0)(1.49)2
=1.12 m from edge of table
t = 1.49 s SHARED BY X AND Y
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Trajectory
Shape of “flight path” is parabola
Eliminate time, t, from y(t) using x(t) withcomponents in terms of initial angle.
222 cos2
)(tan)( xv
gxxy
iii ��
�
����
�−=
θθ
What is this? Give me any X coordinate,
I’ll tell you the Y coordinate
Trajectory Example
θθθθi=32.5°°°°
vi=40.2 m/s
x=173 m
How high is the cannonball when it hits the wall?
222 cos2
)(tan)( xv
gxxy
iii ��
�
����
�−=
θθ =107 m
g=9.8 m/s2
We have X, find Y.
Trajectory—Caution!
θθθθi=32.5°°°°
vi=15.2 m/s
x=173 m
How high is the cannonball when it hits the wall?
222 cos2
)(tan)( xv
gxxy
iii ��
�
����
�−=
θθ =-652 m??
vi=40.2 m/s
Range
gv ii θsin2
Depends on “hang time”: R = x(ttot) = vxittot
How do we find total time in flight ttot?Find time for y(t) to get back to yi.
y(ttot) = yi = yi + vyittot – ½g(ttot ) 2
ttot = 0 or ttot =
gv
R ii θ2sin2
=Derivate to find MMax R for θθθθ= 45º
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Range-Caution! gv
R ii θ2sin2
=
θθθθi=32.5°°°°
vi=15.2 m/s
Range equations work only on a FLAT surface
θθθθi=32.5°°°°
vi=15.2 m/s
Assumption in derivation was that projectile lands at same level!
gv
t iitot
θsin2=
Must solve this problem like ball on table!
Range
gv ii θsin2
Range demo
R = x(ttot) = vxittot
How do we find total time in flight ttot?Find time for y(t) to get back to yi.
y(ttot) = yi = yi + vyittot – ½g(ttot ) 2
ttot = 0 or ttot =
gv
R ii θ2sin2
= Maximum R for θθθθ= 45º
Examples:
�How long does it take to get to the top ofa trajectory?
�How far from the base does the cube get shot?
�Will the moving cart catch the ball? (demo and Q5)
gv
tt iitotpeak
θsin21 ==
Next Time:
�Accelerating at constant speed�Centripetal, Radial, Tangential…�Swimming across fast rivers
