Motion in a Planein a Plane - Physics & Astronomyphysics.gsu.edu/apalkov//lecture2211_4.pdf ·...
Transcript of Motion in a Planein a Plane - Physics & Astronomyphysics.gsu.edu/apalkov//lecture2211_4.pdf ·...
Motion in a PlaneMotion in a Plane
Readings: Chapter 41
Readings: Chapter 4
Kinematics in Two DimensionsΔAverage velocity: avg
rvt
Δ=Δ
rΔavgv
t=Δ
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Kinematics in Two Dimensions: Instantaneous Velocityy
rv Δ=Δ 0tt Δ →Δ
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Kinematics in Two Dimensions
rv Δ=
dsvd
=0t
vt Δ →Δdt
(motion along a line)
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Kinematics in Two Dimensions: Instantaneous Acceleration
Average acceleration:
avgvat
Δ=Δ
Instantaneous acceleration:
vΔ
0t
vat Δ →
Δ=Δ
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Kinematics in Two Dimensions: Instantaneous Acceleration
Motion along a line: acceleration results in change of speed (theMotion along a line: acceleration results in change of speed (the magnitude of velocity)
Motion in a plane: acceleration can change the speed (the magnitude of velocity) and the direction of velocity.
0
vat Δ
Δ=Δ 0tt Δ →Δ
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A Projectile is an object that moves in two dimensions (in a plane) under the influence of only the gravitational force.
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A Projectile is an object that moves in two dimensions with free fall acceleration
a g=
0xa = - motion along x-axis – with zero acceleration – constant velocity
constantxv =
29.8yma gs
= − = −
x
- motion along y-axis – free fall motion with constant acceleration
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constant cosv v v θ= = =
Example: Find the distance AB
cosx v t v t θ= =
29.8 ma g= − = −
,constant cosx i x iv v v θ= = = , cosi x ix v t v t θ= =
iv v gt= − sinv v θ=29.8ya gs ,y i yv v gt
2ABt
Point A - 0i Ay y= =2t
, sini y iv v θ=
, 2AB
final i y ABt
y v t g= −Point B - 0final By y= = then , 2AB
i y ABt
v t g=
2 2 siniv v θ,2 2 sini y iAB
v vt
g gθ
= =
then
A B
then
2 2
cos
2AB i AB
i i
x v t
v v
θ
θ θ θ
= =
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A B 2sin cos sin 2i iv v
g gθ θ θ= =
2
sin 2iAB
vx θ=
g
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Motion in a CircleMotion in a Circle
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Kinematics in Two Dimensions: Uniform Circular Motion: Motion with Constant Speed
The speed (magnitude of velocity) is the same
Period:
2 rπ2 rTvπ
=
θ
The position of the particle is characterized by angle (or by arc length)
Arc length:
where angle is in RADIANS
s rθ=
θ13
g0
03601 57.32
radπ
= =
Example:
What is the arc length for if0 0 045 90 10θ = 1r m=What is the arc length for if
0 00
245 45 0.78360 4
rad rad radπ π= = =
45 ,90 ,10θ = 1r m=
360 40 0
0
290 90 1.57360 2
rad rad radπ π= = =
0 00
210 10 0.174360 18
rad rad radπ π= = =
Arc length: s rθ=rπ
45 0.784rs mπ
= =
90 1.572rs mπ
= =
10 0.17418
rs mπ= =
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90 2
Kinematics in Two Dimensions: Uniform Circular Motion: Motion with Constant Speed
The speed (magnitude of velocity) is the same s vt=tθ ω= tθ ω
vr
ω =2 rTvπ
=
Then
2π2Tπω =
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Uniform Circular Motion s vt= 2Tπω =
tθ ω=
vr
ω =
Then
cos cos( )x r r tθ ω= = ( )
sin sin( )y r r tθ ω= =
2 2 2 2 2 2 2
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2 2 2 2 2 2 2sin ( ) cos ( )x y r t r t rω ω+ = + =
tθ ω=
0ω >0ω <
0ω =
Positive angle
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Positive angle
vΔ
Uniform Circular Motion: Acceleration
rΔ
0t
vat Δ →
Δ=Δ 0t
rvt Δ →
Δ=Δ
1 1r v tΔ = Δ
2 2r v tΔ = Δ2 2r v tΔ Δ2 1 2 1
2( )v v r r
at t− Δ − Δ
= =Δ Δ( )t tΔ Δ
Direction of acceleration is the same as direction of vector (toward the center)
→
2 1CB r r→
= Δ − ΔThe magnitude of acceleration:
2( 2 )CB φ θΔ Δ Δ Δ Δ18
2
2 2 2 2 2
( 2 )( ) ( ) ( ) ( ) ( )CB r r r v t t va v
t t t t t rφ π α θ ω
ωΔ Δ − Δ Δ Δ
= = = = = = =Δ Δ Δ Δ Δ
Uniform Circular Motion: Acceleration vr
ω =
centripetal acceleration
The magnitude of acceleration:
22va r v
rω ω= = =
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