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TRANSCRIPT
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Newton’s Second Law:Newton s Second Law:
Uniform Circular Motion
Readings: Chapter 51
Readings: Chapter 5
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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=
θ2
g0
03601 57.32
radπ
= =
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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 0.17418rs mπ= =
3
90 2
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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 =
4
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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
5
2 2 2 2 2 2 2sin ( ) cos ( )x y r t r t rω ω+ = + =
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v∆
Uniform Circular Motion: Acceleration
r∆
0t
vat ∆ →
∆=∆ 0t
rvt ∆ →
∆=∆
1 1r v t∆ = ∆
2 2r v t∆ = ∆2 2r v t∆ ∆2 1 2 1
2( )v v r r
at t− ∆ − ∆
= =∆ ∆( )t t∆ ∆
Direction of acceleration is the same as direction of vector (toward the center)
→
2 1CB r r→
= ∆ − ∆The magnitude of acceleration:
2( 2 )CB φ θ∆ ∆ ∆ ∆ ∆6
2
2 2 2 2 2
( 2 )( ) ( ) ( ) ( ) ( )CB r r r v t t va vt t t t t r
φ π α θ ωω
∆ ∆ − ∆ ∆ ∆= = = = = = =
∆ ∆ ∆ ∆ ∆
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Uniform Circular Motion: Acceleration vr
ω =
centripetal acceleration
The magnitude of acceleration:
22va r v
rω ω= = =
7
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Uniform Circular Motion: Dynamics vr
ω =
The magnitude of acceleration:netF
22va r v
rω ω= = =
F
netF
Direction of acceleration: toward the center
netF
The second Newton’s law:
8netF ma=
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Uniform Circular Motion: Dynamics2v
The magnitude of the net force
2va r vr
ω ω= = =
The magnitude of the net force2
netvF mr
=
Direction of the net force: toward the center
r
netF n T w ma= + + =
0n w− =2vT ma m= =
9
T ma mr
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Uniform Circular Motion: Dynamics
tFnetFn
fnetF n f w ma= + + =
f0n w− =2vf ma m= =
wf ma m
r
maxf n w mgµ µ µ= = =
then2vm mgµ≤m mgr
µ≤
v rgµ≤
10
gµ
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Fictitious Forces
The force that seems to push the object outside of the circle is called centrifugal force
The direction of centrifugal force is opposite to the direction of acceleration
The centrifugal force exists only inside the systems which is moving in a circle (inside the car)
The centrifugal force is not a REAL FORCE
The magnitude of centrifugal force is
2vF m=11
cfF m r=
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Fictitious Forces: generalized second Newton’s law
In the reference frame of the car (inside the car) there are no motion, so there are no acceleration,there are no motion, so there are no acceleration, but THERE IS CENTRIFUGAL FORCE
nn
f
0net cfF n f w F= + + + =
cfF
w
0n w− =
0f F− = w0cff F− =
2
fvF m=
12
cfF m r