Classical Mechanics Lecture 9springer/phys1500/Lectures/PHYS...Classical Mechanics Lecture 9 Today's...
Transcript of Classical Mechanics Lecture 9springer/phys1500/Lectures/PHYS...Classical Mechanics Lecture 9 Today's...
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Classical Mechanics Lecture 9
Today's Concepts and Examples:a) Energy and Frictionb) Potential energy & force
Midterm 2 will be held on March 13. Covers units 4-9
Unit 9 Homework DueThursday March 12 11:30 PM.No extension!
Mechanics Lecture 9, Slide 1
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Homework 8. Awesome Job!
Average = 92%
Mechanics Lecture 8, Slide 2
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Practice Exams
Phys 1500 Exams https://utah.instructure.com/courses/320947/files- Spring 2013: http://www.physics.utah.edu/~springer/phys1500/exams/MidtermExam2.pdf- Solutions: http://www.physics.utah.edu/~springer/phys1500/exams/MidtermExam2Soln.pdf- Long Sample: https://utah.instructure.com/courses/320947/files/45779670/download?wrap=1
Phys 2210 Exams- Practice : http://www.physics.utah.edu/~woolf/2210_Jui/rev2.pdf- Spring 2015: http://www.physics.utah.edu/~woolf/2210_Jui/ex2.pdf
Mechanics Lecture 8, Slide 3
http://www.physics.utah.edu/%7Espringer/phys1500/exams/MidtermExam2.pdfhttp://www.physics.utah.edu/%7Espringer/phys1500/exams/MidtermExam2Soln.pdfhttps://utah.instructure.com/courses/320947/files/45779670/download?wrap=1http://www.physics.utah.edu/%7Ewoolf/2210_Jui/rev2.pdfhttp://www.physics.utah.edu/%7Ewoolf/2210_Jui/ex2.pdf
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Main Points
Mechanics Lecture 8, Slide 4
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Main Points
Mechanics Lecture 8, Slide 5
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Incline with Friction: Work-Kinetic Energy
Using Work-Kinetic Energy Theorem
Same result as using Newton’s Law
Mechanics Lecture 8, Slide 6
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H
N1
mg
N2
mg
µmg
must be negative
m
Work by Friction :Wfriction
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Checkpoint
What is the total macroscopic work done on the block by all forces during this process?
A) mgH B) –mgH C) µk mgD D) 0
Mechanics Lecture 9, Slide 8
D
m
H
0000
=⇒=∆
===∆
tot
f
i
tot
WKKK
WK
Mechanics Lecture 8, Slide 8
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Mechanics Lecture 8, Slide 9
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Force from Potential Energy:1D
Mechanics Lecture 8, Slide 10
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Force from Potential Energy in 3-d
Gradient operator
Mechanics Lecture 8, Slide 11
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Potential Energy vs. Force
dxxdUxF )()( −=
Mechanics Lecture 9, Slide 12
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Potential Energy vs. Force
2
21)( kxxU −=
dxxdUxF )()( −= kx−=
Mechanics Lecture 9, Slide 13
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Potential Energy vs. Force
Mechanics Lecture 9, Slide 14
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Potential Energy vs. Force
Mechanics Lecture 9, Slide 15
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Demo
Potential Energy vs. Force
dxxdUxF )()( −=
Mechanics Lecture 9, Slide 16
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Equilibrium
Mechanics Lecture 8, Slide 17
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Equilibrium points
Mechanics Lecture 8, Slide 18
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Equilibrium points
Mechanics Lecture 8, Slide 19
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Equilibrium points
Mechanics Lecture 8, Slide 20
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Block on Incline
1cos xTxdTWtension ∆=⋅= ∫ θ
mxTv
xTWvvmK
xTWWWWW
f
netif
tensiontensionnormalfrictionnet
1
122
1
cos2
cos)(21
cos
∆=
∆==−=∆
∆==++=
θ
θ
θ
Mechanics Lecture 8, Slide 21
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Block on Incline
θµ cosxmgxdfW kkfriction ∆−=⋅= ∫
θsinxmgxdWWgravity ∆−=⋅= ∫
Mechanics Lecture 8, Slide 22
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Block on Incline
θµθθ
θµθθθ
θ
cossincos
cossincoscos
cos2;0
12
2221
1
1
2
mgmgTxTx
xmgxmgxTxTxTK
mxTvv
KWWWW
xTxdTW
k
k
if
frictiongravitytensionnet
tension
−−∆−
=∆
∆−∆−∆=∆−∆−=∆
∆==
∆=++=
∆=⋅= ∫
Mechanics Lecture 8, Slide 23
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Block on Incline
?0 =⇒
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Energy Conservation Problems in general
)()( tUtKUKUKE ffiimechanical +=+=+=
For systems with only conservative forces acting
0=∆ mechanicalE
Emechanical is a constant
Mechanics Lecture 8, Slide 25
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Gravitational Potential Energy
rEr
Err
−
Mr
Mrr
−
Mechanics Lecture 8, Slide 26
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Gravitational Potential Problems
gravitymechanical hUhmvUKE )()(21 2 +=+=
conservation of mechanical energy can be used to “easily” solve problems.
Define coordinates: where is U=0?
M
M
E
Etotal
M
MMMoon
E
EEEarth
rrmGM
rrmGMrU
rmGMrU
rmGMrU
−−
−−=
−=
−=
)(
)(
)(
0)( →−=r
mGMrU E as ∞→r
rEr
Err
−
Mr
Mrr
−
Add potential energy from each source.
Mechanics Lecture 8, Slide 27
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Trip to the moon
)2
1(21
21
21
222
2
iE
E
E
E
Ei
E
E
Ei
Ef
f
E
E
Ei
ffii
vGMRR
RGMv
GM
RmGMmv
mGMR
RmGM
RmGMmv
UKUK
−=
−
−=
−
−=
−=−
+=+
f
Ef
f
E
Ei
ii
RmGMU
KR
mGMU
mvK
−=
=
−=
=
0
21 2
Mechanics Lecture 8, Slide 28
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Trip to the moon
%02.0000204.0
)01659.0)(01232.0()()(
)()()(
)(
)(
==
=⇒
≈
−=
−=
−−=
−=
EE
EM
EF
FMEE
EM
E
E
FME
M
EE
EM
FME
MFM
F
EFE
RURU
RR
RdMRM
RM
RdM
RURU
RdmGMRU
RmGMRU Can ignore effect
of moon for this problem at level of precision for SmartPhysics
Mechanics Lecture 8, Slide 29
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Trip to the moon
( )
ME
M
E
ME
m
E
Ei
fME
m
f
E
ME
m
E
Ei
ddmGMcmGMb
dmGM
RmGMmva
bdxcbadaxcxbxbdaxadxcxxdbxdax
xdxcxxdb
xx
xdc
xb
xdxd
xdc
xba
RdmGM
RmGM
dmGM
RmGMmv
=−=−=
−−=
=−−++−
=−+−−
+−=−
−+−
=
−
+−−
=−
+=
−−−=−−
2
2
2
2
21
0)(0
)()()(
)()()(
21
…or you can practice solving the quadratic equation with many terms!!!
Mechanics Lecture 8, Slide 30
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Trip to the moon
M
m
ME
Ef
ff
EM
m
E
Ei
ii
RmGM
dmGMU
mvK
dmGM
RmGMU
mvK
−−=
=
−−=
=
→
→
2
2
21
21 Can NOT ignore
effect of moon for this problem since the rocket is AT the moon in the end !!!!
McentersME
EcentersEM
RddRdd
−=−=
→
→
Mechanics Lecture 8, Slide 31
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Trip to the moon
−−+−=
++−−=
++−−=
−−=−−
+=+
→→
→→
→→
→→
)1(21
22221
212
21
21
2
2222
2
22
ME
Em
ME
E
EME
Em
Ei
Eif
Mi
m
MEi
E
EMi
m
Ei
Eif
M
m
ME
E
EM
m
E
Eif
M
m
ME
Ef
EM
m
E
Ei
ffii
RMRM
dR
dMRM
RvGMvv
RvGM
dvGM
dvGM
RvGMvv
RGM
dGM
dGM
RGMvv
RmGM
dmGMmv
dmGM
RmGMmv
UKUK
Mechanics Lecture 8, Slide 32
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Trip to the moon
020608.12
0453.0
0167.0
000207.0
)1(21
2
2
=
=
=
=
−−+−=
→
→
→→
Ei
E
ME
Em
ME
E
EME
Em
ME
Em
ME
E
EME
Em
Ei
Eif
RvGM
RMRM
dR
dMRM
RMRM
dR
dMRM
RvGMvv
Mechanics Lecture 8, Slide 33
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Trip to the moon
xx
Mechanics Lecture 8, Slide 34
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Block on Incline 2
1cos xTxdTWtension ∆=⋅= ∫ θ
11 xmgxdfW kkfriction ∆−=⋅= ∫ µ
Mechanics Lecture 8, Slide 35
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Block on Incline 2
mxmgxTv
xmgxTWvvmK
xmgxTWWWWWW
kf
knetif
ktensionfrictiontensionnormalfrictionnet
)cos(2
cos)(21
cos
111
11122
111
∆−∆=
∆−∆==−=∆
∆−∆=+=++=
µθ
µθ
µθ
Mechanics Lecture 8, Slide 36
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Block on Incline 2
θµθθ
θµθθθ
θ
cossincos
cossincoscos
cos2;0
12
2221
1
1
2
mgmgTxTx
xmgxmgxTxTxTK
mxTvv
KWWWW
xTxdTW
k
k
if
frictiongravitytensionnet
tension
−−∆−
=∆
∆−∆−∆=∆−∆−=∆
∆==
∆=++=
∆=⋅= ∫
Mechanics Lecture 8, Slide 37
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Block on Incline 2
2sin xmgxdWWgravity ∆=⋅= ∫ θ
Mechanics Lecture 8, Slide 38
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Clicker Question A.B.C.
D.
0% 0%0%0%
Suppose the potential energy of some object U as a function of xlooks like the plot shown below.
Where is the force on the object zero?A) (a) B) (b) C) (c) D) (d)
U(x)
x
(a) (b) (c) (d) dxxdUxF )()( −=
Mechanics Lecture 8, Slide 39
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Clicker Question A.B.C.
D.
0% 0%0%0%
Suppose the potential energy of some object U as a function of xlooks like the plot shown below.
Where is the force on the object in the +x direction?A) To the left of (b) B) To the right of (b) C) Nowhere
U(x)
x
(a) (b) (c) (d) dxxdUxF )()( −=
Mechanics Lecture 8, Slide 40
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Clicker Question A.B.C.
D.
0% 0%0%0%
Suppose the potential energy of some object U as a function of xlooks like the plot shown below.
Where is the force on the object biggest in the –x direction?A) (a) B) (b) C) (c) D) (d)
U(x)
x
(a) (b) (c) (d) dxxdUxF )()( −=
Mechanics Lecture 8, Slide 41
Classical Mechanics �Lecture 9Homework 8. Awesome Job!Practice ExamsMain PointsMain PointsIncline with Friction: Work-Kinetic EnergyWork by Friction :Wfriction