classical mechanics lecture 18 today’s concepts: a) static equilibrium b) potential energy &...
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Classical Mechanics Lecture 18
Today’s Concepts:a) Static Equilibriumb) Potential Energy & Stability
Mechanics Lecture 18, Slide 1
Schedule
Mechanics Lecture 14, Slide 2
One unit per One unit per lecture!lecture!
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Lectures will Lectures will only contain only contain summary, summary, homework homework problems, problems, clicker clicker questions, questions, Example exam Example exam problems….problems….
Midterm 3 Wed Dec 11 Midterm 3 Wed Dec 11
footprint
footprint
Stability & Potential Energy
Mechanics Lecture 18, Slide 11
Unstable if CM is “outside of footprint”
CM must be lifted for the object to topple over
If CM is inside the footprint
Object is stable
Statics Problems
Mechanics Lecture 18, Slide 13
r
r
gMr
Into the screen
Tr
Tr
TrT
Out of the screen
02
sin0
0
LMgTL
weighttensionhinge
http://hyperphysics.phy-astr.gsu.edu/hbase/torcon.html
Clicker QuestionA. B. C. D. E.
Mechanics Lecture 18, Slide 15
20% 20% 20%20%20%
A (static) mobile hangs as shown below. The rods are massless and have lengths as indicated. The mass of the ball at the bottom right is 1 kg.
What is the total mass of the mobile?
A) 4 kg B) 5 kg
C) 6 kg
D) 7 kg
E) 8 kg 1 kg
1 m 3 m
1 m 2 m
Clicker QuestionA. B. C.
Mechanics Lecture 18, Slide 16
33% 33%33%
In which of the static cases shown below is the tension in the supporting wire bigger? In both cases M is the same, and the blue strut is massless.
A) Case 1 B) Case 2 C) Same
M
300
L/2
T2
Case 2
M
300
L
T1
Case 1
ML
T1
d1
Case 1
Balancing Torques
11dTMgL sin1LT 222dT
LMg sin
22
LT
sin1
MgT
sin2
MgT
It’s the same. Why?
Mechanics Lecture 18, Slide 17
M
T2 d2
Case 2
L/2
CM
Mg
T1T2 T1 T2 TSame distance from CM:
T Mg/2So:
Mechanics Lecture 18, Slide 20
T1 T2MgBalance forces:
Homework Problem
What is the moment of inertia of the beam about the rotation axis shown by the blue dot?
A)
B)
C)
L
Clicker Question
Mechanics Lecture 18, Slide 23
d
M2
12
1MLI
2MdI
22
12
1MdMLI
The center of mass of the beam accelerates downward. Use this fact to figure out how T1 compares to weight of the beam?
A) T1 Mg
B) T1 Mg
C) T1Mg
Clicker Question
Mechanics Lecture 18, Slide 24
T1
Mgy
x
ACM
d
M
The center of mass of the beam accelerates downward. How is this acceleration related to the angular acceleration of the beam?
A) ACM d
B) ACMd /
C) ACM / d
Clicker Question
Mechanics Lecture 18, Slide 25
T1
Mgy
x
d
M
ACM
Apply
Apply
Plug this into the expression for T1
ACM d
Use ACM dto find
Iext
Mechanics Lecture 18, Slide 26
T1
Mgy
x
d
M
ACM
CMext MAF
CMMATMg 1
CMMAMgT 1
d
AIIMgd CM
I
dMgACM
2
After the right string is cut, the meterstick swings down to where it is vertical for an instant before it swings back up in the other direction.
What is the angular speed when the meter stick is vertical?
Mechanics Lecture 18, Slide 27
M
y
x
d
T
Conserve energy:2
2
1 IMgd
I
Mgd2
y
xMg
dACM 2d
Centripetal acceleration
Mechanics Lecture 18, Slide 28
T
Applying CMext MAF
dMMgT 2
dMMgT 2
General Case of a Person on a Ladder
Bill (mass m) is climbing a ladder (length L, mass M) that leans against a smooth wall (no friction between wall and ladder). A frictional force f between the ladder and the floor keeps it from slipping. The angle between the ladder and the wall is .
How does f depend on the angle of the ladder and Bill’s distance up the ladder?
L
m
f
y
x
MBill
Mechanics Lecture 18, Slide 30
Balance forces:
x: Fwall f
y: N Mg mg
Balance torques:
Mg
mg
L/2
f y
xN
Fwall
daxis
cot2
Mg
L
dmgFwall
Mechanics Lecture 18, Slide 31
wallF f
cot2
Mg
L
dmgf
cosmgd cos2
LMg sinLFwall 0
f
d
M
m
Lets try it out
Climbing further up the ladder makes it more likely to slip:
Making the ladder more vertical makes it less likely to slip:
This is the General Expression:
Mechanics Lecture 18, Slide 32
cot2
Mg
L
dmgf
cot2
Mg
L
dmgf
Moving the bottom of the ladder further from the wall makes it more likely to slip:
If its just a ladder…
Mechanics Lecture 18, Slide 33
cot2
Mgf
In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the ground the biggest?
A) Case 1 B) Case 2 C) Same
CheckPoint
Case 1 Case 2Mechanics Lecture 18, Slide 34
In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the ground the biggest?
A) Case 1 B) Case 2 C) Same
Case 1 Case 2A) Because the bottom of the ladder is further away from the wall.
B) The angle is steeper, which means there is more normal force and thus more friction.
C) Both have same mass.
Mechanics Lecture 18, Slide 35
CheckPoint
CheckPoint
Suppose you hang one end of a beam from the ceiling by a rope and the bottom of the beam rests on a frictionless sheet of ice. The center of mass of the beam is marked with an black spot. Which of the following configurations best represents the equilibrium condition of this setup?
A) B) C)
Mechanics Lecture 18, Slide 36
CheckPointWhich of the following configurations best represents the equilibrium condition of this setup?
A) B) C)
If the tension has any horizontal component, the beam will accelerate in the horizontal direction.
Objects tend to attempt to minimize their potential energy as much as possible. In Case C, the center of mass is lowest.
Mechanics Lecture 18, Slide 37
Homework
Mechanics Lecture 18, Slide 39
Choose axis for torque calculation
Decompose forces into x,y components
Draw Free Body Diagram
0
0,
xTN
ixF
Sum x-component forces
Sum y-component forces
0
0,
RLy FFTf
iyFDefine torques
00
00
90sin
0180sin
90sin
T
F
WFWF
NL
fLfL
T
RF
LLF
rodN
rodrodf
R
L
Sum of Torques
0
0
Lrod
TFFNf
WFfLRL
tan
sin
cos
x
y
y
x
T
T
TT
TT
Relationship between Tx and Ty
Definition of coefficient of friction
N
f
Mechanics Lecture 18, Slide 40
tan/
0
0
0
xy
Lrod
RLy
x
TT
WFfL
FFTf
TN
2sign
RL
WFF
tan
02
0
0
x
y
signrod
signy
x
T
T
WW
fL
WTf
TN
tan
)2
1(
tan
)2
1(
tan/
)2
1(2
2
signrod
x
signrod
yx
signrodrod
signsigny
rod
sign
x
WL
W
TN
WL
W
TT
WL
WW
L
WWT
WL
Wf
TN
)2(
tan
)2
1(
tan2
WL
W
WLW
WL
W
N
f
rodsign
rod
rod
sign