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Physics 1710—Warm-up Quiz
Why does a diver rotate faster when she tucks in her
arms and legs?
A.
B.
C.
D.
E.
She increases her angular momentum.
She increases her moment of inertia.
She decreases her moment of inertia.
She pushes against her inertia.
None of the above
65%
24%
6%
A
4%
B
C
D
2%
E
Physics 1710—Chapter 13 App: E & E
Analysis:
• Like an ice skater. Why does an ice skater increase her
angular velocity without the benefit of a torque?
L = r x p= r x ( m v)
= r x ( m r x ⍵)
Li = mi ri 2 ⍵ z
Lz = (∑i mi ri 2 ) ⍵z
Lz = Iz ⍵z ; & ⍵z = Lz / Iz
Therefore, a decrease in I ( by reducing r) will result in an
increase in ⍵ even if dL/dt = 0!
Physics 1710—Chapter 13 App: E & E
Rotating Platform Demonstration
Physics 1710—Chapter 13 App: E & E
Analysis:
•Why does an ice skater increase her angular velocity
without the benefit of a torque?
L=rxp
= r x ( m v)
= r x ( m r x ⍵)
Li = mi ri 2 ⍵
Lz = (∑i mi ri 2 ) ⍵
Lz = I ⍵; & ⍵ = Lz / I
Therefore, a decrease in I ( by reducing r) will result in an
increase in ⍵.
Physics 1710—Chapter 13 App: E & E
How does a ladder stay up?
Think!
No Talking!
Confer!
Peer Instruction Time
Physics 1710—Chapter 13 App: E & E
1′ Lecture
•Static equilibrium (no translational or rotational
acceleration) requires that all forces and torques to
balance.
• If the acceleration due to gravity is the same for all
particles comprising a body the center of mass is the
center of gravity.
CM = CG if g the same for all mi .
• The moduli of elasticity (Y, E, B) characterizes the
stress-strain relation:
• stress= modulus • strain; σ = Y ε
Physics 1710—Chapter 13 App: E & E
One more turn on L—Angular Momentum:
L=rxp
The angular momentum is the vector product of the
moment arm and the linear momentum.
∑ T = d L/dt
The net torque is equal to the time rate of change in the
angular momentum.
Physics 1710—Chapter 12 App: E & E
Second Law of Motion
F=ma
Or F = dp/dt
Then:
r x F = d (r x p)/dt
Torque = τ = d L/dt
L = r x p is the “angular momentum.”
Physics 1710—Chapter 12 App: E & E
Second Law of Motion
Torque = τ = d L/dt
If τ = 0, then
L is a constant.
L = constant means angular momentum is
conserved.
Physics 1710—Chapter 12 App: E & E
Second Law of Motion & Gyroscopic Precession
L=Iω
Torque = τ = d L/dt
= d( I ω)/dt
r
τ = r x F = d L/dt
= I (dω/dt);
τ
F = mg
dω/dt = I-1 τ
Physics 1710—Chapter 13 App: E & E
What will happen to a tilted spinning top when it is
supported on one end only? Why?
Think!
Peer Instruction Time
No Talking!
Confer!
Physics 1710—Chapter 11 App: E & E
What will happen to a tilted, spinning top when
it is supported on one end only? Why?
A.
B.
C.
D.
E.
It will fall over because of gravity.
It will spin faster because of I is
changing.
It will precess clockwise due to
torque.
It will precess counterclockwise
due to torque.
None of the above.
54%
30%
11%
5%
0%
A
B
C
D
E
Physics 1710—Chapter 11 App: E & E
Experiment:
Spinning Top
•Observe the direction of rotation (ω)
• Observe the motion (dω/dt)
Physics 1710—Chapter 11 App: E & E
Second Law of Motion & Gyroscopic Precession
L=Iω
Torque = τ = d L/dt
= d( I ω)/dt
r
τ = r x F = d L/dt
= I (dω/dt);
τ
F = mg
dω/dt = I-1 τ
Physics 1710—Chapter 11 App: E & E
Torque and the Right Hand Rule:
r
X
F
Physics 1710—Chapter 11 App: E & E
Gyroscopic Precession
Torque = τ = d L/dt
= d( I ω)/dt
= I ( dω/dt)
(Top view)
Physics 1710—Chapter 11 App: E & E
Fundamental Angular Momentum
Fundamental unit of angular momentum = ℏ
ℏ = 1.054 x 10 -34 kg‧m/s2
ICM⍵ ≈ ℏ
⍵ ≈ ℏ / ICM
= 1.054 x 10 -34 kg‧m/s2 / (1.95 x 10 -46 kg‧m)
= 5.41 x 10 11 rad/s
Physics 1710—Chapter 11 App: E & E
MRI (a.k.a. NMR: Nuclear Magnetic Resonance)
Magnetic Resonance Imaging
Magnetic
Field
Torque=?
Hydrogen atoms
Precession!
MRI permits imaging of soft tissue due
precession of the hydrogen in the water
of the human body.
Physics 1710—Chapter 11 App: E & E
Teeter-totter: How does it balance?
Fsupport = - Fg
τnet = Στi
F1
F2
For equilibrium:
τnet = Στi = 0
Fnet = ΣFi = 0
Fg=F1+F2
Physics 1710—Chapter 11 App: E & E
How does a ladder stay up?
Good ideas?
Physics 1710—Chapter 11 App: E & E
Legends of the Fall—How a ladder stays up
Fnet = 0
Tnet = 0
Fnet = 0
Physics 1710—Chapter 11 App: E & E
Equilibrium: (equi= equal
“=“; libium = scales “ ♎”)
Equilibrium implies “balanced.”
Fnet = d P/dt
In equilibrium Fnet = 0
τnet = d L/dt
In equilibrium τnet = 0
Physics 1710—Chapter 11 App: E & E
Can an object be in equilibrium when the center
of mass lies outside of the object?
A.
B.
C.
Yes.
No.
Depends.
56%
26%
18%
A
B
C
Physics 1710—Chapter 11 App: E & E
Solution:
In equilibrium
Hollow Cone
Fnet = 0
τnet = 0
Is it “stable?”
i.e. does it recover
from a small
displacement ?
Physics 1710—Chapter 11 App: E & E
Center of Gravity
If the acceleration due to gravity is the same for all parts
of a body, then the center of mass corresponds to the
center of gravity.
CG =CM if g uniform
Proof: Ti = Σxi Fgi = Σxi gmgi = Mg Σxi mgi /M= Fcg xcm
Physics 1710—Chapter 11 App: E & E
Center of Gravity (comparison)
CM
Ball
Fg = - mg
CM
CG
Moon
Fg = - GmM/r2
Physics 1710—Chapter 11 App: E & E
Elasticity
Definitions:
•
Stress σ : the deforming force per unit area.
•
Strain ε : the unit deformation.
Stress = modulus x strain
σ = F/A = Y ε
Physics 1710—Chapter 11 App: E & E
Elasticity
Stress σ – Strain ε “Curve”
Elastic limit
Stress σ (N/m2)
•
σ=Yε
Strain ε = ΔL/L (%)
Failure
Physics 1710—Chapter 11 App: E & E
Elasticity
•
•
Stress σ : the deforming force per unit area.
Strain ε : the unit deformation.
Tensile/Compressive Stress
Young’s Modulus E
Stress = modulus x strain
σ = F/A = E ε = E ΔL/L
L
ΔL
σ=Eε
Physics 1710—Chapter 11 App: E & E
Elasticity
•
•
Stress σ : the deforming force per unit area.
Strain ε : the unit deformation.
Shear Modulus G
Stress = modulus x strain
σ = F/A = G ε = G Δx/L
L
Δx
σ
Physics 1710—Chapter 11 App: E & E
Elasticity
•
•
Stress σ : the deforming force per unit area.
Strain ε : the unit deformation.
Hydraulic Stress:
Bulk Modulus B
Stress = modulus x strain
σ = F/A = p = B ε = B ΔV/V
ΔV
V
p
Physics 1710—Chapter 11 App: E & E
Elasticity
•
•
Stress: the deforming force per unit area.
Strain: the unit deformation.
–
–
–
–
–
Tensile: “stretch”
Compressive: “squeeze”
Shear: “lean”
Hydraulic: pressure
Yield: permanently deformed
Physics 1710—Chapter 11 App: E & E
Summary
•
Static equilibrium implies that all forces and
torques balance.
•
The center of mass is often the center of
gravity.
•
The moduli of elasticity characterizes the
stress-strain relation:
•
stress= modulus x strain
Stress = modulus x strain
σ = F/A = Y ε
Physics 1710—Chapter 11 App: E & E
Physics 1710—Chapter 11 App: E & E
Why does the platform spin faster when he
brings his arms in?
Think!
Peer Instruction Time
No Talking!
Confer!
Physics 1710—Chapter 11 App: E & E
Why does the platform spin faster when he brings his
arms in?
A.
B.
C.
D.
E.
He increases his angular
momentum.
He increase his moment of
inertia.
He decrease his moment of
inertia.
He pushes against the
inertia of the weights.
None of the above
0%
0 of 1
10
Answer Now !
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Physics 1710—Chapter 11 App: E & E
10
Where should the fulcrum be place to balance
the teeter-totter?
A.
B.
C.
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Answer Now !
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Physics 1710—Chapter 11 App: E & E
10
Which way will the torque ladder move?
A. Clockwise
B. Counterclockwise
C. Will stay balanced
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C
Answer Now !
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