Download Physic 231 Lecture 21

Physic 231 Lecture 21
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Main points of last lecture:
Analogies between rotational and
translational motion
Solutions involving:
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Main points of today’s lecture:
Angular momentum
L = Iω
Tensile stress and strain
•
Shear stress and strain:
I
τ = Iα
KErot
1 2
= Iω
2
L = Iω
•
Rolling motion:
1
1
KEtotal = Mv 2 + Iω 2
2
2
∆F
∆L
=Y
A
L0
∆F
∆x
=S
A
h
• Bulk stress and strain:
∆F
∆V
= ∆P = B
A
V
• Pressure in fluids:
P=
F
; Pbot = Ptop + ρgh
A
Conceptual quiz
•
A solid disk and a ring roll down an incline.The ring is slower than the
disk if
– a) mring = mdisk, where m is the inertial mass.
– b) rring = rdisk, where r is the radius.
– c) mring = mdisk and rring = rdisk.
– d) The ring is always slower regardless of the relative values of m
and r.
v=
Mgh
1 I 
1
M
+
 2
2 R 2 
ring : I = MR 2
1
disk : I = MR 2
2
ring : v = gh
4
disk : v =
gh
3
quiz
•
A diver leaps from the 10 m platform and executes a triple forward
somersault dive. While in the tuck, the diver has a moment of inertia of
about 3.3 kg⋅m2 and during the final layout, the diver stretches her
body and increases her moment of inertia to 10 kg⋅m2. If the diver
achieves an angular velocity of 2 rev/s while in the tuck, what is the
angular velocity during the layout?
– a) 6 rev/s
I0
– b) 66 rev/s
ω f = ω0
If
– c) 0.66 rev/s
– d 0.17 rev/s
3 .3
ωf =
10
2rad / s = 0.66rad / s
Stress and strain in solids
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All materials deform when under stress, ie.
when a force is applied to them. The atoms in a
solid are held in place by forces that act like
springs.
•
If you pull on a bar of cross-sectional area A
with a force F, the bar will stretch a distance
∆L. In this case, the force is perpendicular to
the surface, and the force F and distance ∆L are
related by Young’s modulus Y:
F
∆L F
∆L
=Y
; = Stress;
= Strain
A
L0 A
L0
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The force will be linearly proportional to the
strain up to the elastic limit. Beyond this point, the
bar will be permanently deformed. Stress units
Pa=N/m2.
Example
•
Bone has a Young’s modulus of about 18 x 109 Pa. Under
compression, it can withstand a stress of about 160 x 106 Pa before
breaking. Assume that a femur (thigh bone) is 0.50 m long and
calculate the amount of compression this bone can withstand before
breaking.
– a) 4.4 mm
– b) 6.4 mm
– c) 1.6 cm
– d) 2.0 cm
∆L
F
= 160 x106 N / m 2 = Y
L0
A
F L0
0.5m
6
2
⇒ ∆L =
= 160 x10 N / m
= 4.4mm
9
2
AY
18 x10 N / m
Shear
•
You can also exert a force on an object with a
force that is parallel to its surface. Friction and
viscous drag are two example of such forces,
which act on the surface of the object and can
cause a shear stress.
•
Such shear forces cause the kind of
deformation indicated in the figure. Over a
distance h, from one surface, there is a shift ∆x
in the direction of the applied force. This shift
is governed by the shear modulus S.
F
∆x F
∆x
=S
; = Shear stress;
= Shear strain
h A
h
A
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The shear modulus governs how easily an
object can be bent.
Example
•
A child slides across a floor in a pair of rubber-soled shoes. The
friction force acting on each foot is 20 N, the footprint area of each
foot is 14 cm2, and the thickness of the soles is 5.0 mm. Find the
horizontal distance traveled by the sheared face of the sole. The shear
modulus of the rubber is 3.0 x 106 Pa.
F
∆x
=S
h
A
F h
20 N
0.005m
−5
⇒ ∆x =
=
=
2
.
4
x
10
m
2
6
2
A S .0014m 3.0 x10 N / m
Pressure in fluids
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Consider a vessel filled with liquid and a smaller rectangular
volume V of the same liquid contained within the vessel. If
the liquid is at rest, there can be no shear force on V, because
it would cause the liquid to move, but there is a pressure
force perpendicular to
v all 6 surfaces.
In equilibrium : ∑ F = 0
x : Fleft − Fright = 0; Fleft = Pleft Ax ; Fright = Pright Ax ⇒ Pleft = Pright
y
z
x
z : F front − Fback = 0; F front = Pfront Az ; Fback = Pback Az ⇒ Pfront = Pback
W
y : Fbot − Ftop − W = 0; Fbot = Pbot Ay ; Ftop = Ptop Ay ⇒ Pbot = Ptop +
Ay
• More detailed considerations show that pressure always exerts a force
perpendicular to a surface and that pressure is the same everywhere at the same
height, i.e. Pfront =Pback, and Pleft =Pright and Pfront= Pright, etc.
• Defining density ρ of the liquid by ρ=M/V we can simplify the dependence of P
on height as follows:
W = Mg = ρVg = ρAy hg
Pbot = Ptop +
ρAy hg
Ay
⇒ Pbot = Ptop + ρgh
Pascal’s principal
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Because F=PA, one can make hydraulic
machines that can generate large forces.
Units: One atmosphere (1 atm) =
– 76.0 cm of mercury
– 1.013 x 105 Pa = 1.013 x 105 N/m2
– 14.7 lb/in
Quiz: A simple hydraulic lift is employed to lift a truck that weighs
30000N. If A1=1 cm2 and A2= 1000 cm2, what force F1 is needed to lift
the truck?
– a) 3N
P1 = P2
– b) 300N
F1 F2
F2
2 30000 N
=
⇒
F
=
A
=
1
cm
= 30 N
1
1
– c) 30N
2
A1 A2
A2
1000cm
– d) .0.33N
Pressure and Depth equation
Pbot = Ptop + ρgh
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If the top is open, Pbot is normal
atmospheric pressure
– 1.013 x 105 Pa = 14.7 lb/in2
The pressure does not depend
upon the shape of the container
Bulk modulus
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If you drop a metal block in the ocean, it will sink to bottom where the
pressure is much greater than that of the atmosphere. This will
compress the block; the amount of compression is governed by the
bulk modulus B.
∆P = B
•
∆V
∆V
; ∆P = pressure change;
= Volume strain
V
V
Example: A steel block with a volume of 8 cm3 is dropped in the ocean
and comes to rest at the bottom, which is 3 km below the surface. 1)
What is the pressure at the bottom of the ocean? 2) By what amount
∆V is the volume decrease. (The Bulk modulus for steel is 1.6x1011
N/m2. The density of water is 1000kg/m3. 1 atm. =1.01x105 N/m2)
∆V
5
2
b)
P
B
∆
=
– 1a) 2x10 N/m
V
– 1b) 4x105 N/m2
a) Pbot = Ptop + ρgh6
V
Pbot − Ptop
∆V =
– 1c) 2x10 N/m2
5
3
2
B
Pbot = 1–.0 x1d)
Pa +7 1000
10 3x10
N/m2kg / m 9.8m / s 3000m
3
(
Pbot = 3.0 x107 N / m 2
)
(
)
8cm
7
2
3
x
10
N
/
m
1.6 x1011 N / m 2
= 1.5 x10−3 cm 3
=