Physic 231 Lecture 21 - Michigan State Universitylynch/Physics 231 lecture21.pdfwhen a force is...
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Transcript of Physic 231 Lecture 21 - Michigan State Universitylynch/Physics 231 lecture21.pdfwhen a force is...
• Main points of last lecture:• Analogies between rotational and
translational motion• Solutions involving:
• Rolling motion:
• Main points of today’s lecture:• Angular momentum
• Tensile stress and strain
• Shear stress and strain:
• Bulk stress and strain:
• Pressure in fluids:
Physic 231 Lecture 21
ω
ω
ατ
IL
IKE
II
rot
=
=
=
2
21
22
21
21 ωIMvKEtotal +=
hxS
AF ∆=
∆
ghPPAFP topbot ρ+== ;
0LLY
AF ∆=
∆
VVBP
AF ∆
=∆=∆
ωIL =
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.
2
2
2
MR21I :
MRI :
21
21
=
=
+
=
disk
ring
RIM
Mghv
ghv
ghvring
34 :disk
:
=
=
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– b) 66 rev/s– c) 0.66 rev/s– d 0.17 rev/s
sradsrad
II
f
ff
/66.0/210
3.3
00
==
=
ω
ωω
Stress and strain in solids• 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.
StrainLLStress
AF
LLY
AF
=∆
=∆
=00
; ;
• 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:
• 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
mmmNx
mmNxYL
AFL
LLYmNx
AF
4.4/1018
5.0/10160
/10160
29260
0
26
===∆⇒
∆==
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.
trainShearhxtressShear
AF
hxS
AF s ;s ; =
∆=
∆=
• 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.
• 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.
mxmNx
mm
NSh
AFx
hxS
AF
5262 104.2
/100.3005.0
0014.20 −===∆⇒
∆=
Pressure in fluids• 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, becauseit would cause the liquid to move, but there is a pressure force perpendicular to all 6 surfaces.
y
zx
backfrontzbackbackzfrontfrontbackfront PPAPFAPFFF =⇒===− ; ;0 :z
• 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:
ghPPA
hgAPP
hgAVgMgW
topboty
ytopbot
y
ρρ
ρρ
+=⇒+=
===
rightleftxrightrightxleftleftrightleft PPAPFAPFFFxF
=⇒===−
=∑; ;0:
0 :mequilibriu Inv
ytopbotytoptopybotbottopbot A
WPPAPFAPFWFF +=⇒===−− ; ;0 :y
• 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
Pascal’s principal
• 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– b) 300N– c) 30N– d) .0.33N
Ncm
NcmAFAF
AF
AF
PP
301000300001 2
2
2
211
2
2
1
1
21
===⇒=
=
Pressure and Depth equation
• 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
ghPP topbot ρ+=
Bulk modulus
• 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 thebulk modulus B.
strain Volume ;change pressure ; =∆
=∆∆
=∆VVP
VVBP
( )
33
27211
3
105.1
/103/106.1
8
b)
cmx
mNxmNx
cm
PPBVV
VVBP
topbot
−=
=
−=∆
∆=∆
• 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)– 1a) 2x105 N/m2
– 1b) 4x105 N/m2
– 1c) 2x106 N/m2
– 1d) 3x107 N/m2 ( )27
235
/100.3
3000/8.9/1000100.1
a)
mNxPmsmmkgPaxP
ghPP
bot
bot
topbot
=
+=
+= ρ