Physics 121,Lecture 19,Pg 1
Physics 121,Lecture 19
Today’s Agenda
? Announcements
? Homework 8,due Friday Nov,11 @ 6:00 PM.
?Chap,7,# 7,22,28,33,35,44,45,50,54,61,and 65.
? Today’s topics
?Kinetic energy & rotation
?Rolling motion
?Angular momentum
?Fluid and solids
Physics 121,Lecture 19,Pg 2
Summary
(with comparison to 1-D kinematics)
Angular Linear
c o n s ta n t??
??
? ? ? 0 ??t
? ? ? ?? ? ?0 0 212t t
co n st a n ta ?
v v at? ?0
x x v t at? ? ?0 0 212
And for a point at a distance R from the rotation axis:
x = R????????????v = ?R ??????????a = ?R
Physics 121,Lecture 19,Pg 3
Rotation & Kinetic Energy...
? The kinetic energy of a rotating system looks similar to that of a
point particle:
Point Particle Rotating System
??
K ? 12 I?2
I ? ? m ri i
i
2
??
K ? 12 mv 2
v is,linear” velocity
m is the mass.
? is angular velocity
I is the moment of inertia
about the rotation axis.
Physics 121,Lecture 19,Pg 4
Rolling Motion
? Now consider a cylinder rolling at a constant speed,
K MVT O T CM CM? ?12 122 2I ?
VCM CM
The cylinder is rotating about CM and its CM is moving at constant
speed (vCM),Thus its total kinetic energy is given by,
Physics 121,Lecture 19,Pg 5
Rolling Motion
? Consider again a cylinder rolling at a constant speed,
VCM
P
Q
CM
?
At any instant the cylinder is rotating about point P,Its kinetic
energy is given by its rotational energy about that point,
KTOT = 1/2 IP ?2
Physics 121,Lecture 19,Pg 6
Rolling Motion
? We can find IP using the parallel axis theorem
K MVT O T CM CM? ?12 122 2I ?
VCM
P
Q
CM
?
IP = ICM + MR2
KTOT = 1/2 (ICM + MR2 ) ?2
KTOT = 1/2 ICM ?2 + 1/2 M (R2?2 ) = 1/2 ICM ?2 + 1/2 M vCM2 !
Physics 121,Lecture 19,Pg 7
Rolling Motion
? Cylinders of different I rolling down an inclined plane:
h
v = 0
??= 0
K = 0
R
?K = - ?U = Mgh
v = ?R
M K MV
T O T CM CM? ?12 122 2I ?
Physics 121,Lecture 19,Pg 8
Rolling...
? If there is no slipping (due to friction):
v 2v
In the lab reference frame
v
In the CM reference frame
v
Where v = ?R
?
Physics 121,Lecture 19,Pg 9
Rolling...
??
K T O T ? 12 MR 2 ? 2 ? 12 Mv 2 ? 12 ? 1? ? Mv 2
Use v= ?R and I = cMR2,
So,( ) M ghMv1
2
1 2 =+ 11gh2v +=
The rolling speed is always lower than in the case of simple
Sliding since the kinetic energy is shared between CM motion
and rotation.
hoop,c=1
disk,c=1/2
sphere,c=2/5
etc...
c c
c c
K MVT O T CM CM? ?12 122 2I ?
Physics 121,Lecture 19,Pg 10
Example, Rolling Motion
? A cylinder is about to roll down an inclined plane,What is its speed at
the bottom of the plane?
M
?
h
M v?
Ball has radius R
Physics 121,Lecture 19,Pg 11
Example, Rolling Motion
? Use conservation of energy.
Ei = Ui + 0 = Mgh
Ef = 0 + Kf = 1/2 Mv2 + 1/2 I ?2
= 1/2 Mv2 + 1/2 (1/2MR2)(v/R)2
Mgh = 1/2 Mv2 + 1/4 Mv2
v2 = 4/3 g h
v = ( 4/3 g h ) 1/2
Physics 121,Lecture 19,Pg 12
Lecture 19,ACT 1
Rolling Motion
? A race !!
Two cylinders are rolled down a ramp,They have the same
radius but different masses,M1 > M2,Which wins the race to
the bottom?
A) Cylinder 1
B) Cylinder 2
C) It will be a tie
M1
?
h
M?
M2
Active Figure
Physics 121,Lecture 19,Pg 13
Remember our roller coaster.
Perhaps now we can get the ball to go
around the circle without anyone dying.
Note:
Radius of loop = R
Radius of ball = r
Physics 121,Lecture 19,Pg 14
How high do we have to start the ball?
Use conservation of energy.
Also,we must remember that the minimum
speed at the top is vtop = (gR)1/2
E1 = mgh + 0 + 0
E2 = mg2R + 1/2 mv2 + 1/2 I?2
= 2mgR + 1/2 m(gR) + 1/2 (2/5 mr2)(v/r)2
= 2mgR + 1/2 mgR + (2/10)m (gR)
= 2.7 mgR
1
2
Physics 121,Lecture 19,Pg 15
How high do we have to start the ball?
E1 = mgh + 0 + 0
E2 = 2.7 mgR
mgh = 2.7 mgR
h = 2.7 R
h = 1.35 D
(The rolling motion added an extra 2/10 R to the
height)
1
2
Physics 121,Lecture 19,Pg 16
Rotational Dynamics:
more about Torque
? Suppose a force acts on a mass constrained to move in a circle,
Consider its acceleration in the tangential direction at some
instant:
?at = ?r
? Now use Newton’s 2nd Law in the tangential
direction:
?Ft= mat = m?r
? Multiply by r, r Ft = mr2 ?
? But t???r Ft and I= mr2
t???I ?
r
at
?
F
m
rt
Ft
Physics 121,Lecture 19,Pg 17
System of particles
? So for a collection of many particles
arranged in a rigid configuration:
r1
r2r3
r4
m4
m1
m2
m3
?F4
F1
F3 F2
r F m ri i
i i ii i,?
?? ? ? 2
ti I
t ?ii? ? I
? Since the particles are connected rigidly,
they all have the same ?,
?t I=tot
Physics 121,Lecture 19,Pg 18
Comment on t = I?
? When we write t = I? we are really talking about
the z component of a more general vector
equation,(we normally choose the z-axis to be
the the rotation axis.)
tz = Iz?z
? We usually omit the
z subscript for simplicity.
z
?z
tz
Iz
Physics 121,Lecture 19,Pg 19
Example
? To loosen a stuck nut,a man pulls at an angle of 45o on
the end of a 50cm wrench with a force of 200 N,
?What is the magnitude of the torque on the nut?
?If the nut suddenly turns freely,what is the angular
acceleration of the wrench? (The wrench
has a mass of 3 kg,and its shape
is that of a thin rod).
L=0.5m
F=200N
45o
Physics 121,Lecture 19,Pg 20
Example,solution
L=0.5m
F=200N
45o
?
? Torque t = Lfsin ? = (0.5m)(200N)(sin(45)) = 70.7 Nm
? If the nut turns freely (no more friction),t = I?
? We know t and we want ?,so we need to figure out I.
? ? ? ?I,,? ? ?13 13 3 0 5 252 2 2ML kg m k gm
??= 283 rad/s2
So ??= t?/ I = (70.7Nm) / (.25kgm2)
Physics 121,Lecture 19,Pg 21
Angular Momentum
,angular momentum
r
m
F
??
t ? I? ? I ? ? ? 0? t?????? ?????? ? I? ? I? 0? t
?IL?
t
L
?
???
i n t e r v a l t i m e
m o m e n t u ma n g u l a r i n C h a n g et
? An object as mass mass m
rotating in a circular path
under the action of a
constant torque:
Physics 121,Lecture 19,Pg 22
Angular Momentum
? The torque acting on an object is equal to the time
rate of change of the object’s momentum
? The angular momentum of an system is conserved
when the net external torque action on the system is
zero,That is,when ?t = 0,the initial angular
momentum equals the final angular momentum.
t
L
?
???
i n t e r v a l t i m e
m o m e n t u ma n g u l a r i n C h a n g et
? ?? 0II ffii t?? if
Lf = Li
Physics 121,Lecture 19,Pg 23
Angular momentum of a rigid body
about a fixed axis:
? In general,for an object rotating about a fixed (z) axis we
can write LZ = I?
? The direction of LZ is given by the
right hand rule (same as ?).
? We will omit the,Z” subscript for simplicity,
and write L = I? L Z ? I ?
?
z
Physics 121,Lecture 19,Pg 24
Example,Two Disks
? A disk of mass M and radius R rotates around the z axis
with angular velocity ?0,A second identical disk,initially
not rotating,is dropped on top of the first,There is friction
between the disks,and eventually they rotate together with
angular velocity ?F.
?0
z
?F
z
Physics 121,Lecture 19,Pg 25
Example,Two Disks
? First realize that there are no external torques acting on the
two-disk system.
?Angular momentum will be conserved !
? Initially,the total angular momentum
is due only to the disk on the bottom:
?0
z
L MRIN I ? ?I 1 1 2 012? ?
2
1
Physics 121,Lecture 19,Pg 26
Example,Two Disks
? First realize that there are no external torques acting on the
two-disk system.
?Angular momentum will be conserved !
? Finally,the total angular momentum is due
to both disks spinning:
?F
z
L MRF IN F? ? ?I I1 1 2 2 2? ? ?21
Physics 121,Lecture 19,Pg 27
Example,Two Disks
? Since LINI = LFIN
?0
z
?F
z
LINI LFIN
1
2
2
0
2MR MR
F? ??
? ?F ? 12 0An inelastic collision,
since E is not
conserved (friction) !
Physics 121,Lecture 19,Pg 28
Demonstration of
Conservation of Angular Momentum
? Figure Skating,
?A
z
?B
z
Arm Arm
IA < IB
?A > ?B
LA = LB
Physics 121,Lecture 19,Pg 29
Example,Throwing ball from stool
? A student sits on a stool which is free to rotate,The
moment of inertia of the student plus the stool is I,She
throws a heavy ball of mass M with speed v such that its
velocity vector passes a distance d from the axis of
rotation,
?What is the angular speed ?F of the student-stool
system after she throws the ball?
top view,before after
d
vM
I I
?F
Physics 121,Lecture 19,Pg 30
Example,Throwing ball from stool...
? Conserve angular momentum (since there are no external
torques acting on the student-stool system):
?LBEFORE = 0,Lstool = I?F
?LAFTER = 0 = Lstool - Lball,Lball = Iball?ball = Md2 (v/d) = M d v
?0 = I?F - M d v
top view,before after
d
vM
I I
?F
? F M v d? I
Physics 121,Lecture 19,Pg 31
Gyroscopic Motion:
? Suppose you have a spinning gyroscope in the
configuration shown below:
? If the left support is removed,what will happen
?
pivotsupport
g
Physics 121,Lecture 19,Pg 32
Gyroscopic Motion...
? Suppose you have a spinning gyroscope in the
configuration shown below:
? If the left support is removed,what will happen?
?The gyroscope does not fall down !
?There is an external torque
?
pivot
g
Physics 121,Lecture 19,Pg 33
Fluids, Chapter 9
? At ordinary temperature,matter exists in one of three states
?Solid - has a shape and forms a surface
?Liquid - has no shape but forms a surface
?Gas - has no shape and forms no surface
? What do we mean by,fluids”?
?Fluids are,substances that flow”….,substances that
take the shape of the container”
?Atoms and molecules are free to move.
?No long range correlation between positions.
Physics 121,Lecture 19,Pg 34
Fluids
? What parameters do we use to describe fluids?
?Density
V?
?? m? units,
kg/m3 = 10-3 g/cm3
?(water) = 1.000 x103 kg/m3 = 1.000 g/cm3
?(ice) = 0.917 x103 kg/m3 = 0.917 g/cm3
?(air) = 1.29 kg/m3 = 1.29 x10-3 g/cm3
?(Hg) = 13.6 x103 kg/m3 = 13.6 g/cm3
Physics 121,Lecture 19,Pg 35
Fluids
nF ?pA? A
n
? Any force exerted by a fluid is perpendicular to a surface of
contact,and is proportional to the area of that surface.
?Force (a vector) in a fluid can be expressed in terms of
pressure (a scalar) as:
A
Fp
?
??
units,
1 N/m2 = 1 Pa (Pascal)
1 bar = 105 Pa
1 mbar = 102 Pa
1 torr = 133.3 Pa
1atm = 1.013 x105 Pa
= 1013 mbar
= 760 Torr
= 14.7 lb/ in2 (=PSI)
? What parameters do we use to describe fluids?
?Pressure
Physics 121,Lecture 19,Pg 36
?Young’s modulus,measures the resistance of a solid to a
change in its length.
?Shear modulus,measures the resistance to motion of
the planes of a solid sliding past each other.
?Bulk modulus,measures the resistance of solids or
liquids to changes in their volume.
Some definitions
L ?L
F
F1 F2 V
V - ?V
F
? Elastic properties of solids,
elasticity in length
elasticity of shape (ex,pushing a book)
volume elasticity
Physics 121,Lecture 19,Pg 37
Fluids
? Bulk Modulus
)V/V(
pB
??
??
?LIQUID,incompressible (density almost constant)
?GAS,compressible (density depends a lot on pressure)
??? ??? ??? ??? ??? ???? ???????
Bulk modulus (Pa=N/m2)
H2O SteelGas (STP) Pb
Physics 121,Lecture 19,Pg 38
? When the pressure is much less
than the bulk modulus of the fluid,
we treat the density as constant
independent of pressure:
incompressible fluid
? For an incompressible fluid,the
density is the same everywhere,
but the pressure is NOT!
Pressure vs,Depth
Incompressible Fluids (liquids)
? Consider an imaginary fluid volume (a cube,face area A)
?The sum of all the forces on this volume must be ZERO as
it is in equilibrium,F2 - F1 - mg = 0
y
1
y
2
A
p
1
p
2
F
1
F
2
mg
0
p
ApApFF 1212 ???
Ag)yy(mg 12 ??? )yy(gpp 1212 ????
Physics 121,Lecture 19,Pg 39
? If the pressures were different,fluid would flow in the tube!
? However,if fluid did flow,then the system was NOT in equilibrium
since no equilibrium system will spontaneously leave equilibrium.
Pressure vs,Depth
? For a fluid in an open container
pressure same at a given depth
independent of the container p ( y )
y
? Fluid level is the same
everywhere in a connected
container,assuming no surface
forces
? Why is this so? Why does the
pressure below the surface
depend only on depth if it is in
equilibrium?
? Imagine a tube that would connect two regions at the same depth.
Physics 121,Lecture 19,Pg 40
Lecture 19,ACT 2
Pressure
? What happens with two fluids Consider a U
tube containing liquids of density ?1 and ?2 as
shown:
?Compare the densities of the liquids:
A) ?1 < ?2 B) ?1 = ?2 C) ?1 > ?2
?1
?2
dI
Physics 121,Lecture 19,Pg 41
Pascal’s Principle
? So far we have discovered (using Newton’s Laws):
?Pressure depends on depth,?p = ?g?y
? Pascal’s Principle addresses how a change in pressure is
transmitted through a fluid.
Any change in the pressure applied to an enclosed fluid is
transmitted to every portion of the fluid and to the walls
of the containing vessel.
? Pascal’s Principle explains the working of hydraulic lifts
?i.e,the application of a small force at one place can result
in the creation of a large force in another.
?Does this,hydraulic lever” violate conservation of energy?
?Certainly hope not.,Let’s calculate.
Physics 121,Lecture 19,Pg 42
Pascal’s Principle
? Consider the system shown:
?A downward force F1 is applied
to the piston of area A1.
?This force is transmitted through
the liquid to create an upward
force F2.
?Pascal’s Principle says that
increased pressure from F1
(F1/A1) is transmitted throughout
the liquid.
F F
1
2
d
2
d
1
A A 2
1
2
2
1
1
A
F
A
F ?
1
212
A
AFF ?
? F2 > F1, Have we violated conservation of energy
Physics 121,Lecture 19,Pg 43
Pascal’s Principle
? Consider F1 moving through a
distance d1.
?How large is the volume of the
liquid displaced?
F F
1
2
d
2
d
1
A A
21
111 dV A??
12 VV ??? 2112 A
Add ?
? Therefore the work done by F1 equals the work done by F2
We have NOT obtained,something for nothing”.
1211121222 WA
Ad
A
AFdFW ???
?This volume determines the
displacement of the large piston.
Physics 121,Lecture 19,Pg 44
A1 A
10
A2 A
10
A1 A20
M
M
MdC
dB
dA
A) dA = (1/2)dB B) dA = dB C) dA = 2dB
? If A10 = 2′A20,compare dA and dC.
A) dA = (1/2)dC B) dA = dC C) dA = 2dC
Lecture 19,ACT 3
Hydraulics
? Consider the systems shown to the right.
?In each case,a block of mass M is
placed on the piston of the large
cylinder,resulting in a difference dI in
the liquid levels.
? If A2 = 2′A1,compare dA and dB.
Physics 121,Lecture 19,Pg 45
Recap for today:
? Todays’ topics
?Kinetic energy & rotation
?Rolling motion
?Angular momentum
?Fluid and solids
? Homework 8,due Friday Nov,11 @ 6:00 PM.
?Chap,7,# 7,22,28,33,35,44,45,50,54,61,and 65,
Physics 121,Lecture 19
Today’s Agenda
? Announcements
? Homework 8,due Friday Nov,11 @ 6:00 PM.
?Chap,7,# 7,22,28,33,35,44,45,50,54,61,and 65.
? Today’s topics
?Kinetic energy & rotation
?Rolling motion
?Angular momentum
?Fluid and solids
Physics 121,Lecture 19,Pg 2
Summary
(with comparison to 1-D kinematics)
Angular Linear
c o n s ta n t??
??
? ? ? 0 ??t
? ? ? ?? ? ?0 0 212t t
co n st a n ta ?
v v at? ?0
x x v t at? ? ?0 0 212
And for a point at a distance R from the rotation axis:
x = R????????????v = ?R ??????????a = ?R
Physics 121,Lecture 19,Pg 3
Rotation & Kinetic Energy...
? The kinetic energy of a rotating system looks similar to that of a
point particle:
Point Particle Rotating System
??
K ? 12 I?2
I ? ? m ri i
i
2
??
K ? 12 mv 2
v is,linear” velocity
m is the mass.
? is angular velocity
I is the moment of inertia
about the rotation axis.
Physics 121,Lecture 19,Pg 4
Rolling Motion
? Now consider a cylinder rolling at a constant speed,
K MVT O T CM CM? ?12 122 2I ?
VCM CM
The cylinder is rotating about CM and its CM is moving at constant
speed (vCM),Thus its total kinetic energy is given by,
Physics 121,Lecture 19,Pg 5
Rolling Motion
? Consider again a cylinder rolling at a constant speed,
VCM
P
Q
CM
?
At any instant the cylinder is rotating about point P,Its kinetic
energy is given by its rotational energy about that point,
KTOT = 1/2 IP ?2
Physics 121,Lecture 19,Pg 6
Rolling Motion
? We can find IP using the parallel axis theorem
K MVT O T CM CM? ?12 122 2I ?
VCM
P
Q
CM
?
IP = ICM + MR2
KTOT = 1/2 (ICM + MR2 ) ?2
KTOT = 1/2 ICM ?2 + 1/2 M (R2?2 ) = 1/2 ICM ?2 + 1/2 M vCM2 !
Physics 121,Lecture 19,Pg 7
Rolling Motion
? Cylinders of different I rolling down an inclined plane:
h
v = 0
??= 0
K = 0
R
?K = - ?U = Mgh
v = ?R
M K MV
T O T CM CM? ?12 122 2I ?
Physics 121,Lecture 19,Pg 8
Rolling...
? If there is no slipping (due to friction):
v 2v
In the lab reference frame
v
In the CM reference frame
v
Where v = ?R
?
Physics 121,Lecture 19,Pg 9
Rolling...
??
K T O T ? 12 MR 2 ? 2 ? 12 Mv 2 ? 12 ? 1? ? Mv 2
Use v= ?R and I = cMR2,
So,( ) M ghMv1
2
1 2 =+ 11gh2v +=
The rolling speed is always lower than in the case of simple
Sliding since the kinetic energy is shared between CM motion
and rotation.
hoop,c=1
disk,c=1/2
sphere,c=2/5
etc...
c c
c c
K MVT O T CM CM? ?12 122 2I ?
Physics 121,Lecture 19,Pg 10
Example, Rolling Motion
? A cylinder is about to roll down an inclined plane,What is its speed at
the bottom of the plane?
M
?
h
M v?
Ball has radius R
Physics 121,Lecture 19,Pg 11
Example, Rolling Motion
? Use conservation of energy.
Ei = Ui + 0 = Mgh
Ef = 0 + Kf = 1/2 Mv2 + 1/2 I ?2
= 1/2 Mv2 + 1/2 (1/2MR2)(v/R)2
Mgh = 1/2 Mv2 + 1/4 Mv2
v2 = 4/3 g h
v = ( 4/3 g h ) 1/2
Physics 121,Lecture 19,Pg 12
Lecture 19,ACT 1
Rolling Motion
? A race !!
Two cylinders are rolled down a ramp,They have the same
radius but different masses,M1 > M2,Which wins the race to
the bottom?
A) Cylinder 1
B) Cylinder 2
C) It will be a tie
M1
?
h
M?
M2
Active Figure
Physics 121,Lecture 19,Pg 13
Remember our roller coaster.
Perhaps now we can get the ball to go
around the circle without anyone dying.
Note:
Radius of loop = R
Radius of ball = r
Physics 121,Lecture 19,Pg 14
How high do we have to start the ball?
Use conservation of energy.
Also,we must remember that the minimum
speed at the top is vtop = (gR)1/2
E1 = mgh + 0 + 0
E2 = mg2R + 1/2 mv2 + 1/2 I?2
= 2mgR + 1/2 m(gR) + 1/2 (2/5 mr2)(v/r)2
= 2mgR + 1/2 mgR + (2/10)m (gR)
= 2.7 mgR
1
2
Physics 121,Lecture 19,Pg 15
How high do we have to start the ball?
E1 = mgh + 0 + 0
E2 = 2.7 mgR
mgh = 2.7 mgR
h = 2.7 R
h = 1.35 D
(The rolling motion added an extra 2/10 R to the
height)
1
2
Physics 121,Lecture 19,Pg 16
Rotational Dynamics:
more about Torque
? Suppose a force acts on a mass constrained to move in a circle,
Consider its acceleration in the tangential direction at some
instant:
?at = ?r
? Now use Newton’s 2nd Law in the tangential
direction:
?Ft= mat = m?r
? Multiply by r, r Ft = mr2 ?
? But t???r Ft and I= mr2
t???I ?
r
at
?
F
m
rt
Ft
Physics 121,Lecture 19,Pg 17
System of particles
? So for a collection of many particles
arranged in a rigid configuration:
r1
r2r3
r4
m4
m1
m2
m3
?F4
F1
F3 F2
r F m ri i
i i ii i,?
?? ? ? 2
ti I
t ?ii? ? I
? Since the particles are connected rigidly,
they all have the same ?,
?t I=tot
Physics 121,Lecture 19,Pg 18
Comment on t = I?
? When we write t = I? we are really talking about
the z component of a more general vector
equation,(we normally choose the z-axis to be
the the rotation axis.)
tz = Iz?z
? We usually omit the
z subscript for simplicity.
z
?z
tz
Iz
Physics 121,Lecture 19,Pg 19
Example
? To loosen a stuck nut,a man pulls at an angle of 45o on
the end of a 50cm wrench with a force of 200 N,
?What is the magnitude of the torque on the nut?
?If the nut suddenly turns freely,what is the angular
acceleration of the wrench? (The wrench
has a mass of 3 kg,and its shape
is that of a thin rod).
L=0.5m
F=200N
45o
Physics 121,Lecture 19,Pg 20
Example,solution
L=0.5m
F=200N
45o
?
? Torque t = Lfsin ? = (0.5m)(200N)(sin(45)) = 70.7 Nm
? If the nut turns freely (no more friction),t = I?
? We know t and we want ?,so we need to figure out I.
? ? ? ?I,,? ? ?13 13 3 0 5 252 2 2ML kg m k gm
??= 283 rad/s2
So ??= t?/ I = (70.7Nm) / (.25kgm2)
Physics 121,Lecture 19,Pg 21
Angular Momentum
,angular momentum
r
m
F
??
t ? I? ? I ? ? ? 0? t?????? ?????? ? I? ? I? 0? t
?IL?
t
L
?
???
i n t e r v a l t i m e
m o m e n t u ma n g u l a r i n C h a n g et
? An object as mass mass m
rotating in a circular path
under the action of a
constant torque:
Physics 121,Lecture 19,Pg 22
Angular Momentum
? The torque acting on an object is equal to the time
rate of change of the object’s momentum
? The angular momentum of an system is conserved
when the net external torque action on the system is
zero,That is,when ?t = 0,the initial angular
momentum equals the final angular momentum.
t
L
?
???
i n t e r v a l t i m e
m o m e n t u ma n g u l a r i n C h a n g et
? ?? 0II ffii t?? if
Lf = Li
Physics 121,Lecture 19,Pg 23
Angular momentum of a rigid body
about a fixed axis:
? In general,for an object rotating about a fixed (z) axis we
can write LZ = I?
? The direction of LZ is given by the
right hand rule (same as ?).
? We will omit the,Z” subscript for simplicity,
and write L = I? L Z ? I ?
?
z
Physics 121,Lecture 19,Pg 24
Example,Two Disks
? A disk of mass M and radius R rotates around the z axis
with angular velocity ?0,A second identical disk,initially
not rotating,is dropped on top of the first,There is friction
between the disks,and eventually they rotate together with
angular velocity ?F.
?0
z
?F
z
Physics 121,Lecture 19,Pg 25
Example,Two Disks
? First realize that there are no external torques acting on the
two-disk system.
?Angular momentum will be conserved !
? Initially,the total angular momentum
is due only to the disk on the bottom:
?0
z
L MRIN I ? ?I 1 1 2 012? ?
2
1
Physics 121,Lecture 19,Pg 26
Example,Two Disks
? First realize that there are no external torques acting on the
two-disk system.
?Angular momentum will be conserved !
? Finally,the total angular momentum is due
to both disks spinning:
?F
z
L MRF IN F? ? ?I I1 1 2 2 2? ? ?21
Physics 121,Lecture 19,Pg 27
Example,Two Disks
? Since LINI = LFIN
?0
z
?F
z
LINI LFIN
1
2
2
0
2MR MR
F? ??
? ?F ? 12 0An inelastic collision,
since E is not
conserved (friction) !
Physics 121,Lecture 19,Pg 28
Demonstration of
Conservation of Angular Momentum
? Figure Skating,
?A
z
?B
z
Arm Arm
IA < IB
?A > ?B
LA = LB
Physics 121,Lecture 19,Pg 29
Example,Throwing ball from stool
? A student sits on a stool which is free to rotate,The
moment of inertia of the student plus the stool is I,She
throws a heavy ball of mass M with speed v such that its
velocity vector passes a distance d from the axis of
rotation,
?What is the angular speed ?F of the student-stool
system after she throws the ball?
top view,before after
d
vM
I I
?F
Physics 121,Lecture 19,Pg 30
Example,Throwing ball from stool...
? Conserve angular momentum (since there are no external
torques acting on the student-stool system):
?LBEFORE = 0,Lstool = I?F
?LAFTER = 0 = Lstool - Lball,Lball = Iball?ball = Md2 (v/d) = M d v
?0 = I?F - M d v
top view,before after
d
vM
I I
?F
? F M v d? I
Physics 121,Lecture 19,Pg 31
Gyroscopic Motion:
? Suppose you have a spinning gyroscope in the
configuration shown below:
? If the left support is removed,what will happen
?
pivotsupport
g
Physics 121,Lecture 19,Pg 32
Gyroscopic Motion...
? Suppose you have a spinning gyroscope in the
configuration shown below:
? If the left support is removed,what will happen?
?The gyroscope does not fall down !
?There is an external torque
?
pivot
g
Physics 121,Lecture 19,Pg 33
Fluids, Chapter 9
? At ordinary temperature,matter exists in one of three states
?Solid - has a shape and forms a surface
?Liquid - has no shape but forms a surface
?Gas - has no shape and forms no surface
? What do we mean by,fluids”?
?Fluids are,substances that flow”….,substances that
take the shape of the container”
?Atoms and molecules are free to move.
?No long range correlation between positions.
Physics 121,Lecture 19,Pg 34
Fluids
? What parameters do we use to describe fluids?
?Density
V?
?? m? units,
kg/m3 = 10-3 g/cm3
?(water) = 1.000 x103 kg/m3 = 1.000 g/cm3
?(ice) = 0.917 x103 kg/m3 = 0.917 g/cm3
?(air) = 1.29 kg/m3 = 1.29 x10-3 g/cm3
?(Hg) = 13.6 x103 kg/m3 = 13.6 g/cm3
Physics 121,Lecture 19,Pg 35
Fluids
nF ?pA? A
n
? Any force exerted by a fluid is perpendicular to a surface of
contact,and is proportional to the area of that surface.
?Force (a vector) in a fluid can be expressed in terms of
pressure (a scalar) as:
A
Fp
?
??
units,
1 N/m2 = 1 Pa (Pascal)
1 bar = 105 Pa
1 mbar = 102 Pa
1 torr = 133.3 Pa
1atm = 1.013 x105 Pa
= 1013 mbar
= 760 Torr
= 14.7 lb/ in2 (=PSI)
? What parameters do we use to describe fluids?
?Pressure
Physics 121,Lecture 19,Pg 36
?Young’s modulus,measures the resistance of a solid to a
change in its length.
?Shear modulus,measures the resistance to motion of
the planes of a solid sliding past each other.
?Bulk modulus,measures the resistance of solids or
liquids to changes in their volume.
Some definitions
L ?L
F
F1 F2 V
V - ?V
F
? Elastic properties of solids,
elasticity in length
elasticity of shape (ex,pushing a book)
volume elasticity
Physics 121,Lecture 19,Pg 37
Fluids
? Bulk Modulus
)V/V(
pB
??
??
?LIQUID,incompressible (density almost constant)
?GAS,compressible (density depends a lot on pressure)
??? ??? ??? ??? ??? ???? ???????
Bulk modulus (Pa=N/m2)
H2O SteelGas (STP) Pb
Physics 121,Lecture 19,Pg 38
? When the pressure is much less
than the bulk modulus of the fluid,
we treat the density as constant
independent of pressure:
incompressible fluid
? For an incompressible fluid,the
density is the same everywhere,
but the pressure is NOT!
Pressure vs,Depth
Incompressible Fluids (liquids)
? Consider an imaginary fluid volume (a cube,face area A)
?The sum of all the forces on this volume must be ZERO as
it is in equilibrium,F2 - F1 - mg = 0
y
1
y
2
A
p
1
p
2
F
1
F
2
mg
0
p
ApApFF 1212 ???
Ag)yy(mg 12 ??? )yy(gpp 1212 ????
Physics 121,Lecture 19,Pg 39
? If the pressures were different,fluid would flow in the tube!
? However,if fluid did flow,then the system was NOT in equilibrium
since no equilibrium system will spontaneously leave equilibrium.
Pressure vs,Depth
? For a fluid in an open container
pressure same at a given depth
independent of the container p ( y )
y
? Fluid level is the same
everywhere in a connected
container,assuming no surface
forces
? Why is this so? Why does the
pressure below the surface
depend only on depth if it is in
equilibrium?
? Imagine a tube that would connect two regions at the same depth.
Physics 121,Lecture 19,Pg 40
Lecture 19,ACT 2
Pressure
? What happens with two fluids Consider a U
tube containing liquids of density ?1 and ?2 as
shown:
?Compare the densities of the liquids:
A) ?1 < ?2 B) ?1 = ?2 C) ?1 > ?2
?1
?2
dI
Physics 121,Lecture 19,Pg 41
Pascal’s Principle
? So far we have discovered (using Newton’s Laws):
?Pressure depends on depth,?p = ?g?y
? Pascal’s Principle addresses how a change in pressure is
transmitted through a fluid.
Any change in the pressure applied to an enclosed fluid is
transmitted to every portion of the fluid and to the walls
of the containing vessel.
? Pascal’s Principle explains the working of hydraulic lifts
?i.e,the application of a small force at one place can result
in the creation of a large force in another.
?Does this,hydraulic lever” violate conservation of energy?
?Certainly hope not.,Let’s calculate.
Physics 121,Lecture 19,Pg 42
Pascal’s Principle
? Consider the system shown:
?A downward force F1 is applied
to the piston of area A1.
?This force is transmitted through
the liquid to create an upward
force F2.
?Pascal’s Principle says that
increased pressure from F1
(F1/A1) is transmitted throughout
the liquid.
F F
1
2
d
2
d
1
A A 2
1
2
2
1
1
A
F
A
F ?
1
212
A
AFF ?
? F2 > F1, Have we violated conservation of energy
Physics 121,Lecture 19,Pg 43
Pascal’s Principle
? Consider F1 moving through a
distance d1.
?How large is the volume of the
liquid displaced?
F F
1
2
d
2
d
1
A A
21
111 dV A??
12 VV ??? 2112 A
Add ?
? Therefore the work done by F1 equals the work done by F2
We have NOT obtained,something for nothing”.
1211121222 WA
Ad
A
AFdFW ???
?This volume determines the
displacement of the large piston.
Physics 121,Lecture 19,Pg 44
A1 A
10
A2 A
10
A1 A20
M
M
MdC
dB
dA
A) dA = (1/2)dB B) dA = dB C) dA = 2dB
? If A10 = 2′A20,compare dA and dC.
A) dA = (1/2)dC B) dA = dC C) dA = 2dC
Lecture 19,ACT 3
Hydraulics
? Consider the systems shown to the right.
?In each case,a block of mass M is
placed on the piston of the large
cylinder,resulting in a difference dI in
the liquid levels.
? If A2 = 2′A1,compare dA and dB.
Physics 121,Lecture 19,Pg 45
Recap for today:
? Todays’ topics
?Kinetic energy & rotation
?Rolling motion
?Angular momentum
?Fluid and solids
? Homework 8,due Friday Nov,11 @ 6:00 PM.
?Chap,7,# 7,22,28,33,35,44,45,50,54,61,and 65,
