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Transcript of Physics 207: Lecture 16, Pg 1 Physics 207, Lecture 16, Oct. 29 Agenda: Chapter 13 l Center of Mass l...
Physics 207: Lecture 16, Pg 1
Physics 207, Physics 207, Lecture 16, Oct. 29Lecture 16, Oct. 29
Agenda: Chapter 13Agenda: Chapter 13 Center of Mass Center of Mass TorqueTorque Moment of InertiaMoment of Inertia Rotational EnergyRotational Energy Rotational MomentumRotational Momentum
Assignment: Assignment: Wednesday is an exam review session, Exam will be Wednesday is an exam review session, Exam will be
held in rooms B102 &held in rooms B102 & B130 in Van Vleck at 7:15 PM MP Homework 7, Ch. 11, 5 problems, MP Homework 7, Ch. 11, 5 problems,
NOTE: Due Wednesday at 4 PMNOTE: Due Wednesday at 4 PM MP Homework 7A, Ch. 13, 5 problems, available soonMP Homework 7A, Ch. 13, 5 problems, available soon
Physics 207: Lecture 16, Pg 2
Chap. 13: Rotational DynamicsChap. 13: Rotational Dynamics
Up until now rotation has been only in terms of circular motion with ac = v2 / R and | aT | = d| v | / dt
Rotation is common in the world around us. Many ideas developed for translational motion are
transferable.
Physics 207: Lecture 16, Pg 3
Conservation of angular momentum has consequencesConservation of angular momentum has consequences
How does one describe rotation (magnitude and direction)?
Physics 207: Lecture 16, Pg 4
Rotational Dynamics: A child’s toy, a physics Rotational Dynamics: A child’s toy, a physics playground or a student’s nightmareplayground or a student’s nightmare
A merry-go-round is spinning and we run and jump on it. What does it do?
We are standing on the rim and our “friends” spin it faster. What happens to us?
We are standing on the rim a walk towards the center. Does anything change?
Physics 207: Lecture 16, Pg 7
Summary Summary (with comparison to 1-D kinematics)(with comparison to 1-D kinematics)
Angular Linear
constant
0 t
0 021
2t t
constanta
at 0vv
200 2
1v attxx
And for a point at a distance R from the rotation axis:
x = R v = Ra = R
Physics 207: Lecture 16, Pg 12
System of Particles (Distributed Mass):System of Particles (Distributed Mass): Until now, we have considered the behavior of very simple
systems (one or two masses). But real objects have distributed mass ! For example, consider a simple rotating disk and 2 equal
mass m plugs at distances r and 2r.
Compare the velocities and kinetic energies at these two points.
1 2
Physics 207: Lecture 16, Pg 13
System of Particles (Distributed Mass):System of Particles (Distributed Mass): An extended solid object (like a disk) can be thought of as
a collection of parts. The motion of each little part depends on where it is in the
object!
The rotation axis matters too!
1 K= ½ m v2 = ½ m (r)2
2 K= ½ m (2v)2 = ½ m (2r)2
Physics 207: Lecture 16, Pg 14
System of Particles: Center of MassSystem of Particles: Center of Mass
If an object is not held then it rotates about the center of mass. Center of mass: Where the system is balanced !
Building a mobile is an exercise in finding
centers of mass.
m1m2
+m1 m2
+
mobile
Physics 207: Lecture 16, Pg 15
System of Particles: Center of MassSystem of Particles: Center of Mass
How do we describe the “position” of a system made up of many parts ?
Define the Center of MassCenter of Mass (average position):For a collection of N individual pointlike particles whose
masses and positions we know:
M
mN
iii
1CM
rR
(In this case, N = 2)
y
x
rr2rr1
m1m2RRCM
Physics 207: Lecture 16, Pg 16
Sample calculation:Sample calculation:
Consider the following mass distribution:
(24,0)(0,0)
(12,12)
m
2m
m
RCM = (12,6)
kji CM CM CM1
CM ZYXM
mN
iii
r
R
XCM = (m x 0 + 2m x 12 + m x 24 )/4m meters
YCM = (m x 0 + 2m x 12 + m x 0 )/4m meters
XCM = 12 meters
YCM = 6 meters
Physics 207: Lecture 16, Pg 17
System of Particles: Center of MassSystem of Particles: Center of Mass
For a continuous solid, convert sums to an integral.
y
x
dm
rr
where dm is an infinitesimal
mass element.
Mdmr
dmdmr
CM
R
Physics 207: Lecture 16, Pg 18
Rotational Dynamics: What makes it spin?Rotational Dynamics: What makes it spin?
TOT = |r| |FTang| ≡ |r| |F| sin
Torque is the rotational equivalent of forceTorque is the rotational equivalent of forceTorque has units of kg m2/s2 = (kg m/s2) m = N m
A constant torque gives constant angular acceleration iff the mass distribution and the axis of rotation remain constant.
FFTangential aa
r
FFrandial
FF
A force applied at a distance from the rotation axis
Physics 207: Lecture 16, Pg 19
Lecture 16, Lecture 16, Exercise 1Exercise 1TorqueTorque
A. Case 1
B. Case 2
C. Same
In which of the cases shown below is the torque provided by the applied force about the rotation axis biggest? In both cases the magnitude and direction of the applied force is the same.
Remember torque requires F, r and sin or the tangential force component times perpendicular distance
L
LF F
axis
case 1 case 2
Physics 207: Lecture 16, Pg 20
Lecture 16, Lecture 16, Exercise 1Exercise 1TorqueTorque
In which of the cases shown below is the torque provided by the applied force about the rotation axis biggest? In both cases the magnitude and direction of the applied force is the same.
Remember torque requires F, r and sin or the tangential force component times perpendicular distance
(A)(A) case 1
(B)(B) case 2
(C) (C) same L
LF F
axis
case 1 case 2
Physics 207: Lecture 16, Pg 21
Rotational Dynamics: What makes it spin?Rotational Dynamics: What makes it spin?
TOT = |r| |FTang| ≡ |r| |F| sin
Torque is the rotational equivalent of forceTorque is the rotational equivalent of forceTorque has units of kg m2/s2 = (kg m/s2) m = N m
FFTangential aa
r
FFrandial
FF
A force applied at a distance from the rotation axis
TOT = r FTang = r m a = r m r = m r2
For every little part of the wheel
Physics 207: Lecture 16, Pg 22
TOT TOT = m r= m r22 and inertiaand inertia
This is the rotational version of FTOT = ma
Moment of inertia, Moment of inertia, I ≡ m rm r2 2 , (here , (here II is just a point is just a point on the wheel) is the rotational equivalent of mass.on the wheel) is the rotational equivalent of mass.
If I is big, more torque is required to achieve a given angular acceleration.
FFTangential aa
r
FFrandial
FF
The further a mass is away from this axis the greater the inertia (resistance) to rotation
Physics 207: Lecture 16, Pg 23
Calculating Moment of InertiaCalculating Moment of Inertia
where r is the distance from the mass to the axis of rotation.
N
iiirm
1
2I
Example: Calculate the moment of inertia of four point masses
(m) on the corners of a square whose sides have length L,
about a perpendicular axis through the center of the square:
mm
mm
L
Physics 207: Lecture 16, Pg 24
Calculating Moment of Inertia...Calculating Moment of Inertia...
For a single object, I depends on the rotation axis! Example: I1 = 4 m R2 = 4 m (21/2 L / 2)2
L
I = 2mL2I2 = mL2
mm
mm
I1 = 2mL2
Physics 207: Lecture 16, Pg 25
Lecture 16, Lecture 16, Exercise 2Exercise 2Moment of InertiaMoment of Inertia
A triangular shape is made from identical balls and identical rigid, massless rods as shown. The moment of inertia about the a, b, and c axes is Ia, Ib, and Ic respectively.
Which of the following is correct:
((A)A) Ia > Ib > Ic
(B)(B) Ia > Ic > Ib
(C)(C) Ib > Ia > Ic
a
b
c
i
iirm 2I
Physics 207: Lecture 16, Pg 27
Calculating Moment of Inertia...Calculating Moment of Inertia...
For a discrete collection of point masses we found:
For a continuous solid object we have to add up the mr2 contribution for every infinitesimal mass element dm.
An integral is required to find I :
N
iiirm
1
2I
r
dm
dmr 2I
Physics 207: Lecture 16, Pg 29
Moments of Inertia...Moments of Inertia...
Some examples of I for solid objects:
Solid sphere of mass M and radius R,
about an axis through its center.
I = 2/5 M R2
R
See Table 13.3, Moments of Inertia
Thin spherical shell of mass M and radius R, about an axis through its center.
Use the table…
R
Physics 207: Lecture 16, Pg 31
Rotation & Kinetic EnergyRotation & Kinetic Energy Consider the simple rotating system shown below.
(Assume the masses are attached to the rotation axis by massless rigid rods).
The kinetic energy of this system will be the sum of the kinetic energy of each piece:
K = ½m1v1+ ½m2v2
+ ½m3v3+ ½m4v4
rr1
rr2rr3
rr4
m4
m1
m2
m3
4
1
221 v
iiimK
Physics 207: Lecture 16, Pg 32
Rotation & Kinetic EnergyRotation & Kinetic Energy
Notice that v1 = r1 , v2 = r2 , v3 = r3 , v4 = r4
So we can rewrite the summation:
We recognize the quantity, moment of inertia or I, and
write:
rr1
rr2rr3
rr4
m4
m1
m2
m3
24
1
221
4
1
2221
4
1
221 ][ rrv
iii
iii
iii mmmK
221 IK
Physics 207: Lecture 16, Pg 33
Lecture 16, Lecture 16, Exercise 2Exercise 2Rotational Kinetic EnergyRotational Kinetic Energy
A. ¼
B. ½
C. 1
D. 2
E. 4
We have two balls of the same mass. Ball 1 is attached to a 0.1 m long rope. It spins around at 2 revolutions per second. Ball 2 is on a 0.2 m long rope. It spins around at 2 revolutions per second.
What is the ratio of the kinetic energy
of Ball 2 to that of Ball 1 ?
221 IK
i
iirm 2I
Ball 1 Ball 2
Physics 207: Lecture 16, Pg 34
Lecture 16, Lecture 16, Exercise 2Exercise 2Rotational Kinetic EnergyRotational Kinetic Energy
K2/K1 = ½ m r22 / ½ m r1
2 = 2 / 2 = 4
What is the ratio of the kinetic energy of Ball 2 to that of Ball 1 ?
(A) 1/4 (B) 1/2 (C) 1 (D) 2 (E) 4
Ball 1 Ball 2
Physics 207: Lecture 16, Pg 35
Rotation & Kinetic Energy...Rotation & Kinetic Energy...
The kinetic energy of a rotating system looks similar to that of a point particle:
Point ParticlePoint Particle Rotating System Rotating System
i
iirm 2I
221 IK2
21 vmK
v is “linear” velocity
m is the mass.
is angular velocity
I is the moment of inertia
about the rotation axis.
Physics 207: Lecture 16, Pg 36
Moment of Inertia and Rotational EnergyMoment of Inertia and Rotational Energy
Notice that the moment of inertia I depends on the distribution of mass in the system. The further the mass is from the rotation axis, the
bigger the moment of inertia.
For a given object, the moment of inertia depends on where we choose the rotation axis (unlike the center of mass).
In rotational dynamics, the moment of inertia I appears in the same way that mass m does in linear dynamics !
221 IK
iiirm 2I So where
Physics 207: Lecture 16, Pg 38
Work & Kinetic Energy:Work & Kinetic Energy:
Recall the Work Kinetic-Energy Theorem: K = WNET
This is true in general, and hence applies to rotational motion as well as linear motion.
So for an object that rotates about a fixed axis:
NET22
21 I WK if
Physics 207: Lecture 16, Pg 45
Connection with CM motionConnection with CM motion
If an object of mass M is moving linearly at velocity VCM without rotating then its kinetic energy is
CM21
RK I
2CM2
12CM2
1 VK MI What if the object is both moving linearly and rotating?
If an object of moment of inertia ICM is rotating in place about its center of mass at angular velocity then its kinetic energy is
2CM2
1T VK M
Physics 207: Lecture 16, Pg 59
Angular Momentum:Angular Momentum:
0p
dt
d
F
We have shown that for a system of particles,
momentum is conserved if
What is the rotational equivalent of this?
angular momentum
is conserved if
vm
p
IL
0dtLd
Physics 207: Lecture 16, Pg 60
Example: Two DisksExample: 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 207: Lecture 16, Pg 61
Example: Two DisksExample: Two Disks A disk of mass M and radius R rotates around the z axis
with initial angular velocity 0. A second identical disk, at rest, 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
No External Torque so Lz is constant
Li = Lf I i i = I f ½ mR2 0 = ½ 2mR2 f
Physics 207: Lecture 16, Pg 62
Lecture 16, Oct. 29Lecture 16, Oct. 29
Assignment: Assignment: Wednesday is an exam review session, Exam will be held Wednesday is an exam review session, Exam will be held
in rooms B102 &in rooms B102 & B130 in Van Vleck at 7:15 PM MP Homework 7, Ch. 11, 5 problems, MP Homework 7, Ch. 11, 5 problems,
NOTE: Due Wednesday at 4 PMNOTE: Due Wednesday at 4 PM MP Homework 7A, Ch. 13, 5 problems, available soonMP Homework 7A, Ch. 13, 5 problems, available soon
Physics 207: Lecture 16, Pg 63
Angular Momentum:Angular Momentum:Definitions & DerivationsDefinitions & Derivations
We have shown that for a system of particles
Momentum is conserved if
What is the rotational equivalent of this?
dtdp
F
EXT 0EXT F
prL
Fr The rotational analog of force FF is torque
Define the rotational analog of momentum pp to be
angular momentum, L or
p = mv
Physics 207: Lecture 16, Pg 64
Recall from Chapter 9: Linear MomentumRecall from Chapter 9: Linear Momentum
Newton’s 2nd Law:
FF = maa
)v(v
mdtd
dtd
m
Units of linear momentum are kg m/s.
pp ≡ mvv (pp is a vector since vv is a vector)
So px = mvx etc.
Definition: Definition: For a single particle, the momentum pp is defined as:
dt
dp
F
Physics 207: Lecture 16, Pg 65
Linear Momentum and Angular MomentumLinear Momentum and Angular Momentum
Newton’s 2nd Law:
FF = maa
)v(v
mdtd
dtd
m
Units of angular momentum are kg m2/s.
So from:So from:
dt
dp
F
vp
m
IL
dt
dL
Frdt
d
L
Physics 207: Lecture 16, Pg 66
Putting it all togetherPutting it all together
dtdLEXT EXTEXT Fr prL
EXTddt
L0
In the absence of external torquesIn the absence of external torques
Total angular momentum is conservedTotal angular momentum is conserved
dtFrd L
prdtFr
L
Frdtd
L
pr
L
Physics 207: Lecture 16, Pg 67
Angular momentum of a rigid bodyAngular momentum of a rigid bodyabout a fixed axis:about a fixed axis:
kv i
iiii
iiiii
i rmm vrprLz
Consider a rigid distribution of point particles rotating in the x-y plane around the z axis, as shown below. The total angular momentum around the origin is the sum of the angular momentum of each particle:
i rr1
rr3
rr2
m2
m3
j
m1
vv2
vv1
vv3
We see that LL is in the z direction.
Using vi = ri , we get
IL
(since ri , vi , are perpendicular)
k L 2
z i
iirm
Physics 207: Lecture 16, Pg 68
Example: Two DisksExample: 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 207: Lecture 16, Pg 69
Example: Two DisksExample: Two Disks A disk of mass M and radius R rotates around the z axis
with initial angular velocity 0. A second identical disk, at rest, 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
No External Torque so Lz is constant
Li = Lf I i i = I f ½ mR2 0 = ½ 2mR2 f
Physics 207: Lecture 16, Pg 70
Demonstration:Demonstration:Conservation of Angular MomentumConservation of Angular Momentum
Figure Skating :
A
z
B
z
Arm Arm
IA IB
A B
LA = LB
No External Torque so Lz is constant even if internal work done.
Physics 207: Lecture 16, Pg 71
Demonstration:Demonstration:Conservation of Angular MomentumConservation of Angular Momentum
Figure Skating :
A
z
B
z
Arm Arm
IA < IB
A > B
½ IAA2
> ½ IB B2 (work needs to be done)
IAA= LA = LB = IBB
No External Torque so Lz is constant even if internal work done.
Physics 207: Lecture 16, Pg 72
Angular Momentum ConservationAngular Momentum Conservation
A freely moving particle has a well defined angular momentum about any given axis.
If no torques are acting on the particle, its angular momentum remains constant (i.e., will be conserved).
In the example below, the direction of LL is along the z axis, and its magnitude is given by LZ = pd = mvd.
y
x
vv
dm
Physics 207: Lecture 16, Pg 73
Example: Bullet hitting stickExample: Bullet hitting stick
A uniform stick of mass M and length D is pivoted at the center. A bullet of mass m is shot through the stick at a point halfway between the pivot and the end. The initial speed of the bullet is v1, and the final speed is v2.
What is the angular speed F of the stick after the collision? (Ignore gravity)
v1 v2
MF
before after
mD
D/4
Physics 207: Lecture 16, Pg 74
Example: Bullet hitting stickExample: Bullet hitting stick
What is the angular speed F of the stick after the collision? (Ignore gravity).
Process: (1) Define system (2) Identify Conditions
(1) System: bullet and stick (No Ext. torque, L is constant)
(2) Momentum is conserved (Istick = I = MD2/12 )
Linit = Lbullet + Lstick = m v1 D/4 + 0 = Lfinal = m v2 D/4 + I f
v1 v2
MF
before after
mD
D/4
Physics 207: Lecture 16, Pg 75
Example: Throwing ball from stoolExample: Throwing ball from stool A student sits on a stool, initially at rest, but which is
free to rotate. The moment of inertia of the student plus the stool is I. They throw a heavy ball of mass M with speed v such that its velocity vector moves a distance d from the axis of rotation. What is the angular speed F of the student-stool
system after they throw the ball ?
Top view: before after
d
vM
I I
F
Physics 207: Lecture 16, Pg 76
Example: Throwing ball from stoolExample: Throwing ball from stool
What is the angular speed F of the student-stool system after they throw the ball ?
Process: (1) Define system (2) Identify Conditions
(1) System: student, stool and ball (No Ext. torque, L is constant)
(2) Momentum is conserved
Linit = 0 = Lfinal = m v d + I f
Top view: before after
d
vM
I I
F
Physics 207: Lecture 16, Pg 77
An example: Neutron Star rotationAn example: Neutron Star rotation
period of pulsar is 1.187911164 s
Neutron star with a mass of 1.5 solar masses has a diameter of ~11 km.
Our sun rotates about once every 37 days
f / i = Ii / If = ri2 / rf
2 = (7x105 km)2/(11 km)2 = 4 x 109
gives millisecond periods!
Physics 207: Lecture 16, Pg 78
Angular Momentum as a Fundamental QuantityAngular Momentum as a Fundamental Quantity
The concept of angular momentum is also valid on a submicroscopic scale
Angular momentum has been used in the development of modern theories of atomic, molecular and nuclear physics
In these systems, the angular momentum has been found to be a fundamental quantity Fundamental here means that it is an intrinsic
property of these objects
Physics 207: Lecture 16, Pg 79
Fundamental Angular MomentumFundamental Angular Momentum
Angular momentum has discrete values These discrete values are multiples of a fundamental
unit of angular momentum The fundamental unit of angular momentum is h-bar
Where h is called Planck’s constant
s
m kg10054.1
234
h
),...3,2,1( nnL
Physics 207: Lecture 16, Pg 80
Intrinsic Angular MomentumIntrinsic Angular Momentum
photon
Physics 207: Lecture 16, Pg 81
Angular Momentum of a MoleculeAngular Momentum of a Molecule
Consider the molecule as a rigid rotor, with the two atoms separated by a fixed distance
The rotation occurs about the center of mass in the plane of the page with a speed of
Physics 207: Lecture 16, Pg 82
Angular Momentum of a Molecule Angular Momentum of a Molecule (It heats the water in a microwave over)(It heats the water in a microwave over)
L
CM/ IE = h2/(82I) [ J (J+1) ] J = 0, 1, 2, ….
Physics 207: Lecture 16, Pg 83
Center of Mass Example: Astronauts & RopeCenter of Mass Example: Astronauts & Rope
Two astronauts are initially at rest in outer space and 20 meters apart. The one on the right has 1.5 times the mass of the other (as shown). The 1.5 m astronaut wants to get back to the ship but his jet pack is broken. There happens to be a rope connected between the two. The heavier astronaut starts pulling in the rope.
(1) Does he/she get back to the ship ?
(2) Does he/she meet the other astronaut ?
M = 1.5mm
Physics 207: Lecture 16, Pg 84
Example: Astronauts & RopeExample: Astronauts & Rope(1) There is no external force so if the larger astronaut
pulls on the rope he will create an impulse that accelerates him/her to the left and the small astronaut to the right. The larger one’s velocity will be less than the smaller one’s so he/she doesn’t let go of the rope they will either collide (elastically or inelastically) and thus never make it.
M = 1.5mm
Physics 207: Lecture 16, Pg 85
Example: Astronauts & RopeExample: Astronauts & Rope
(2) However if the larger astronaut lets go of the rope he will get to the ship. (Too bad for the smaller astronaut!)
In all cases the center of mass will remain fixed because they are initially at rest and there is no external force.
To find the position where they meet all we need do is find the Center of Mass
M = 1.5mm