Chapter 12: Rotation of Rigid Bodies - Santa Rosa …lwillia2/40/40ch12_s14.pdf · Chapter 12:...
Transcript of Chapter 12: Rotation of Rigid Bodies - Santa Rosa …lwillia2/40/40ch12_s14.pdf · Chapter 12:...
Chapter 12: Rotation of Rigid Bodies
Center of Mass
Moment of Inertia
Torque
Angular Momentum
Rolling
Statics
Translational vs Rotational
2
/
/
1/ 2
m
x
v dx dt
a dv dt
F ma
p mv
KE mv
Work Fd
P Fv
2
/
/
1/ 2
I
d dt
d dt
I
L I
KE I
Work
P
2
c
t
s r
v r
a r
a r
Fr
L pr
Connection
More…
Center of Mass The geometric ‘center’ or average location of the mass.
Rotational & Translational Motion Objects rotate about their Center of Mass.
The Center of Mass Translates as if it were a point particle.
CMCM
rv
d
dt
The Center of Mass Translates as if it were a point particle
and, if no external forces act on the system, momentum is
then conserved. This means:
EVEN if the bat EXPLODED into a thousand pieces, all the
pieces would move so that the momentum of the CM is
conserved – that is, the CM continues in the parabolic
trajectory!!!! THIS IS VERY VERY IMPORTANT!
Center of Mass
System of Particles Center of Mass
• A projectile is fired into the air and suddenly explodes
• With no explosion, the projectile would follow the dotted line
• After the explosion, the center of mass of the fragments still follows the dotted line, the same parabolic path the projectile would have followed with no explosion!
If no external forces act on the system, then the velocity of the CM doesn’t change!!
Center of Mass: Stability
If the Center of Mass is above the base
of support the object will be stable. If
not, it topples over.
This dancer balances en pointe by having her center of mass directly over her toes, her base of support.
Balance and Stability
Slide 12-88
Center of Mass
CM i i
i
m x
xM
The geometric ‘center’ or average location of the mass.
System of Particles: Extended Body:
CM
1dm
M r r
CM Lecture Problem
1. Identical particles are placed at the 50-cm and 80-cm
marks on a meter stick of negligible mass. This rigid
body is then mounted so as to rotate freely about a
pivot at the 0-cm mark on the meter stick. A) What is
the CM of the system? B) What is the torque acting on
the system? C) What is the rotational inertia of the
system about the end? D) If this body is released from
rest in a horizontal position, what is the angular
acceleration at the release? E) What is the angular
speed of the meter stick as it swings through its lowest
position?
Prelab
A meter stick has a mass of 75.0 grams and has two masses attached to it – 50.0 grams at the 20.0cm mark and 100.0 grams at the 75.0 cm mark.
(a) Find the center of mass of the system - that is, at what mark on the meter stick should the fulcrum be placed so that the system balances?
(b) A fulcrum is then placed at the CM. Sketch and label the system.
(c) Show that the net torque about the cm is zero.
(d) Calculate the rotational inertia of the system about the CM axis. Box answers, make it neat.
Center of Mass The geometric ‘center’ or average location of the mass.
Extended Body:
CM
1dm
M r r
Center of Mass of a Solid Object
Divide a solid object into
many small cells of mass m.
As m 0 and is replaced
by dm, the sums become
Before these can be integrated:
dm must be replaced by
expressions using dx and dy.
Integration limits must be established.
Slide 12-41
Example 12.2 The Center of Mass of a Rod
Slide 12-42
Example 12.2 The Center of Mass of a Rod
Slide 12-43
Extended Body:
Example
CM
1dm
M r r
You must generate an expression for the density and the mass differential, dm, from geometry and by analyzing a strip of the sign. We assume the sign has uniform density. If M is the total mass then the total volume and density is given by:
2( )
1
2
M Mydm dV ytdx ytdx dx
ababt
1,
12
2
MV abt
abt
Where a, b and t are the width, height and thickness of the sign, respectively. Then the mass element for the strip shown is:
CM
1 x x dm
M
3
2
0 0
1 2 2 2 2( )
3 3
aaMy b x
x dx x x dx aM ab ab a a
Newton’s 2nd Law for Rotation
I
The net external torques acting on an object
around an axis is equal to the rotational
inertia times the angular acceleration.
The rotational equation is limited to rotation about a single
principal axis, which in simple cases is an axis of symmetry.
Inertia thing
Acceleration thing
Force thing
Torque: Causes Rotations
The moment arm, d, is
the perpendicular
distance from the axis
of rotation to a line
drawn along the
direction of the force
sinFr Fd
lever arm: sind r
The horizontal component of F (F cos ) has no tendency to produce a rotation
Torque: Causes Rotations
sinFr Fd
The direction convention is: Counterclockwise rotations are positive. Clockwise rotations are negative.
If the sum of the torques is zero, the system
is in rotational equilibrium.
Newton’s 1st Law for Rotation
250 3 750girl N m Nm
500 1.5 750boy N m Nm
0
Torque
Is there a difference in torque? (Ignore the mass of the rope)
NO!
In either case, the lever arm is the same!
What is it? 3m
CM Lecture Problem
1. Identical particles are placed at the 50-cm and 80-cm
marks on a meter stick of negligible mass. This rigid
body is then mounted so as to rotate freely about a
pivot at the 0-cm mark on the meter stick. A) What is
the CM of the system? B) What is the torque acting
on the system? C) What is the rotational inertia of the
system about the end? D) If this body is released from
rest in a horizontal position, what is the angular
acceleration at the release? E) What is the angular
speed of the meter stick as it swings through its lowest
position?
Newton’s 2nd Law for Rotation
I
The net external torques acting on an object
around an axis is equal to the rotational
inertia times the angular acceleration.
The rotational equation is limited to rotation about a single
principal axis, which in simple cases is an axis of symmetry.
Inertia thing
Acceleration thing
Force thing
Torque is a Vector!
= r x F
The direction is given by the right hand rule where the fingers
extend along r and fold into F. The Thumb gives the direction of .
The Vector Product
• The magnitude of C is
AB sin and is equal to the
area of the parallelogram
formed by A and B
• The direction of C is
perpendicular to the plane
formed by A and B
• The best way to determine
this direction is to use the
right-hand rule
ˆ ˆ ˆy z z y x z z x x y y xA B A B A B A B A B A B A B i j k
COMPARE!
CROSS PRODUCT
F and d must be
mutually perpendicular! sin
r F
Fr Fd
sin
L r p
L mvr
cosW F d Fd
DOT PRODUCT
F and d must be
mutually PARALLEL!
CROSS PRODUCT
L and p must be
mutually perpendicular!
Two vectors lying in the xy plane are
given by the equations A = 5i + 2j and
B = 2i – 3j. The value of AxB is
a. 19k
b. –11k
c. –19k
d. 11k
e. 10i – j
ˆ ˆ ˆy z z y x z z x x y y xA B A B A B A B A B A B A B i j k
CM Lecture Problem
1. Identical particles are placed at the 50-cm and 80-cm
marks on a meter stick of negligible mass. This rigid
body is then mounted so as to rotate freely about a
pivot at the 0-cm mark on the meter stick. A) What is
the CM of the system? B) What is the torque acting
on the system? C) What is the rotational inertia of the
system about the end? D) If this body is released from
rest in a horizontal position, what is the angular
acceleration at the release? E) What is the angular
speed of the meter stick as it swings through its lowest
position?
Net external torques
Find the Net Torque
( 20 )(.5 ) (35 cos30)(1.10 ) N m N m
1 1 2 2 Fd F d
( 20 )(.5 ) (35 )(1.10 sin60) N m N m
OR
23.3 CCW Nm
F and d must be
mutually
perpendicular!
sinFr Fd
Newton’s 2nd Law for Rotation
I
The net external torques acting on an object
around an axis is equal to the rotational
inertia times the angular acceleration.
The rotational equation is limited to rotation about a single
principal axis, which in simple cases is an axis of symmetry.
Inertia thing
Acceleration thing
Force thing
NEW: Rotational Inertia
The resistance of an object to rotate. The further away the mass is from the axis of rotation, the
greater the rotational inertia.
Rotational Inertia
2
i iI m r 2I r dm
System of Particles: Extended Body:
The resistance of an object to rotate.
Moments of Inertia of Various
Rigid Objects
Rotational Inertia
Depends on the axis.
For a system of point particles:
Where r is the distance to the axis of rotation.
2 i i iI m r I
Calculate Rotational Inertia
SI units are kg.m2
CM Lecture Problem
1. Identical particles are placed at the 50-cm and 80-cm
marks on a meter stick of negligible mass. This rigid
body is then mounted so as to rotate freely about a
pivot at the 0-cm mark on the meter stick. A) What is
the CM of the system? B) What is the torque acting on
the system? C) What is the rotational inertia of the
system about the end? D) If this body is released
from rest in a horizontal position, what is the angular
acceleration at the release? E) What is the angular
speed of the meter stick as it swings through its lowest
position?
2I r dm
Moment of Inertia of a Rigid
Object: The Rod
Calculate: Moment of
Inertia of a Rigid Object
22 I r dm R dm
The trick is to write dm in terms of the density:
dm dV
2 2 4
0
12
2
R
I r dm r Lr dr LR
2 dV rL drDivide the cylinder into concentric shells with
radius r, thickness dr and length L:
21
2zI MR2But /M R L
Newton’s 2nd Law for Rotation
I
The net external torques acting on an object
around an axis is equal to the rotational
inertia times the angular acceleration.
The rotational equation is limited to rotation about a single
principal axis, which in simple cases is an axis of symmetry.
Inertia thing
Acceleration
thing Force thing
Tangential and Angular
Acceleration
d
dtt
dva
dt
( )
d r
dt
ta r
d
rdt
II
2
0
1 2 2
2 /t t t
I
2
0
1:
2Use t t
A 50 N m torque acts on a wheel of moment of inertia 150 kg m2.
If the wheel starts from rest, how long will it take the wheel to
make a quarter turn (90 degrees)?
2
2 / 23.1
50 /150
radt s
N m kg m
24.3
.314
m kg
R m
) ?
) ?
a
b
1 2
( 90 125 ).314
11
F r F r
N N m
Nm
2
=1
mR2
I
2
11=
1/2 24.3kg(.314m)
Nm
2= 9.2 /rad s
CM Lecture Problem
1. Identical particles are placed at the 50-cm and 80-cm
marks on a meter stick of negligible mass. This rigid
body is then mounted so as to rotate freely about a
pivot at the 0-cm mark on the meter stick. A) What is
the CM of the system? B) What is the torque acting on
the system? C) What is the rotational inertia of the
system about the end? D) If this body is released
from rest in a horizontal position, what is the
angular acceleration at the release? E) What is the
angular speed of the meter stick as it swings through its
lowest position?
Superposition of Inertia
The Parallel Axis Thm
The theorem states I = ICM + MD 2 I is about any axis parallel to the axis through the center of mass of the object ICM is about the axis through the center of mass D is the distance from the center of mass axis to the arbitrary axis
iI ISuperposition: Inertia ADD
Moment of Inertia for a Rod Rotating
Around One End
The moment of inertia of the
rod about its center is
The position of the CM is
D=½ L
Therefore,
21
12CMI ML
2
CM
2
2 21 1
12 2 3
I I MD
LI ML M ML
It is easier to rotate a rod about its center than about an end.
Rotational Inertia: Parallel Axis Theorem
A uniform rod (mass = 2.0 kg, length = 0.60 m) is free to
rotate about a frictionless pivot at one end. The rod is
released from rest in the horizontal position. What is the
magnitude of the angular acceleration of the rod at the
instant it is 60° below the horizontal?
a. 15 rad/s2
b. 12 rad/s2
c. 18 rad/s2
d. 29 rad/s2
e. 23 rad/s2
© 2013 Pearson Education, Inc.
Rotational Energy
A rotating object has kinetic energy because all particles in the object are in motion.
The kinetic energy due to rotation is called rotational kinetic energy.
Adding up the individual kinetic energies, and using vi = ri :
Slide 12-47
© 2013 Pearson Education, Inc.
Define the object’s moment of inertia:
The units of moment of inertia are kg m2. Moment of inertia depends on the axis of
rotation. Mass farther from the rotation axis contributes more
to the moment of inertia than mass nearer the axis. This is not a new form of energy, merely the familiar
kinetic energy of motion written in a new way.
Then the rotational kinetic energy is simply
Rotational Energy
Slide 12-48
CM Lecture Problem
1. Identical particles are placed at the 50-cm and 80-cm
marks on a meter stick of negligible mass. This rigid
body is then mounted so as to rotate freely about a
pivot at the 0-cm mark on the meter stick. A) What is
the CM of the system? B) What is the torque acting on
the system? C) What is the rotational inertia of the
system about the end? D) If this body is released from
rest in a horizontal position, what is the angular
acceleration at the release? E) What is the angular
speed of the meter stick as it swings through its
lowest position?
Rotational Inertia
Which reaches the bottom first?
(Same mass and radius)
Why Solid Cylinder?
2CMI mr
21
2CMI mr
PROVE IT!
Rolling
The red curve shows the path moved by a point on the rim of the object. This path is called a cycloid
The green line shows the path of the center of mass of the object which moves in linear motion.
Fig. 10.28, p.317
All points on a rolling object move in a direction perpendicular to an axis through the instantaneous point of contact P. In other words, all points rotate about P. The center of mass of the object moves with a velocity vCM, and the point P ’ moves with a velocity 2vCM. (Why 2?)
Rolling without Slipping
For pure rolling motion, (no slipping) as the cylinder rotates through an angle , its center moves a linear distance s = R with speed vCM. At any instant, the point on the rim located at point P is at rest relative to the surface since no slipping occurs.
Fig. 10.29, p.318
The motion of a rolling object can be modeled as a combination of pure translation and pure rotation.
Translation: CM
moves with vCM.
Rotation: All points
rotate about P with
angular speed .
Rolling Without Slipping
Friction
If an object rolls without slipping
then the frictional force that causes
the rotation is a static force.
If no slipping occurs, the point of
contact is momentarily at rest and
thus friction is static and does no
work on the object. No energy is
dissipated and Mechanical Energy is
Conserved. If the object slips, then
friction is not static, does work on
the object and dissipates energy.
21
2 PK I
2 21 1
2 2 CM CMK I mv
Rolling Without Slipping
Kinetic Energy
m 2
CMUse Parallel-axis Th : PI I MR
2 2
CM
1( )
2 K I MR
The total kinetic energy of a rolling object is the sum of the rotational
kinetic energy about the center of mass and the translational kinetic
energy of the center of mass.
2 21 1
2 2 CM CMMgh Mv I
i fE E
2 21 1( )
2 2 CM
CM CM
vMv I
R
Rolling Without Slipping
Conservation of Energy
2 2 21 1 2( )( )
2 2 5 CM
CM
vMv MR
R
2 21 1
2 5 CM CMMv Mv 27
10 CMMv
10
7CMv gh
21
2Mgh Mv
i fE E
Rolling Without Slipping
Conservation of Energy
10
7CMv gh
Compare to Slipping Only:
2CMv gh
Some of the gravitational
potential energy goes into
rolling the sphere so the
translational velocity of the
cm is less when rolling!
What about little ball big ball?
21
2Mgh Mv
i fE E
Rolling Without Slipping
Conservation of Energy
10
7CMv gh
Compare to Slipping Only:
2CMv gh
What about the
acceleration down the
incline?
Rolling down:
Does change while in flight?
Rolling Down the Incline
Rotational Inertia
Which reaches the bottom first?
(Same mass and radius)
Why Solid Cylinder?
2CMI mr
21
2CMI mr
PROVE IT!
Total Energy
2 2
0
1 1
2 2mgh mv I
trans rotE KE KE PE
0
2
2
/
mghv
m I r
2 21 1( )
2 2
vmv I
r
2
2
/
mghv
m I r
0v gh
0
4
3v gh
General equation
for the total final
velocity at the
end of the ramp:
Solid Disk has the
greatest velocity at the
bottom of the ramp!
Note: the velocity is
independent of the radius!
Total Kinetic Energy A 1.0-kg wheel in the form of a
solid disk rolls without slipping
along a horizontal surface with a
speed of 6.0 m/s. What is the total
kinetic energy of the wheel?
2 21 1
2 2mv I total trans rotK K K
2 2 21 1 1( )( )
2 2 2
vmv mr
r
2 21 1
2 4mv mv
23(1 )(6 / )
4kg m s23
4mv 27J
What I to use?
A tennis ball starts from rest and
rolls without slipping down the hill.
Treat the ball as a solid sphere. Find
the range x.
Rolling Ball Problem
2 2 21 1 2( )( )
2 2 5
vmv Mr
r
2 21 1
2 2 CM CMmgh mv I
10
7v gh
cosxR v t v t
210 sin
2v t gt
2 sinvt
g
cosxR v t v t 2 2sin cosv
g
(10 / 7 )2sin cosgh
g
10 sin 2
7
hR
g
= .25m
21
2y yy v t a t
Does change while in flight?
Angular Momentum
L I
finalL I
If no NET external Torques act on a
system then
Angular Momentum is Conserved.
IinitialL
Angular Momentum
L I
Angular Momentum
IL
A Skater spins with an angular speed of 2 rev/s. If she
brings her arms in and decreases her rotational inertia by
a factor of 5, what is her new angular speed in rev/s?
f f fL I 0 0 0L I
0 0 0 f f fL LI I
00f
f
I
I
2 // 5
Irev s
I
10 /rev s
Angular Momentum for a
Point Particle
r v L I
2 vmr mvr
r
L mvr
Single mass, a distance r from the axis of rotation.
68.37 10
625.1 10
8450 /
?
r x mP
r x mA
v m sP
vA
What is the velocity
of the satellite at apogee?
68.37 10
625.1 10
8450 /
?
r x mP
r x mA
v m sP
vA
Is the angular momentum of the system
CONSERVED?
A PL L
68.37 10
625.1 10
8450 /
?
r x mP
r x mA
v m sP
vA
A A P Pmr v mr v
PA P
A
rv v
r /2820m s
Angular Momentum is a Vector
Angular Speed is a Vector
When a rigid object rotates about an axis, the angular
momentum L is in the same direction as the angular velocity ,
according to the expression L = I, both directions given by the
RIGHT HAND RULE.
Angular Momentum is a Vector
The angular momentum L of a particle of mass m and linear momentum p located at
the vector position r is a vector given by L = r p. The value of L depends on the
origin about which it is measured and is a vector perpendicular to both r and p.
Only the perpendicular
component of p
contributes to L.
sin
L r p
L mvr
Torque Changes Angular Momentum
ext
L
t
S and L must be measured about the same origin
This is valid for any origin fixed in an inertial frame
Sext= dL/dt
Looks like
SFext= dp/dt
A particle whose mass is 2 kg moves in
the xy plane with a constant speed of
3 m/s along the direction r = i + j. What is
its angular momentum (in kg · m2/s)
relative to the origin?
a. 0 k
b. 6k
c. –6k
d. 6 k
e. –6 k
p
= 0
Only the
perpendicular
component of p
contributes to L!
sinL mvr
Torque Changes Angular Momentum
ext
L
t
The torque is perpendicular to both the applied force and the
lever arm, but parallel to the angular speed, angular acceleration
and angular momentum.
S and L must be measured about the same origin
This is valid for any origin fixed in an inertial frame
Sext= dL/dt
= r x F = dL/dt
COMPARE!
CROSS PRODUCT
F and d must be
mutually perpendicular! sin
r F
Fr Fd
sin
L r p
L mvr
cosW F d Fd
DOT PRODUCT
F and d must be
mutually PARALLEL!
CROSS PRODUCT
L and p must be
mutually perpendicular!
Colliding Probems
sin
L r p
L mvr
A particle of mass m = 0.10 kg and speed v0 = 5.0
m/s collides and sticks to the end of a uniform solid
cylinder of mass M = 1.0 kg and radius R = 20 cm.
If the cylinder is initially at rest and is pivoted about
a frictionless axle through its center, what is the
final angular velocity (in rad/s) of the system after
the collision?
a. 8.1
b. 2.0
c. 6.1
d. 4.2
e. 10
sin
L r p
L mvr
System of Particles Center of Mass
tot CMp vM
CMCM
( / )
v
rv
i i i i
i i
d m r M md
dt dt M
CM toti i i
i i
M m v v p p
tot,final tot,initial CM Mp p v
CM
1 v vi i
i
mM
CM tot
1
Mv p
Note: If the velocity of the CM is zero, the total momentum is zero!
If the sum of the net external torques is zero,
the system is in rotational equilibrium.
Chapter 12: Statics
Newton’s 1st Law: Conditions for Equilibrium
0
If the sum of the net external forces is zero,
the system is in translational equilibrium.
SF = 0
The ladder is 8m long and
weighs 355 N. The weight of the
firefighter is 875N and he stands
2.30m from the center of mass
of the ladder. If the wall is
frictionless, find the minimum
coefficient of friction of the
floor so that the ladder doesn’t
slip.
The Ladder Problem
0F 0
What Keeps a Spinning Gyroscope
from Falling Down?
The “Couple” are
equal and opposite
forces so
SF = 0.
Since the normal
force is through the
axis of rotation P
(the support) it
produces no torque.
Only the weight of
the CM produces a
torque with lever
arm r that changes
the angular
momentum, L.
Precession Keeps it from Falling!
L
t
r L is in the direction of !
This causes the Precession!
P
Spinning Wheel What happens if you rotate the wheel while sitting in a spin stool?
The stool will spin in the direction of the torque.
Conservation of Angular Momentum!
The Gyroscopic Effect Once you spin a gyroscope, its axle wants to keep pointing
in the same direction. If you mount the gyroscope in a set
of gimbals so that it can continue pointing in the same
direction, it will. This is the basis of the gyro-compass
which is used in navigation systems.
What is the best gyroscope on
Earth?
Earth’s Precession
The Earth’s
precession is due to
the gravitational tidal
forces of the Moon
and Sun applying
torque as they attempt
to pull the Equatorial
Bulge into the plane
of Earth’s orbit.
The Earth’s
Precession causes the
position of the North
Pole to change over a
period of 26,000
years.
TOTAL Angular Momentum Orbital and Spin Angular Momentum
The TOTAL angular momentum includes both the angular
momentum due to revolutions (orbits) and rotations (spin).
It is the TOTAL angular momentum that is conserved.
Earth- Moon System:
Total Angular Momentum is Conserved!
•Earth Rotation Slowing due to friction of ocean on bottom
•.0023 s per century: 900 Million yrs ago, Earth day was 18 hrs!
•Decrease of Earth’s spin angular momentum, increases the
orbital angular momentum of the Moon by increasing the
distance, r, in order to keep L conserved!
•Earth is slowing down and Moon is moving further away!
Bouncing laser
beams off the
Moon
demonstrates
that it slowly
moving away
from the Earth
~.25 cm/month
Angular 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
– It is a part of their nature
Orbital Angular Momentum
& its Z-component are Quantized
h( 1) ( 0,1,2,... 1)
2L l l l n
h ( ,.. 1,0,1,2,... )
2Z l lL m m l l
Angular momentum magnitude is quantized:
Angular momentum direction is quantized:
Note: for n = 1, l = 0. This means that the ground
state angular momentum in hydrogen is zero, not
h/2, as Bohr assumed. What does it mean for
L = 0 in this model?? Standing Wave
Intrinsic Spin is Quantized
h 1( 1) ( )
2 2S s s s
h 1 1 ( , )
2 2 2Z s sS m m
Fine Line Splitting: B field due to intrinsic spin interacts with B
field due to orbital motion and produces additional energy states.
Spin magnitude is quantized:
Spin direction is quantized:
Cosmic Rotations: Conservation
of Angular Momentum
Does the Universe have NET
Angular Momentum? According the General Relativity,
The Universe is ISOTROPIC – it looks the
same in every direction. Net Angular
Momentum would define a direction.
Isotropic Universe