Chapter 9-10 rotation of rigid bodies kinematics dynamics 1. model 2.Angular quantity 1. model...
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Transcript of Chapter 9-10 rotation of rigid bodies kinematics dynamics 1. model 2.Angular quantity 1. model...
Chapter 9-10 rotation of rigid bodies
kinematicskinematics dynamicsdynamics
1. model2.Angular quantity
1. model2.Angular quantity
Instantaneouseffect
Instantaneouseffect Time effectTime effectSpace effectSpace effect
theorem of rotation
theorem of rotation
1.Theorem ofangular momentum2.conservation of angular momentum
1.Theorem ofangular momentum2.conservation of angular momentum
1.work 2.kinetic and potential energy3.conservationof energy
1.work 2.kinetic and potential energy3.conservationof energy
Rotational motionRotational motion
Key terms:rigid bodyrotation translationradianrad angular velocityangular accelerationangular speedmoment of inertiarotational kinetic energyparallel-axis theorem
Chapter 9-10 rotation of rigid bodies
Key terms:torqueline of action lever armrotational analog of Newton’s second lawcombined translation and rotationangular momentumconservation of angular momentum
Chapter 9-10 rotation of rigid bodies
Types of Motion
Translation
The linear position of a body changes with respect to a fixed frame, but its angular orientation remains unchanged.
Rotation
The angular orientation of a body changes about a fixed frame of reference, but its linear position remains fixed.
Complex
Simultaneous combination of translation and rotation motions
1. Ideal Model-----Rigid Body
we neglect the deformations and assume that the body has a perfectly definite and unchanging shape and size.
Definite shape and definite size.
The distance between any two points on a rigid body remain unchanging.
A rigid body can be considered as a special collection of particles with mass continuously distributed.
A rigid body in general can have rotational as well as translational motion.
y
xo
rP
2. Angular Quantities
let’s think first about a rigid body that rotates about a fixed axis.
let’s think first about a rigid body that rotates about a fixed axis. (The axis passes through point O and is perpendicular to the x-y plane)
We can use the linear quantities ( r, v, a) as well as the angular quantities ( , , ) to describe the motion of the rigid body.
y
xo
rP
2.1 Angular Coordinate
radr
s
Counterclockwise(+)
s
12
2.2 Angular Velocity
s/raddt
d
rv
2.3 Angular Acceleration
22
2
s/raddt
d
dt
d
If is a constant (=c):t0
20 t2/1t
20
22
rs
2.4 Relating Linear and Angular Kinematics
y
xo
rP
s
v
We can use linear quantities ( r, v, a) or angular quantities ( , , ) to describe the rotational motion. The relationships between them are:
r2
2
n rr
a
rdt
dat
The linear quantities of any point on the rigid body are different, while angular quantities are the same.
3. Theorem of Rotation
Fr
3.1 torque (/moment) of a force
d
o
mi
r
F
x
y
z sinrF FdVector: right-hand rule.
axis//F,r//F,rF:discussion
force times lever-arm, tends to produce rotation.
i
iii
i FrFrFr
2111
Net torque is the vector sum of individual torques:
Example: a force on particle
find the torque on particle if the fixed axis
is in original point.
)( jyixF
Solution:
0)()1(
rrFr
0)()()2(
jyixjyixFr
o
mi
ir
3.2 Rotation Analog of Newton’s Second Law
exF
inf
The net torque on mi to point O:
Proof: A rigid body can be looked as a special collection of particles.
)ji(j
jiiiii frFr
The net torque on the rigid body to point O:
i )ji(j
jiii
iii
i frFr
0fri )ji(j
jii
z
I
i
iii
i Fr
i
iiii
iii rmramr )()( tan, ii
iii
ra
amF
tan,
tan,tan,
i
i2i )mr(
I
i )ji(j
jiii
iii
i frFr 0
)(
i jij
jii fr
i
,taniiFr
“”s are the same for every part of the rigid body.
Rotational analog of Newton’s law for a rigid body.
o
mi
ir
exF
z
i
ii mrI 2
Means external torque, the torque offered by a pair of internal force equals to zero
)1
moment of inertial, measurement of rotational inertial
)2 The angular acceleration of rigid body with fixed axis
0)3 The rigid body keep rest or rotate in uniform angular velocity
Caution:
0
1
, IGiven
I)4
I
i
2iirmI
1) A few point masses: i
2iirmI
2) Continuous distribution of mass: dmrI 2
3.3 Calculation of Moment of Inertial
Moment of inertial depends on the distribution of mass in the system.
Moment of inertia depends on the location and orientation of the axis.
Example: An engineer is designing a one-piece machine part consisting of three heavy connectors linked by light molded struts. The connectors can be considered as massive particles connected by massless rods.
a) What is the moment of inertia of this body about an axis through point A, perpendicular to the plane of the diagram?
b) What is the moment of inertia about an axis coinciding with rod BC?
(see page 277)
i
2iirmI 0.30m
0.40m
0.50m
mA=0.30kg
mB=0.10kg
mC=0.20kg
B
AC
Solution:
i
2iia rmI
a)
22 )m40.0)(kg20.0()m50.0)(kg10.0( 2m.kg057.0
i
2iib rmI 2)m40.0)(kg30.0( 2m.kg048.0
Solution:
a)
b)
0.30m
0.40m
0.50m
mA=0.30kg
mB=0.10kg
mC=0.20kg
B
AC
Example: A slender uniform rod with mass M and length L. It might be a baton held by a twirler in a marching band. Compute its moment of inertia about an axis through center of mass point O.
dmrI 2dmrI 2
dmx 2
2
L
2
L2dxx
L
M
2ML12
1
Cdx
dmx
xo
Solution:
22cm MRdmRI
Uniform thin loop (M ,R):
Uniform thin disc(M, R)
cI2r 2R
M
rdr2
2MR2
1
R
0
R
r dr
2cmp MdII
Icm: about axis through CM
M: total mass
d : distance between CM and point p.
2) Parallel-Axis Theorem: (p281)
CM
Icm
O
Ip
p d
0M
xdmxc o
M
ydmyc
2cmp MdII
dmbyaxdmrI p ])()[( 222
dmbaydmbxdmadmyxI p )(22)( 2222
o
Example: If the rod rotates about the axis through one end, Use parallel-axis theorem:
2cm MdII
22 ML4
1ML
12
1
2ML3
1
3.3 Application of rotational analog of Newton’s second law
ICaution:
1) The left is vector sum of external torque about a fixed axis, and the direction is parallel to axis.
2) “ I ” is the Moment of inertial
3) “” means the angular acceleration to fixed axis, same direction with torque
Example: In judo, a weaker and smaller fighter who understands physics can defeat a stronger and larger fighter who doesn’t. The fact is demonstrated by the basic ‘ hip throw ’, in which a fighter rotates his opponent around his hip.
IFd 1
ImgddF 21
FF
Solution:
Example: The massive shield door at a neutron test facility at Lawrence Livermore Laboratory is the world’s heaviest hinged door M=44000kg, I=8.7*104 kgm2, and face width 2.4m find the smallest force to move the door from rest through an angle of 900 in 30s.
y
O
Example: We wrap a light, flexible cable around a solid cylinder with mass M and radius R. The cylinder rotates with negligible friction about a stationary horizontal axis. We tie the free end of the cable to an object of mass m and release the object with no initial velocity. Find the acceleration of the object of mass m and the angular acceleration of the cylinder
M
R
m
h
N
MgTT
mg
Solution:
For m: )1(maTmg For M: )2()MR
2
1(TR 2
)3(Ra
Example: An Atwood’s machine. There is no slipping between the cord and the surface of the wheel. Find the linear accelerations of blocks A and B, the angular acceleration of the wheel C, and the tension in each side of the cord
a) If weight of cylinder is negligible ;b) If weight of cylinder is not negligible.
C
mAmB
R
m2gm1g
T2T1
T’2 T’
1
T
y
OSolution:
a) The weight of cylinder is negligible …
b)If weight of cylinder is not negligible:
For m1: )1(amTgm y1111
For m2: )2(amTgm y2222
For m1: )1(amTgm y1111 For m2: )2(amTgm y2222
For M: )3()MR2
1(RTRT 2'
1'2
)4(aa y2y1 )5(Ra y1
From (1),(2),(3),(4),(5) :
g2Mmm
mmaa
21
21y2y1
g2Mmm
)2Mm2(mT,g
2Mmm
)2Mm2(mT
21
122
21
211
C
m1m2
R
m2gm1g
T2T1
T’2 T’
1
T
y
O
IL
mg cos2
d
dI
dt
dI
Lmg cos
2
0
2
cos2
dIdL
mg
2
2
1)sin1(
2 I
Lmg
Solution:
Example: A grindstone 1.0 m in diameter, of mass 50 kg, is rotating at 900 rev.min-1. A tool is pressed against the rim with a normal force of 200N, and the grindstone comes to rest in 10s. Find the coefficient of friction between the tool and the grindstone. Neglect friction in the bearings.
Solution: t,I 0 ,Rfk kk Nf
s10t,0
)s/rad(3060/2900
t
0
)s/rad(3t/)( 20t
2MR2
1I 59.ok
Example: A wheel rotates about a fixed axis from rest. A constant torque,=20N.m, acts on it for 10 seconds. =100rev/min at t=10s. Then it takes 100 seconds for the wheel to come to stop again. Find the moment of inertial of the wheel with respect to the axis.
Solution: =I, = o+ t
20=I 1 , 1= /t1 (for: o=0) 20- r=I1 , 1= /t1 (for: o=0) (1)
When release external torque: - r=I2 , 2=- /t2 (2)
t1=10s , t2=100s , =(100×2)/60=10.5rad/s,
From (1) 、 (2): I=17.3kg.m2 。
acted by external torque:
r
iv
Pmi
4. Energy in Rotational Motion
O
The kinetic energy of mi :
2iii vm
2
1K 22
iirm2
1 iii rv
The rigid body : 2
i
2ii
i
2ii
ii )rm(
2
1vm
2
1K
i
iirmI 2
2I2
1K
4.1 Rotational Kinetic Energy
Example: Each of the helicopter rotor blades is 5.2m long, m=240kg. The rotor is rotating at 350rev/min.
Find 1) the rotational inertial of the rotor assemble about the axis of rotation.
2)what is the total kinetic energy of rotation.
4.2 Gravitational Potential Energy
i
ii gymU i
cmii MyymcmMgy
The gravitational potential energy is the same as though all the mass were concentrated at the center of mass of the rigid body.
4.3 Work and Power in Rotational Motion
sdFdW
2
1
dW
Work of the torque
Suppose a force F acts on the rigid body. The rigid body rotates through an infinitesimal angle d about a fixed axis. The work dW done by the force while point P moves distance ds is:
F
dsr
d
P
dsinFr d
work done by a torque
2
1
dW
CAUTION:
2
1
ddt
dI
dt
d
I
21
22 I
2
1I
2
1dI
2
1
2
1
dW
a work done by the forces doesn’t equal to the work done by the torques.
21
22 2
1
2
1 mmrdFW
For a particle:
dt
d
dt
dWP
Power of a torque:
Compare with the work-energy theorem for collection of particles:
21
22 2
1
2
1mvmvWtot
21
22 2
1
2
1 IIWtot
vFp
4.4 Conservation of Mechanical Energy
If the work done by all forces other than conservative force equals zero. The total mechanical energy of the rigid body is conservative.
CEWW innonext 0
y
O
Example: We wrap a light, flexible cable around a solid cylinder with mass M and radius R. The cylinder rotates with negligible friction about a stationary horizontal axis. We tie the free end of the cable to an object of mass m and release the object with no initial velocity. Find the speed of the falling object and the cylinder just as the object strikes the floor.
See page 280, example 9-9.
M
R
m
h
N
MgTT
mg
Solution:
No work is done by friction, so
mgh 2mv2
1)MR
2
1( 2 2)
R
v(
2
1
)1(2
1sin
222 I
Lmg
Lmg
Solution:
)2(3
1 2mlI
We get
Example: A uniform spherical shell of mass M and radius R rotates about a vertical axis on frictionless bearings. A massless cord passes around the equator of the shell, over a pulley .
Find the speed of the object when it falls a distance h from rest.
r
v
R
vps
mghE (1)
(2)
From the conservation law of energy
222
21
21
21
khIMvMgh
rvmrI ,221
mM
khMghv
21
2 2
Solution:
1)
Example: the given is as follows,find the velocity of the object and the acceleration of M 。
k
h
M
222
21
21
21
khIMvMgh
Example: the given is as follows,find the velocity of the object and the acceleration of M 。
Solution:
2)
Make a derivation
khR
aImamg
khvImvamgv
2
2R
Im
khmga
We get:
k
h
M
Alternative solution:
)2()( 21 IRTT
)3(2 khT
)4(Ra
We get:
2R
Im
khmga
k
h
M
)1(1 maTmg
Example: The power output of an automobile engine is advertised to be 200 hp at 6000 rpm. What is the corresponding torque?
1 HP = 746 WSolution:
Whp
WhphpP 51049.1)
1
746(200200
sradrev /628min/6000
mNsrad
smNP
237
/628
/1049.1 5
5. Angular Momentum
i
iivmP
0
Coordinate(x) --- Angular Coordinate()
Distance(s) --- Angular Displacement()
Velocity(v) --- Angular Velocity()
Acceleration(a) --- Angular Acceleration()
Momentum(P) --- ?
v
m
M, L
?v '
5.1 Angular Momentum of a particle
r
mvm
o
x
y
z
d
vmrprL
Unit: kg m2/s
sinrmvL
Direction: determined by right-hand rule for vector products.
Magnitude: mvd
L
o
m
r
F
x
y
z
d
Fr Compare with torque
Caution: The Value of L and both depends on the choice of origin O.
vmrprL
O
R
vm
L=mvR
P
O
mvd
To point O: L=mvd
To point P: L=0
Angular momentum of collection of particles:
i
iii
iii
i vmrprLL
r
mvm
o
x
y
z
d
L
5.2 Angular Momentum of a rigid body
The angular momentum of mi to point O:
iiii vmrL The angular momentum of the rigid body to the fixed axis (z-axis):
ii rv 2iirm
)rm(LLi
2ii
iiz
i
2iirmI
zz IL
iv
ir
im
O
xy
z
CAUTION: Giving angular momentum you must always specify about axis
Frdt
Ld
5.3 The Theorem of Angular Momentum
vmrprL
for one single particle:
dt
Ld
The rate of change of angular momentum of a particle equals the torque of the net Force acting on It.
dt
)pr(d
pdt
rd
dt
pdr
,dt
0
,
vmv
vdt
rd
122
1
2
1LLdLdt
L
L
t
t
dt
)L(di
i
ii
for collection of particles:
i )ji(j
jiii
iii
i frFr
ofri )ji(j
jii
i
iii
i Frdt
Ld
for rigid body:
dt
dLzz zz IL
dt
Id z )( zI
1122
2
1
2
1
)( IIIddtL
L
t
t
dt
Ld
Lddt
2
1
t
t 2
1
L
L
12 LL
If the torque of all the external forces equals 0:
,0i
i 12 LL
------conservation of the angular momentum for collection of particles or one single particle.
1
t
t
P
P 2 PPpddtF2
1
2
1
Compare with conservation of momentum:
,0i
i 12 LL
,0FFi
i �
12 PP�
Example: the disk rotate on the table, angular velocity and friction coefficient is , how much time it will take for the disk to stop its motion
r dr
Solution :
drgrrdmgd 22
RMgdR
3
2
0
12 IIdt
g
RIt
It
4
3
Theorem of angular momentum
When the net external torque acting on a system is zero, the total angular momentum of the system is constant.
5.4 Conservation of Angular Momentum
,0ex ,LL 12
2211 II
Example: The position of a object with mass m is given by r=acost i+bsint j (a, b, and are all constant), find the angular moment of the object and the net torque acting on it.
Solution: vmrL
jtsinbitcosar
jtcosbitsinadt
rdv
kji0tsinbtcosa 0tsinbmtsinam
L
kabm
amrFr
2
2
dt
rda
r2
0)( 2 rr
Example: A small block on a frictionless horizontal surface has a mass of m. it is attached to a massless cord passing through a hole in the surface. The block is originally revolving at a distance of r from the hole with an angular speed 0. Pull the cord from below slowly, shorten the radius of the circle to r/2. How much work was done in pulling the cord?
See page 322 10-33, 10-76, 10-87
F
0
ro
m
Solution: zeroisOtonet
21 LL:Lofonconservati
W
040
2mr 2)2r(m
22)2
r(m
2
1 20
2mr2
1 20
2mr2
3
Example: see page 314 example 10-16
s/m400v
1.00m
0.50m
Solution: Consider the door and bullet as a system, net torque about the axis through O equals zero, so angular momentum is conserved.
2
lmvL:Linitial 1
)II(L:Lfinal rodbul2
21 LL
momentumofonconservati
,)2
l(mI 2
bul ,Ml3
1I 2rod
s/rad40.0
:so
Example: see page 327 10-75
Solution:
A BC
S S
Consider A and B as a system, when remove the accelerating torque from A, angular momentum is conserved.
0AI )II( BA The thermal energy equals the lose of the total rotational kinetic energy:
Q 20AI2
1 2BA )II(
2
1
BA I3
1I,J2400Q
J3200I2
1K
:so
20A0A
20
220 )(
21
21
21
llkmvmv
v
O
l0
l
v0
d
We get : v =4m/s , =300
Solution:
sin00 mvllmv
Example : a block with spring was in a frictionless surface, k=100N/m, a spring was fixed in o point , m=1kg, the natural length of the spring l0=0.2m, the velocity of the block is v0=5m/s, perpendicular to the spring 。 When the spring rotate 900, the length of l=0.5m find the velocity of the block.
RMm
GmvR
MmGmv
321
21 22
0
Conservation of the angular momentum
v
v0
MR
OmA
CO’
3R
Solution:
)43(3sin 2
0
20
GMRv
Rv
We get :
sin30 RmvRmv
Conservation of the energy
Example: a rocket’s mass is m, velocity is v0, oc=3R 。
Find the angle between v and v0 。 (the earth mass is M 、radius=R)
R
Example: A man of mass m runs around the edge of a horizontal turntable that is mounted on a frictionless vertical axis through its center. The velocity of the man, relative to the disk, is v. The turntable is rotating in the opposite direction with an angular velocity of , relative to the earth. The radius of the turntable is R, and its mass is M. Find the final angular velocity of the system if the man comes to rest, relative to the turntable.
Solution: Consider the man and turntable as a system, angular momentum is conserved.
discI 'mandisc )II( Rm
The equation is wrong, because conservation of L is only valid in inertial reference frame
R
vME
R
discI 'mandisc )II(
MEmanI
,mRI 2man ,MR
2
1I 2disc
2/Mm
2/MmR/mv'
EMME D D
Summary:
amF
I
zz
yy
xx
maF
maF
maF
)
dt
dva(,maF
)R
va(,maF
ttt
2
nnn
dmrI
rmI
2
i
2ii
2cmp MdII
Instantaneous Effect
(One particle)
(Rigid body)
rdFWb
atot
inex WW
)UU(W 12con
non,inex WW
Accumulation Effect of Space
(One particle)
(particles)
,0WWif non,inex
dWb
atot (Rigid body)
2cm I
2
1mgyE
21
22 mv
2
1mv
2
1
12 KK i
21ii
i
22ii vm
2
1vm
2
1
)UK()UK( 1122 12 EE
12 EE
21
22 I
2
1I
2
1
时间积累
dtFJ2
1
t
t
(One particle)
12
t
tPPdtFJ
2
1
(particles)
,0Fif
(Rigid body)dt
Ld
(One particle)
dt
Ld
(particles)dt
dLzz
2
1
t
tdt矩冲量
=,0if
12 PP
12 vmvm
i
1iii
2ii vmvm
12 PP
i
1ii
2i LL
221112 IIorLL
12 LL
6* Rigid-body Rotation about a Moving Axis
6.1 Combined translation and rotation: Energy Relations
cmcmv
mi
ri
cmv'
iv
iv
For mi: 'icmi vvv
)vv()vv(m2
1K '
icm'icmii
)vvv2v(m2
1 2'i
'icm
2cmi
For mi: )vvv2v(m2
1K 2'
i'icm
2cmii
For body:
)vm2
1(vvm
)vm2
1(KK
2'ii
'icmi
2cmii
0)vm(vvvm 'iicm
'icmi
2cm
2cm
2'ii
2cmi
I2
1Mv
2
1
)vm2
1()vm
2
1(K
1
2
3
4
R
2. Rolling without slipping
cmv
cmv
cmv
cmv
cmv
cm'1 vv
'2v
'3v
'4v
2v
3v
4v
0v1
Rvcm2
121 I
2
1Mv
2
1K
22cm )MRI(
2
10
2cm
2cm Mv
2
1I
2
1
Center of mass: cmi aMF
Rotation about axis through
CM:
cmI
Example13: See page 303 example 10-6.
Use two ways to find the speed vcm of solid cylinder after it has dropped a distance h.
h
Solution1: 2cm
2cm I
2
1Mv
2
1Mgh
Solution2:
Rvcm
T
Mg
)1(MaTMg cm
)2(ITR cm 2cm MR
2
1I
From (1),(2): mg3
1T,g
3
2acm
gh3
4vha2v cmcm
2cm
Ld
7* Gyroscopes and precession
d
O
p
IL
mg
r
dsinLdL
sinIdt
d mgp
dtdL mg
gmrmg