chapter 5 rotation of a rigid body
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Chapter 5 Rotation of a Rigid Body. §5-1 Motion of a Rigid body. §5-2 Torque The Law of Rotation rotational Inertia. §5-3 Application the Law of Rotation. §5-4 Kinetic Energy and Work in Rotational Motion. - PowerPoint PPT PresentationTRANSCRIPT
Chapter 5 Rotation of a Rigid Body
§5-5 Angular Momentum of a rigid Body Conservation of Angular Momentum
§5-1 Motion of a Rigid body
§5-2 Torque The Law of Rotation rotational Inertia
§5-3 Application the Law of Rotation
§5-4 Kinetic Energy and Work in Rotational Motion
§§5-1 5-1 Motion of a Rigid body
1. Rigid body1. Rigid bodyThe body has a perfectly definite and unchanged The body has a perfectly definite and unchanged shape and size no matter how much external shape and size no matter how much external force acts on it.force acts on it.
2. Motion forms of rigid body2. Motion forms of rigid bodyTranslation( 平动
)Can be regarded as a particleCan be regarded as a particle
The distance between two points in a rigid body maintains constant forever
RotationRotating around axisRotating around axis
Rotating around a pointRotating around a point
Fixed axisFixed axis
Moving axisMoving axis
----Superposition of several rotations Superposition of several rotations around axisaround axis
axis
Translation + rotation
3. Rotation of a rigid body around a fixed axis
Every point of the rigid body moves in a circle
P
Q
axis
,,
They have the same angular displacement, angular speed & angular acceleration.
Select arbitrary point P.P’s rotation can represent the rotation of the rigid body.
P
Q
axis
,, Angular position
Angular displacement
Angular speed
td
d
tt d
d
d
d 2
Angular acceleration
4. Relation between angular and linear quantities
r
v s2
nt a ra
rr
z
x0
v
r P
,,,
nt aavsr ,,,,
§5-2 Torque The Law of Rotation Rotational Inertia
1. Torque
FrM
F
r φ
d
M
is located in the plane perpendicular to the axisF
F
r
No effect at all to rotate the rigid body around the axis
FrM
1Fr
)( 21 FFr
F
F2
1
is not placed in the plane perpendicular to the axisF
Only is useful for the rotation around the axis
2Fr
0
0`
imir
if
iF
i
2. The law of rotation
i
Use Newton’s second law to im
Tangential component
itiiiii amfF )(sinsin
)( iirm
The radial component of force passes through the axis and can’t cause the body to rotate.
iiiiii rfrF sinsin )( 2iirm
Multiply both side by ri ,
External torque Internal torque
dt
dIIM
--Law of rotation
For entire rigid body
iii
iiii
i rfrF sinsin )( 2ii
i
rm
resultant external torque M
Rotational inertia I
resultant internal torque =0
3. Calculation of moment of inertia
The magnitude of moment of inertia depends on the total mass and mass distribution of body, the location and orientation of the axis.
discrete particles
iiirmI 2
It describes the rotating inertia of a rigid body about an axis.
i
iirmI 2 --the rotational inertia (moment of inertia) of the rigid body about the axis
dl dm
—the mass per unit length
the mass distribution over a surface
ds dm —the mass per unit area
the mass distribution over a volume
dvdm —the mass per unit volume
the mass distribution over a line
Continuous distribution of mass
I dmr 2
Exa. A slender rod has mass m , length l . Find its I about some axis follow as
⑴the axis through O and perpendicular with the rod.⑵ the axis through an end of the rod and perpendicular with it.
⑶ The axis at arbitrary distance h from O
m lo
dm
x
dxdm l
m
dx x
Solution
⑴
m l⑵
dmrI 2
dm
x
dxdm
l
m
dxxl
0
2
2
3
1ml
xdxo o
dmrI 20 dxx
l
l 2
2
2 2
12
1ml
m l
h
⑶
dxxIh
l
hlh
2
)2
(
2
xo o
22
12
1mhml
20 mhIIh --Parallel-axis theorem
Perpendicular-axis theorem
x y
z
O
dm
r
mz dmrI 2
m
dmyx )( 22
mm
dmydmx 22
xyz III -- -- Perpendicular-axis theorem
Suppose a thin board is Suppose a thin board is located in located in xOyxOy plane plane
Calculate the Acce. of the blocks and the tension of the rope.
Example are fastened together by a weightless rope across a fixed pulley ( )
21 mm ,
2`1 mm <
mThe pulley has mass and radius . There is a fractional torque exerting on the axis. The rope does not slip over the pulley.
r rMr
1m
2m
§5-3 Application the Law of Rotation
1T 2T
rM
a
a gm1
1T
1ma
gm2
2T
2m a
r
1m
2m
SolutionAccording to New.’s Second Law
1m
2m
amgmT 111 amTgm 222
::
According to Rotational Law
Pulley : IMrTrT r 12
ra 2
2
1mrI
221
12
mmm
rM
gmma
r
)(
)(
)(
22
11
agmT
agmT
We get
Example A uniform circular plate with mass m and radius R is placed on a roughly horizontal plane. At the beginning, the plate rotates with angular speed 0 around the axis across the center of it. Suppose the fractional coefficient between them is . Calculate how many times does the plate rotate before it stops?
Solution
Select mass unit dsdm
2R
m
S
m
rdrds 2
r drR
The friction exerts on dmgdmdfr
Frictional torque correspondingly
rr rdfdM
rdrg2
drrg 22
rr dMM
drrgR 2
02 mgR
3
2
dt
dIM
According to
We have
tdtg
03
2
dt
dmR
2
2
1mgR
3
2
0
02
1
dR
We get04
3 g
Rt
QuestionHow many revolutions does the plate rotate before it stops?
§5-4 §5-4 Kinetic Energy and Work in Rotational Motion
1. The work done by torque
0
0r
F
rd
d
As rdFdA�
dFr sin
MdThe rigid rotating from 21
The work done by the torque 2
1
MdA
2. Kinetic Energy of rotation
ir
iv
im
s KE of rotation im2
2
1iiki vmE
22
2
1 iirm
For the entire rigid body
i
ikk EE 22 )(2
1 i
iirm
i
iirmI 2
Then 2
2
1 JEk
3. The theorem of KE of a RB rotating about a fixed axis
asdt
dIM
MddA d
dt
dI dI
-- KE of rotation
21
22 2
1
2
1 IIA
2
1
2
1
dIMd
-- The theorem of KE of a RB rotating about a fixed axis
4. Gravitational potential energy of RB
…can be seen as the GPE of a particle located on the center of mass with the same mass of m.
Example A uniform thin rod m, l , One end is fixed. Find its of point A as it rotate angular from horizontal line?
?Aa?Av
m 、 l
O
A
A
Av
Solution
mgO
l
21
22 2
1
2
1 IIA as
02
1cos
22
0
Id
lmg
2
3
1mlI lvA
sin3glvA
cos2
3g
dt
vda A
t
sin32
gl
va A
n
We get
Another solutionUse conservation law of energy of rigid body, we can get first.av
Then use dt
vda A
t ,2
l
va A
n
Or use IM We can get
then lat
We can get the Acce. of A finally
im the angular momentum of :
iiii rvmL
Direction
ir
iv
imiL
§5-4 §5-4 Angular Momentum of a rigid Body Conservation of Angular Momentum
1. Angular momentum of a rigid body rotating about a fixed axis
2iirm
Angular momentum of a rigid body rotating about a fixed axis :
I
2. Angular momentum theorem
)( 2ii
i
rmL
Asdt
dLM
Direction: Same as iL
, we have dLMdt
integration 2
1
2
1
L
L
dLdtMt
t
)( Id
22
11
)(
I
IId
1121
2
1
IIMdtt
t
=Constant when M=0IL 3. Conservation Law of Angular Momentum
2
1
t
tMdt ---impulse torque
—Angular momentum theorem
of a rigid body
Example The rotational inertia of a person and round slab is I0. The mass of dumbbell is m . Their rotating angular speed is 0 and the rotating radius of m is r1 at the beginning.
Calculate:The angular speed and the increment of the mechanical energy when the arms of the person contracts from r1 to r2
r
r1
2mm
I0
Person + round slab + dumbbells = system
21 LL
02
12
10 )( mrmrI
)( 22
220 mrmrI
0220
210
2
2 mrI
mrI
Resultant external torque is zero. So its angularmomentum is conservative.
The increment of the mechanical energy
kE
20
210 )2(
2
1 mrI 2220 )2(
2
1 mrI
0kk EE
Example A round platform has mass of M and radius of R. It can rotates around a vertical axis through its center. Suppose all resistant force can be neglected. A girl with mass of m stands on the edge of the platform. At the beginning, the platform and the girl are at rest. If the girl runs one revolution, how much degree does the girl and the platform rounds relative to the ground, respectively?
R
M
m
0 MII
Let : MII , be rotational inertia of the girl and the platform.
,
Resultant external torque of system = 0
0LL
be angular speed of the girl and the platform relative to the ground.
22
2
1mRIMRIM
Solution
then :
Angular momentum is conservative :
And
We get :
M
m2
The girl relative to the platform :
M
mM 2
Let t refer to the time that the girl runs one revolution on the platform, then :
tt
dtM
mMdt
00
2 2
The girl rounds relative to the ground :
mM
Mdt
t
2
20
The platform rounds relative to the ground :
mM
mdt
t
2
20
Example A uniformly thin rod has M,2l . It can rotate in vertical plane around the horizontal axis through its mass center O. At the beginning, the rod is placed along the horizontal position. A small ball with mass of m and speed of u falls to one end of the rod. If the collision between the ball and the rod is elastic. Find the speed of the ball and the angular speed of the rod after they collide each other. m
ou
l2M
v
m
ou
l2MSolution
As mg of the ball << the impulse force between the ball and the rod
Then the resultant external torque of the system with respect to O = 0.
So we can neglect mg during colliding.
22
3
12
12
1mllmI
As the collision is elastic, then the mechanical energy of the system is conservative
222
2
1
2
1
2
1 Imvmu
Jmvlmul
The angular momentum of the system is conservative.
And
mM
mMuv
3
)3(
lmM
mu
)3(
6
Solve above equations, we get
Rod’s
Ball’s