轉動力學實驗 ( 一 ) rotational motion. rigid object ( 剛體 ) a rigid object is one that is...
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轉動力學實驗 (一 )
Rotational Motion
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Rigid Object ( 剛體 ) A rigid object is one that is
nondeformable The relative locations of all
particles making up the object remain constant.
All real objects are deformable to some extent, but the rigid object model is very useful in many situations where the deformation is negligible
Q
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Angular Position
The arc length s and r are related: s = r
ra ( an )dis
r
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Conversions Comparing degrees and radians
Converting from degrees to radians
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Angular Displacement and Angular Speed
Angular Displacement
Average Angular Speed
Instantaneous Angular Speed
( rad/s or s-1 )
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Angular Acceleration Average angular acceleration
Instantaneous angular acceleration
( rad/s2 or s-2 )
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Directions of and
The directions are given by the right-hand rule
out of the plane
into the plane
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Relationship Between Angular and Linear Quantities Displacements
Tangential Speeds
vds
dtd
r rdt
s r
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Relationship Between Angular and Linear Quantities
Tangential Acceleration
Centripetal Acceleration
tvd
dtd
r rdt
ta
2
2vca r
r
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Torque ( 力矩 )
Torque,, is the tendency of a force to rotate an object about some axis.
Torque is a vector = r F sin = F d
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轉動力學實驗原理
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與 轉動平移Ne Newton's 2 wton's 2nd L ndaw Law
平移Newton's 2nd Law
F = ma
轉動 Newton's 2nd Law
I
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Moments of Inertia of Various Rigid Objects
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滑輪轉盤
mT
mg
平移運動 (Translational Motion)
• 砝碼
轉動 ( Rotational Motion)
•滑輪 ( 轉動慣量小,忽略 )
•轉盤
( ) ( ) ( )
(1)
mg T ma
mg - T ma
(2)
力矩
r T I
r T I線軸 轉盤系統
(3)a r砝碼 線軸 轉盤
2
mgr
I mr線軸
轉盤轉盤系統 線軸
理想實驗( 忽略軸承之摩擦力矩 )
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T
m0g
平移運動達到平衡狀態• 利用掛勾與黏土
轉動達到平衡狀態
( ) ( ) 0
0 (1)
0
0
0
g T
g - T
m
m
m m m掛勾 黏土其中
0
0 (2)
淨力矩
0
r T
r T -
r m g =線軸 摩擦力矩
線軸 摩擦力矩
修正軸承之摩擦力矩
m0=掛勾與粘土的質量
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滑輪轉盤
T
(m 砝碼
+m0)g
平移運動 (Translational Motion)
• 砝碼 + 掛勾 + 黏土
轉動 ( Rotational Motion)
•滑輪 ( 轉動慣量小,忽略 )
•轉盤
( ) ( ) ( ) ( ) ( )
( ) ( )
0 0
0 0
m + m g T m + m a
m + m g - T m + m a 砝碼 砝碼
砝碼 砝碼
淨力矩
r T I
r T - I摩擦力矩線軸 轉盤系統
a r砝碼 線軸 轉盤
砝碼 線軸轉盤
轉盤系統 砝碼 線軸02m
m gr
I m r
軸承之摩擦力矩
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角動量守恆實驗
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Angular Momentum( 角動量 )
The instantaneous angular momentum of a particle relative to the origin O is defined as the cross product of the particle’s instantaneous position vector and its instantaneous linear momentum
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Conservation of Angular Momentum
ˆ
2 2κ
L r P = r m v + v = r mr
mr mr Ι
L r Pr P P
= 0
+ r
P + r
= mv +
=
LIF Th
d d d d
dt dt dt dt
v F
v
dL = constant
dtn = e 0
徑向 切向 切向
zz z
Ld= 0 L = constant
dtIF Then = 0
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Angular Momentum of a System of Particles
The total angular momentum of a system of particles is defined as the vector sum of the angular momenta of the individual particles
Differentiating with respect to time
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Angular Momentum of a Rotating Rigid Object
ˆ
ˆ
i i i i ii i i i
i i i i i i ii ii
2i i i
i
1 2 3
2i i
i
= = =
κ
=
=
κ
L L = r P r m v + v
r m v r m r
m r
Rigid b
d
ody
m r Ι
L
dt
dL = c= 0 onst
LIF ant
d n = 0
tThe
徑向 切向
切向 切向
To find the angular momentum of the entire object, add the angular momenta of all the individual particles
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L Ι
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End of Lecture
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Perpendicular-Axis Theorem( 垂直軸定理 )
Iz = Ix + Iy
ri Prove it
x y
2
2 2 2
Z r
r
i i
i
ii i
i i
I = I = m 質點Hint :
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R
M
圓盤
利用垂直軸定理,系統的軸之轉動慣量 = ?
1
4 2M RI =
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Parallel-Axis Theorem( 平行軸定理 )
For an arbitrary axis, the parallel-axis theorem often simplifies calculations
Ip = ICM + MD 2
Ip 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
C.M.
P
D
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C.M.
? ( Parallel-Axis Theo m re )pI
P
C.M.
x
y
x
y
C.M.
P x
y
x
y
△m
2P
P
r
對 m
I = m
轉動慣量質點 的 為
rP
r R
rP
r R
'
Pr
C.M.
P
P
CR .M.
r
對 點 位移向量
對 位移向量
對 點 位移向量
質點 的
質點 的
的
m
m
P'r Rr
=
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C.M.
P x
y
x
y
△m
2P
P
r
對 m
I = m
轉動慣量質點 的 為
rP
r R
rP
r R
P'r Rr
=
2P P P
2
'
' 2
'
'
r r
r
r = r r
+ r2
R R
R R
=
' 2 '2
P
+ R Rr2r
對 質點 的 為
轉動慣量m
I = m
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C.M.
P x
y
x
y
△m1
2
P
r
質點
整個剛體對 的轉 為動慣量
P i ii
ii
I = I = m
r1
r1 R
'
'
'
'
11
2
3
N
2
3
N
r
r
r
r
r
R
R
R
r
r
R
r
質 = 點
質 = 點
質 = 點
質 = 點
1
2
3
N
m
m
m
m
' '
2
2 2
P
r
+ 2r R Rr
整 轉個剛體 動慣量對 的
P i ii i
P i
i
i ii
I = I = m
I = m
△m2
r2
r1
r1 R
r2
r2
r2
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C.M.
P x
y
x
y
' '
2
2 2
P
r
+ 2r R Rr
整 轉個剛體 動慣量對 的
P i ii i
P i
i
i ii
I = I = m
I = m
2' 2 ' 2r R R r
P i i i i ii i i
I = m m m
2R i
i
mC.M.I
2R M
0
IP = ICM + MR2 (Parallel-Axis Theorem)
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Perpendicular-Axis Theorem( 垂直軸定理 )
Iz = Ix + Iy
ri Prove it
x y
2
2 2 2
Z r
r
i i
i
ii i
i i
I = I = m 質點Hint :
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R
M
圓盤
利用垂直軸定理,系統的軸之轉動慣量 = ?
1
4 2M RI =
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Moments of Inertia of Various Rigid Objects
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