轉動力學實驗 ( 一 ) Rotational Motion. Rigid Object ( 剛體 ) A rigid object is one that is nondeformable The relative locations of all particles making up the.

轉動力學實驗 (一 )

Rotational Motion

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

Angular Position

The arc length s and r are related: s = r

ra ( an )dis

r

Conversions Comparing degrees and radians

Converting from degrees to radians

Angular Displacement and Angular Speed

Angular Displacement

Average Angular Speed

Instantaneous Angular Speed

( rad/s or s-1 )

Angular Acceleration Average angular acceleration

Instantaneous angular acceleration

( rad/s2 or s-2 )

Directions of and

The directions are given by the right-hand rule

out of the plane

into the plane

Relationship Between Angular and Linear Quantities Displacements

Tangential Speeds

vds

dtd

r rdt

s r

Relationship Between Angular and Linear Quantities

Tangential Acceleration

Centripetal Acceleration

tvd

dtd

r rdt

ta

2

2vca r

r

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

轉動力學實驗原理

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

Moments of Inertia of Various Rigid Objects

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 )

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

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

角動量守恆實驗

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

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

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

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

L Ι

End of Lecture

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 :

R

M

圓盤

利用垂直軸定理，系統的軸之轉動慣量 = ?

1

4 2M RI =

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

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

=

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

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

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)

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 :

R

M

圓盤

利用垂直軸定理，系統的軸之轉動慣量 = ?

1

4 2M RI =

Moments of Inertia of Various Rigid Objects