Top Banner
RIGID BODY DYNAMICS Mahendra Verma Physics, IITK
26

RIGID BODY DYNAMICS - IIT Kanpur

Feb 06, 2022

Download

Documents

dariahiddleston
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Page 1: RIGID BODY DYNAMICS - IIT Kanpur

RIGID BODY DYNAMICS

Mahendra Verma Physics, IITK

Page 2: RIGID BODY DYNAMICS - IIT Kanpur

Equa%on  of  mo%on  of  a  rigid  body

Page 3: RIGID BODY DYNAMICS - IIT Kanpur

Rolling    Cylinder

Page 4: RIGID BODY DYNAMICS - IIT Kanpur

Unicycle

mWR2↵ = ⌧P � FR

(M +MC)a = F

a =⌧P /R

M +MC +mW

↵ = a/R

Page 5: RIGID BODY DYNAMICS - IIT Kanpur

http://www.tokyocycle.com/bbs/threads/bicycle-physics.4773/

Page 6: RIGID BODY DYNAMICS - IIT Kanpur

Euler  angle

NX

Y

Z

z

x

y

Page 7: RIGID BODY DYNAMICS - IIT Kanpur

Euler  angle

Space axes or

Gyro axes⌦z = (

˙⇣ + ˙� cos ✓)

⌦x

= ✓

⌦y = � sin ✓

Page 8: RIGID BODY DYNAMICS - IIT Kanpur

Body  axes

!z = (

˙⇣ + ˙� cos ✓)

!x

= (

˙✓ cos ⇣ + ˙� sin ✓ sin ⇣)

!y = (� ˙✓ sin ⇣ + ˙� sin ✓ cos ⇣)

Page 9: RIGID BODY DYNAMICS - IIT Kanpur

Torque-­‐free  precession

= 0

= (L cos ✓ sin ✓ � L sin ✓ cos ✓) ˙⇢

Lz = L cos ✓

y

y

Ly = L sin ✓

Lx

= 0

N =

dL

dt= (Lz sin ✓ � Ly cos ✓) ˙⇢

Page 10: RIGID BODY DYNAMICS - IIT Kanpur

˙⇢ = ⌦�

N =

dL

dt= (I3 � I1)⌦ sin ✓ cos ✓ ˆ�

Precessing  cylinder

= (I3⌦ cos ✓ sin ✓ � I1⌦ sin ✓ cos ✓) ˙⇢

Lz = I3 ˙� cos ✓

y

y

y

Lx

= 0

Ly = I2� sin ✓

� = ⌦

theta in left Typo

N =

dL

dt= (Lz sin ✓ � Ly cos ✓) ˙⇢

Page 11: RIGID BODY DYNAMICS - IIT Kanpur

Gyroscope

˙� =

LZ � Lz cos ✓

I1 sin2 ✓

I1¨✓ +

Lz sin ✓ �

(LZ � Lz cos ✓)

sin ✓cos ✓

� LZ � Lz cos ✓

I1 sin2 ✓

�= mgl sin ✓

Lz = I3( ˙⇣ + ˙� cos ✓) = const

= Lz cos ✓ + I1 ˙� sin

2 ✓

Lx

= I1✓

Ly = I1� sin ✓

LZ = Lz cos ✓ + Ly sin ✓ = const

✓dL

dt

x

= I1¨✓ + (Lz

sin ✓ � Ly

cos ✓) ˙� = mgl sin ✓

Page 12: RIGID BODY DYNAMICS - IIT Kanpur

L2x

2I1+

L2y

2I1+

L2z

2I3+mgl cos ✓ = E

1

2

I1( ˙✓2+

˙�2sin

2 ✓) +L2z

2I3+mgl cos ✓ = E

Energy

1

2I1✓

2 + Ue↵(✓) = E0

Ue↵(✓) =(LZ � Lz cos ✓)2

2I1 sin2 ✓

�mgl(1� cos ✓)

E0 = E �mgl � L2z

2I3

t =

pI1I3Lz

t0

a =I3I1

b =LZ

Lz

c =mgl

L2z/I3

E =E0

L2z/I3

1

2

✓d✓

dt0

◆2

+

a

2

✓b� cos ✓

sin ✓

◆2

� c(1� cos ✓) = ¯E

Page 13: RIGID BODY DYNAMICS - IIT Kanpur

m=0  (cylinder)¨✓ +

a

sin

3 ✓(b� cos ✓)(1� b cos ✓) = 0

a=½; b=1/cos(π/6)

Page 14: RIGID BODY DYNAMICS - IIT Kanpur

m=0  (disk)¨✓ +

a

sin

3 ✓(b� cos ✓)(1� b cos ✓) = 0

a=2; b=1/cos(π/10)

Page 15: RIGID BODY DYNAMICS - IIT Kanpur

m  >  0;  b=1.2¨✓ +

a

sin

3 ✓(b� cos ✓)(1� b cos ✓) = c sin ✓

Page 16: RIGID BODY DYNAMICS - IIT Kanpur

m  >  0;  b=0.2¨✓ +

a

sin

3 ✓(b� cos ✓)(1� b cos ✓) = c sin ✓

˙� =

pab� cos ✓

sin

2 ✓

Page 17: RIGID BODY DYNAMICS - IIT Kanpur

Spinning  Top

I 01¨✓ +

Lz sin ✓ �

(LZ � Lz cos ✓)

sin ✓cos ✓

� LZ � Lz cos ✓

I 01 sin2 ✓

�= mgl sin ✓

Page 18: RIGID BODY DYNAMICS - IIT Kanpur

https://www.youtube.com/watch?v=cquvA_IpEsA A Hila Science Video

Page 19: RIGID BODY DYNAMICS - IIT Kanpur

Rolling  coin

f�

z

yZ

b

mg

R

N

¨✓ +

✓I3I 01

+

MR2

I 01

◆✓⇢0

R� sin ✓

◆cos ✓ + sin ✓

�˙�2

=

mgR

I 01sin ✓

Page 20: RIGID BODY DYNAMICS - IIT Kanpur

Euler  Equa%ons

Page 21: RIGID BODY DYNAMICS - IIT Kanpur

A = (A)S +⌦⇥A

Page 22: RIGID BODY DYNAMICS - IIT Kanpur

dL

dt

inertial

=

dL

dt

rotating

+⌦⇥ L = N

I1d!

x

dt+ (I3 � I2)!y

!z

= Nx

I2d!

y

dt+ (I1 � I3)!x

!z

= Ny

I3d!

z

dt+ (I2 � I1)!x

!y

= Nz

Page 23: RIGID BODY DYNAMICS - IIT Kanpur
Page 24: RIGID BODY DYNAMICS - IIT Kanpur
Page 25: RIGID BODY DYNAMICS - IIT Kanpur

http://geophysics.ou.edu/solid_earth/notes/precess/chandler.html

Page 26: RIGID BODY DYNAMICS - IIT Kanpur

Stability analysis of a matchbox