Rotating about a fixed point O: (2)Dongjun Lee Strange Rotation Dynamics in 3D Dongjun Lee Angular Momentum of Planar Rigid Body • Not validfor asymmetrical bodies or three- dimensional motion

Dongjun Lee

Dongjun Lee (이동준) Department of Mechanical Engineering

Seoul National University

Dynamics (동역학)

Lecture 8: Rigid Body Dynamics in 3D

Dongjun Lee

Introduction

• Fundamental relations developed for rigid bodies in 2D are also applicable to rigid bodies in 3D.

• Scalar angular momentum equation for rigid-body in 2D (e.g., H_G=I_Gw) is not validfor general rigid-bodies in 3D  truly nonlinear!

• This chapter concerned with 6-DOF dynamics of rigid-body in 3D, particularly 3-DOF nonlinear angular dynamics, via vector mechanics, impulse-momentum and work-energy principles.

• Dynamics w.r.t. center of massG:

• Dynamics via d’Alembert principle:

• Rotating about a fixed point O:

Dongjun Lee

Strange Rotation Dynamics in 3D

Dongjun Lee

Angular Momentum of Planar Rigid Body

• Not validfor asymmetrical bodies or three- dimensional motion. Still validfor plane motion of bodies symmetric w.r.t. the reference plane.

• Angular momentum of a planar rigid-body:

     

 



















 

G i i

n i

i i i n

i

i i i G

I m r

m r r m v r H

 



























 

Δ

Δ Δ

2

1 1

• Differentiating, we have

where 𝐻 //𝑤//𝛼⃗//𝐾and 𝐼 is scalar/constant.







^MG^{HG IG HG IG}



^MG,j

Dzhanibekov Effect (Tennis Racket Theorem) moment of inertia matrix

Dongjun Lee

Rigid Body Angular Momentum in 3D

   

 

   

 

 

 

Z G XZ Y G XY X G XX

Z i X i Z Y i X i Y Z i Y i X

n i iXiZ i Z

n i iXiY i Y

n

i iY iZ i

X n i

i Z i X X i Z Z i X i Y Y i X Y i n i

Y i i Z Z i i Y i G

I I I

dm r r dm r r dm r r

m r r m r r m r r

m r r r r r r

m r r r r H_X





























































 

 































, , ,

, 2

, 2 ,

1 , ,

1 , ,

1 2 , 2 , 1

, , , , , , 1

, ,

Δ Δ

Δ

Δ

 Δ







• Angular momentum of a rigid-body about G:

      



 

















 ⁿ

i

i i i n

i

i i i

G r v m r r m

H

1 1

Δ

Δ   



 



• If all vectors expressed in an inertial frame {I,J,K

}:

• Moment of inertia matrix 𝐼 (𝑡) about G expressed in {I,J,K} istime-varying.

• Direction 𝑤 is not constant; directions of 𝐻 and 𝑤 are not the same.

Dongjun Lee

Moment of Inertia about Principal Axes

• Angular momentum of a rigid-body about G:

      



 

 



 



 ⁿ

i

i i i n

i

i i i

G r v m r r m

H

1 1

Δ

Δ   



 



• If all vectors expressed in a body-frame {i,j,k

}:

 

z G xz y G xy x G xx

z i x i z y i x i y z i y i x G x

I I I

dm r r dm r r dm r r H





























^{2, 2,}



^{, ,}



^{, ,}

• Moment of inertia matrix 𝐼 (or 𝐼 ) defines transformation: 𝑤 → 𝐻 .

• Two vectors, 𝑤 and 𝐻 , have the same direction, iff 𝑤 along a principal axis of 𝐼 .

• Here, moment of inertia 𝐼 expressed in the body frame {i,j,k} is constant, since the basis vectors {i,j,k} are rotating together with the body.

• Further, if {i,j,k} are along principal axes (i.e., eigenvectors (cf, axisymmetric)):

Dongjun Lee

Rigid Body Angular Momentum about Fixed O

vG

m

L 



linear momentum

G H_Gangular momentumabout

• Angular momentum can be computed about any point O:

• Momenta of a rigid-body are given by:

• If the point O is a fixed point:

 





 ⁿ

i i ii

O r mv

H

1



 

r v m r  r m I w

H ⁿ _O

i

i i i n

i i i O











 

] [ Δ Δ

1 1













 







where 𝐼 is constant, if all vectors (w,HO,MO) are expressed in a body-fixed frame {i,j,k}

© 2019 McGraw-Hill Education.

Concept Question

¹

At the instant shown, the disk shown rotates with an angular velocity ω

₂

with respect to arm ABC, which rotates around the y axis as shown (ω

₁

). Determine the directions of the angular momentum of the disc about point A (choose all that are correct).

+x +y +z

−x −y −z

© 2019 McGraw-Hill Education.

Concept Question

²

At the instant shown, the disk shown rotates with an angular velocity ω

₂

with respect to arm ABC, which rotates around the y axis as shown (ω

₁

). Determine the directions of the angular momentum of the disc about point A (choose all that are correct).

+x +y +z

−x −y −z

© 2019 McGraw-Hill Education.

Concept Question

³

A homogeneous disk of mass m and radius r is mounted on the vertical shaft AB as shown Determine the directions of the angular momentum of the disc about the mass center G (choose all that are correct).

+x +y +z

−x −y −z

© 2019 McGraw-Hill Education.

Concept Question

⁴

A homogeneous disk of mass m and radius r is mounted on the vertical shaft AB as shown Determine the directions of the angular momentum of the disc about the mass center G (choose all that are correct).

+x +y +z

−x −y −z

© 2019 McGraw-Hill Education.

Group Problem Solving

¹

Two L-shaped arms, each weighing 5 kg, are welded at the third points of the 600 mm shaft AB. Knowing that shaft AB rotates at the constant rate ω = 360 rpm, determine the angular momentum of the body about Aand the angle it forms with AB.

Strategy:

• The part rotates about the axis AB. Determine the mass moment of inertia matrix for the part.

• Compute the angular

momentum of the disk using the moments inertia.

© 2019 McGraw-Hill Education.

Group Problem Solving

²

Modeling and Analysis:

Given: m = 5kg, a = 200mm = 0.2 m

2 360

12 rad/s.

60

    

0, 0, 12 rad/s



_x

 

_y

 

_z

 

Find: H_A

There is only rotation about the z-axis, what relationship(s) can you use?

x xz yz z z

G I Gy I G I

H  



H  



H  



Split the part into four different segments, then determine I

_xz

, I

_yz

, and I

_z

. What is the mass of each segment?

• Mass of each of the four segments m   2.5kg.

© 2019 McGraw-Hill Education.

Group Problem Solving

³

Fill in the table below to help you determine I

_xz

and I

_yz

.

Determine I

_z

by using the parallel axis theorem – do parts 1 and 4 first

 

²

^{ }

^{2 2}

2 1

1 4 2

1

z z 12

I_ I_  m a m a  a 

Parts 2 and 3 are also equal to one another

2

2 3

1

z z 3

I_ I_  m a

© 2019 McGraw-Hill Education.

Group Problem Solving

⁴

Calculate H

_x

2

2

2

3 2

3 2.5 0.2 12 2

5.6549 kg m / s





  

   

 

    

 

 

Gx

H m a

Calculate H

_y

2

2

2

1 2

1 2.5 0.2 12 2

1.8850 kg m / s





  

  

      

  

Gy

H m a

Calculate H

_z

2

2

2

10 3

10 2.5 0.2 12 3

12.5664 kg m /s





  

  

 

   

 

Gz

H m a

Total Vector

H_AH_Gr_{G A}/ mvH_G



5.65 kg m / s i (1.885 kg m / s) j²



²



12.57 kg m / s k²



     

HA

© 2019 McGraw-Hill Education.

Group Problem Solving

⁵

Angle Between AB and H

_A

:

k, k H



_AB

 

_AB

 H

_A

 

_A

 H

_AZ

2 2 2 2

5.6549 1.8850 12.5664 13.9085 kg m / s

    

HA

12.5664

cos 13.9085

AZ A

H

  H 

25.4^

Dongjun Lee

Principle of Impulse and Momentum

Sys Momenta₁+ Sys Ext Imp_1-2= Sys Momenta₂

• Linear momentum

• Angular momentum

• Principle of impulse/momentum applied to rigid bodies in 3D in the same way

• Free-body diagram used to further replenish unknown components

• For bodies rotating about a fixed point O, can eliminate the impulse of the reactions at O by writing equations about O.

Dongjun Lee

Kinetic Energy of Rigid Body

• 𝐼 is moment of inertia matrix (varying if expressed in inertial-frame {I,J,K}; constant if in body-frame {I,j,k}; constant/diagonal if along the principal axes):

• Kinetic energy of rigid-body in 3D = translation kinetic energy of G+ rotation kinetic energy about G:

• Kinetic energy of rigid-body in 3Drotating about a fixed point O:

• Kinetic energy of a rigid-body independent on the choice of inertial or body frames, {I,J,K} or {i,j,k}.

© 2019 McGraw-Hill Education.

Concept Question

⁵

At the instant shown, the disk shown rotates with an angular velocity ω

₂

with respect to arm ABC, which rotates around the y axis as shown (ω

₁

). What terms will contribute to the kinetic energy of the disk (choose all that are correct)?

2 1

1

2I^x 1 ₁²

2I^z 1 ₁²

2I^y

2 2

1

2I^z 1 ² ₁²

2mL

© 2019 McGraw-Hill Education.

Concept Question

⁶

At the instant shown, the disk shown rotates with an angular velocity ω

₂

with respect to arm ABC, which rotates around the y axis as shown (ω

₁

). What terms will contribute to the kinetic energy of the disk (choose all that are correct)?

2 1

1

2I^x 1 ₁²

2I^z 1 ₁²

2I^y

2 2

1

2I^z 1 ² ₁²

2mL

Dongjun Lee

Sample Problem 18.2

A homogeneous disk of mass m mounted on axle OG of negligible mass. The disk rotates without slidingcounter-clockwise at ₁about OG. Determine: a) angular velocity of disk, b) its angular momentum about O, c) its kinetic energy, and d) vector and couple at G equivalent to the momenta (mv_G, H_G) of the disk.

SOLUTION:

• Can also use principal axis I about O:

Dongjun Lee

Sample Problem 18.2

A homogeneous disk of mass m mounted on axle OG of negligible mass. The disk rotates without slidingcounter-clockwise at ₁about OG. Determine: a) angular velocity of disk, b) its angular momentum about O, c) its kinetic energy, and d) vector and couple at G equivalent to the momenta (mv_G, H_G) of the disk.

SOLUTION:

The disk rotates about the vertical axis through O as well as about OG.

Combine the rotation components for the angular velocity of the disk.

Compute the angular momentum of the disk using principle axes of inertia and noting that O is a fixed point.

The kinetic energy is computed from the angular velocity and moments of inertia.

The vector and couple at G are also computed from the angular velocity and moments of inertia.

Dongjun Lee

Sample Problem 18.2

18 - 23

SOLUTION:

The disk rotates about the vertical axis through O as well as about OG. Combine the rotation components for the angular velocity of the disk.

j

i 

 

2

1 



 

Noting that the velocity at C is zero,

   

 

L r

k r L

j r i L j i

r

v_{C C}

1 2

1 2

2

0 1

0





















































r L



j

i 

 

1

1 



 

Dongjun Lee

Sample Problem 18.2

18 - 24

Compute the angular momentum of the disk using principle axes of inertia and noting that O is a fixed point.



r L



j

i 

 

1

1 



 

k I j I i I

H_{O x x} _{y y} _{z z}





  



 

  ^{ }



² ₄^{1 2}



^{0 01}

2 41 2

2 1 21





















mr mL I

H

L r mr mL I

H

mr I

H

z z z

y y y

x x x













^{L r}

 ^

^{r L}

^

^j

m i mr

H_O  

2 1 41 1 2

2

21    



The kinetic energy is computed from the angular velocity and moments of inertia.

 

   



2 1 ²



41 2 12 2 21

2 2 2 21

L r r L m mr

I I I

T _{x x y y z z}

























12 2 2 2

81 6 _^









 

 L

mr r T

Dongjun Lee

Sample Problem 18.2

18 - 25



r L



j

i 

 

1

1 



 

The vector and couple at G are also computed from the angular velocity and moments of inertia.

k mr v

m 

1





r L



j mr

i mr

k I j I i I

H_{G x x y y z z}





 

 





















   

2 14 2 1 21



 

 

 j

L i r mr

H_G  

1 2

2 21 

Dongjun Lee

Sample Problem 18.1

Rectangular plate of mass m suspended from two wires is hit at D in a direction perpendicular to the plate. Immediately after impact, determine a) velocity of plate COM G; b) plate angular velocity.

Unknowns: 8, equations 6 + 2

SOLUTION:

Apply the principle of impulse/momentum.

Since initial momenta is zero, the system of impulses equivalent to final momenta.

Assume supporting two wires remain taut

vertical velocity and the rotation about an axis normal to the plate is zero.

Principle of impulse and momentum yields to two Eqs for linear momentum and two equations for angular momentum.

Solve for the two horizontal components of the linear and angular velocity vectors.

Dongjun Lee

Sample Problem 18.1

SOLUTION:

Apply the principle of impulse and momentum. Since the initial momenta is zero, the system of impulses must be equivalent to the final system of momenta.

Assume that the supporting cables remain taut such that the vertical velocity and the rotation about an axis normal to the plate is zero.

k v i v v _x _z



 _xi_yj

Since the x, y, and z axes are principal axes of inertia, j ma i mb j I i I

H_{G x x} _{y y} _x _y







 ²

121 2

121 







Dongjun Lee

Sample Problem 18.1

Principle of impulse and momentum yields two equations for linear momentum and two equations for angular momentum.

Solve for the two horizontal components of the linear and angular velocity vectors.

mvx

 0

0 vx

vz

m t

F 

 Δ

m t F v_z Δ



F t m



k

v 

 Δ



x x

mb H t bF

2

121 21 Δ





mb t

x6FΔ



y y

ma H t aF

2

121 21 Δ









F t ma



y6 Δ





^{ai bj}



mab t

F  

6 Δ 



Dongjun Lee

Sample Problem 18.1



F t m



k

v 

 Δ



^{ai bj}



mab t

F  

6 Δ 



j ma i mb

H_{G x} _y



 ₁₂^{1 2}

2

121 



Reflect and Think:

• Equating the y components of the impulses and momenta and their moments about the z axis, you can obtain two additional equations that yield T_A= T_B= ½W > 0. This verifies the assumption that the wires remain taut.

• If the impulse were at G, this would reduce to a two- dimensional problem (rotating about i).

Dongjun Lee

Dynamics of a Rigid Body in 3D

• Translation (linear) dynamics of COM G:

• Rotational dynamics about COM G:

• Convenient to express rotation dynamics in body-frame {i,j,k} (or even more so in principal axes), with which I^Gxyz_Gbecomes constant.

• With {i,j,k} rotating, the differentiation of 𝐻 = 𝐻 𝚤⃗ + 𝐻 𝚥⃗ + 𝐻 𝑘 is given by:

• So far, consider momenta, impulse and kinetic energy of rigid body in 3D  full dynamics?

Dongjun Lee

Euler’s Equation for 3D Rotation Dynamics

• Differentiation of 𝐻 = 𝐻 𝚤⃗ + 𝐻 𝚥⃗ + 𝐻 𝑘 :

• Rotational dynamics about COM G:

• Further, if {i,j,k} chosen along the principal axes, I_Gbecomes diagonal and

• Along each direction {i,j,k}, can obtain Euler’s equation for rotational dynamics in 3D:

Dongjun Lee

Rotation about a Fixed Point or a Fixed Axis

• For a rigid-body rotation about a fixed point O:

• For a rigid-body rotation about a fixed axis k:

where I^*#Ois constant since expressed in {i,j,k}.

• Rotational dynamics about a fixed axis k (O can be any point on the axis (with M_Oabout O):

Dongjun Lee

Rotation Dynamics about a Fixed Axis

• For a rigid-body rotation about a fixed axis k:

• If symmetric with respect to the xy-plane,

• If not symmetric, the sum of external moments will not be zero even if =0 (i.e., steady-state):

• An asymmetric rotating shaft requires both static (=0) and dynamic (≠0) balancingto avoid excessive vibration and bearing reactions

© 2019 McGraw-Hill Education.

Concept Question

⁷

The device at the right rotates about the x axis with a non-constant angular velocity. Which of the following is true (choose one)?

a) You can use Euler’s equations for the provided x, y, z axes b) The only non-zero moment will be about the x axis

c) At the instant shown, there will be a non-zero y-axis bearing force

d) The mass moment of inertia I

_xx

is zero

© 2019 McGraw-Hill Education.

Concept Question

⁸

The device at the right rotates about the x axis with a non-constant angular velocity. Which of the following is true (choose one)?

a) You can use Euler’s equations for the provided x, y, z axes b) The only non-zero moment will be about the x axis

c) Answer: At the instant shown, there will be a non-zero y-axis bearing force

d) The mass moment of inertia I

_xx

is zero

Dongjun Lee

Sample Problem 18.5

A homogeneous disk of mass m is mounted on an axle OG of negligible mass. The disk rotates without sliding counter-clockwise at the rate ₁about OG. Determine: a) the reaction force from the floor (assume it is vertical);

and 2) reaction force at the pivot O.

SOLUTION:

• Floor reaction force computed from M_O of disk + axle (R at O not needed)

• Reaction force R at O requires other dynamics equations anyway though.

Dongjun Lee

Sample Problem 18.5

A homogeneous disk of mass m is mounted on an axle OG of negligible mass. The disk rotates without sliding counter-clockwise at the rate ₁about OG. Determine: a) the reaction force from the floor (assume it is vertical);

and 2) reaction force at the pivot O.

SOLUTION:

Choose 𝑖, 𝑗, 𝑘 along principal axes for simper expression of 𝑤 and 𝐼 , 𝐼 , 𝐼

Apply D’Alembert or momentum equation to solve equation of motion

Choose origin to avoid constraint force computation

Dongjun Lee

Sample Problem 18.3

Rod AB of weight W = 20 kg is pinned at Ato vertical axle rotating w/ constant angular velocity= 15 rad/s. The rod position is maintained by a horizontal wire BC. Determine tension in the wire and reaction at A.

SOLUTION:

𝑚𝐿if 𝛽 = 0 0if 𝛽 =

Dongjun Lee

Sample Problem 18.3

SOLUTION:

Rod AB of weight W = 20 kg is pinned at Ato vertical axle rotating w/ constant angular velocity= 15 rad/s. The rod position is maintained by a horizontal wire BC. Determine tension in the wire and reaction at A.

0if 𝛽 = 0 𝑜𝑟 𝛽 = (axis-symmetric)

• Can also use:

Dongjun Lee

Sample Problem 18.3

Rod AB of weight W = 20 kg is pinned at Ato vertical axle rotating w/ constant angular velocity= 15 rad/s. The rod position is maintained by a horizontal wire BC. Determine tension in the wire and reaction at A.

• Expressing that the system of external forces is equivalent to the system of effective forces, write vector expressions for the sum of moments about A and the summation of forces.

• Solve for the wire tension and the reactions at A.

Strategy:

• Evaluate the system of effective forces by reducing them to a vector attached at G and couple

a m

G. H

Dongjun Lee

Sample Problem 18.3

SOLUTION:

Evaluate the system of effective forces by reducing them to a vector attached at G and couplema

G. H

 



^{r I}



^{I L I}

a

a _n





 



2

2 21

2

s m 5 . 112

cos













   

   

^I

a

m 

N 2250 5

. 112

20 



k I j I i I

H_{G x x} _{y y} _{z z}





  



0 sin

cos

0 ₂^{1 2}

2 21















z y

x

z y

x mL I I mL

I















i mL

H_G 



cos

2 121





     

 

k

k mL

i mL j i

H H

H_{G GGxyz G}











 





m N 5 . 649 cos

sin

cos sin

cos 0

2 2 121

2 121











































Dongjun Lee

Sample Problem 18.3

• Expressing that the system of external forces is equivalent to the system of effective forces, write vector expressions for the sum of moments about A and the summation of forces.

 

A_eff

A M

M





^{  }

( ) ( ) ( )

( ) ( )

1.732 0.5 196.2 0.866 2250 649.5

1.732 98.1 1948.5 649.5

J TI I J J I K

T K K

? + ? = ? +

- = +

      

 

1557 N

 

^F_eff T =

F





^{ }

1613 196.2 2250

X Y Z

A I+A J+A K- I- J= - I

(

697 N

) (

196.2 N

)

A= - I+ J

Reflect and Think:

• You could have obtained the value of T from H_Aand Eq. (18.28). Even though the rod rotates with a constant angular velocity, the asymmetry of the rod causes a moment about the z axis. Note that we calculated the inertial term 𝐻̇ by adding 𝑟 × 𝑚𝑎 and the couple 𝐻̇

© 2019 McGraw-Hill Education.

Group Problem Solving

⁶

A four-bladed airplane propeller has a mass of 160 kg and a radius of gyration of 800 mm.

Knowing that the propeller rotates at 1600 rpmas the airplane is traveling in a circular path of 600-m radius at 540 km/h, determine the magnitude of the couple exerted by the propeller on its shaftdue to the rotation of the airplane.

Strategy:

• Determine the overall angular velocity of the aircraft propeller.

• Determine the mass moment of inertia for the propeller.

• Calculate the angular momentum of the propeller about its CG.

• Calculate the time rate of change of the angular momentum of the propeller about its CG.

• Calculate the moment that must be applied to the propeller and the resulting moment that is applied on the shaft.

© 2019 McGraw-Hill Education.

Group Problem Solving

⁷

Modeling and Analysis:

Establish axes for the rotations of the aircraft propeller

Determine the x-component of its angular velocity

2 rad 1600 rpm

60 s 167.55 rad/s

x

  ^  ^

 



Determine the y-component of its angular velocity

540 km/h 150 m/s

v 

150 m/s

0.25 rad/s 600 m

y

 v

 

Determine the mass moment of inertia

2 (160 kg)(0.8 m)2 102.4 kg m2

Ixmk   

© 2019 McGraw-Hill Education.

Group Problem Solving

⁸

Angular momentum about G equation:

GIxx Iyy

H i j

Time rate of change of angular momentum about G :

( ) 0 ( )

G G Gxyz  G y  Ixx Iyy

H H Ω H j i j

(102.4 kg m )(167.55 rad/s)(0.25 rad/s)2 G Ix x y   

H k k

(4289 N m) = (4.29 kN m)

G    

H k k

Determine the moment that must be applied to the propeller shaft

(4.29 kN m)

 G  

M H k

Determine the moment that must be applied to the propeller shaft

(4.29 kN m)

on shaft   

M M k

constant rotating

© 2019 McGraw-Hill Education.

Concept Question

⁹

The airplane propeller has a constant angular velocity +ω

_x

. The plane begins to pitch up at a constant angular velocity.

The propeller has zero angular acceleration.

TRUE FALSE

Which direction will the pilot have to steer to counteract the moment applied to the propeller shaft?

a) +x b) +y c) +z

d) −y e) −z

© 2019 McGraw-Hill Education.

Concept Question

¹⁰

The airplane propeller has a constant angular velocity + ω

_x

. The plane begins to pitch up at a constant angular velocity.

The propeller has zero angular acceleration.

TRUE Answer: FALSE

Which direction will the pilot have to steer to counteract the moment applied to the propeller shaft?

a) +x b) Answer: +y c) +z

d) −y e) −z

Dongjun Lee

Helicopter Gyroscopic Precession

https://youtu.be/2tdnqZgKa0E

Dongjun Lee

Aircraft Gyroscope

https://youtu.be/rrEb4_3i3Sw

Gyroscopes are used in the navigation system of the Hubble telescope, and can also be used as sensors (such as in the Segway PT). Modern gyroscopes can also be MEMS (Micro Electro- Mechanical System) devices, or based on fiber optic technology.

gyroscopic rigidity gyroscopic precession

Dongjun Lee

Gyroscope: Eulerian Angles

• A gyroscope consists of a rotor with its mass center fixed in spacebut which can spin freely about its geometric axis and assume any orientation.

• From a reference position with gimbals and a reference diameter of the rotor aligned, the gyroscope may be brought to any orientation through a succession of three steps:

a) rotate outer gimbal by (yaw) about AA’, b) rotate inner gimbal by (pitch) about BB’, c) rotate the rotor by (roll) about CC’.

• , , and  are called the Eulerian Angles and

nutation of rate

spin of rate

precession of

rate



















Dongjun Lee

Dynamics of Gyroscope

• Angular velocity of gyroscope expressed in {i,j,k}:

• Angular momentum equation of the gyroscope:

• Rotation dynamics of the gyroscope in {i,j,k}:

no effect from Ω × 𝐻!

Dongjun Lee

• Steady precession: 𝜃and 𝜙̇, 𝜓̇ are constant, i.e., yawing + spinning w/ fixed tilted angle .

Steady Gyroscopic Precession

• Rotation dynamics of steady precession:

i.e., gyroscope will make precession as the moment M_G

“precedes” the spin momentum 𝐼 𝜓̇𝑘

• When the precession and spin axes are at right angle =90:

To maintain steady precession, moment M_Gshould be perpendicularto the precession and the spin axes.

constant rotating

Dongjun Lee

Gyroscope Precession

Dongjun Lee

Spinning Top: 18.111-18.112

a) show the equation for steady precession is given by

b) if spin is much faster than precession rate, show that

c) two precession rates possible given constant 𝜃 and 𝜓̇

Dongjun Lee

Axisymmetrical Body with No Force

• Consider motion about its mass center of an axisymmetrical body under no force but its own weight(e.g., projectiles, satellites, spacecraft).

constant

0 

 _G

G H

H 

• Z-axis is the fixed direction of 𝐻 , z-axis is the line of symmetryrotating together with the body, x-axis is chosen to lie in the Zz-plane.

x G

x H I

H  sin  x^HG_I^sin 

y

y I

H  0  y^0

z G

z H I

H  cos 

I H_G

z 

  ^cos



 

 tan tan

I I

z x

 





• Angular velocity of the body can be written by:

• This will produce steady precession w/ constant

, 𝜓̇, 𝜙̇since w_y=0 (i.e., constant). Note also:

Dongjun Lee

Axisymmetrical Body with No Force

The motion of a body about a fixed point (or its mass center) can be represented by the motion of a body cone rolling on a space cone. In the case of steady precession the two cones are circular.

• I > I’. Case of a flattened body.  >  and the vector lies outside the angle ZGz. The space cone is inside the body cone; the spin and precession have opposite senses. The precession is said to be retrograde.



 

 tan tan

I I

z

x   



• I < I’. Case of an elongated body.  <  and the vector lies inside the angle ZGz. The space cone and body cone are tangent externally; the spin and precession are both counterclockwise from the positive z axis. The precession is said to be direct.

Dongjun Lee

Axisymmetrical Body with No Force

Dongjun Lee

Axisymmetrical Body with No Force

Two cases of motion of an axisymmetrical body which under no force which involve no precession:

Body set to spin about its axis of symmetry,

aligned are and

0

G x x

H H



  

and body keeps spinning about its axis of symmetry.

Body is set to spin about its transverse axis,

aligned are and

0

G z z

H H



  

and body keeps spinning about the given transverse axis.

Dongjun Lee

Axisymmetrical Body with No Force

Dongjun Lee

Sample Problem 18.6

18 - 60

a) determine angular velocity immediately after impact b) precession axis of the ensuing motion

c) rate of precession and rate of spin