AE 140: Rigid Body Dynamics

Exam #2

April 23, 2014

(30 pt) 1. Derive the equations of motion of a thin disk which spins ( )φ& about its own

axis and precesses ( )ψ& about the vertical axis, as shown. β and ψ& are constant. The rod

of length l is massless. Moments of inertia for a thin disk (about its own center of mass)

are given on the crib sheet.

Write the equations of motion about point “O” at the support.

Define three reference frames. N: Newtonian (fixed in support)

R: fixed in the rod (precesses with rod, but does not spin

with disk)

D: fixed in the disk (spins and precesses)

β

O

nz

l

φ&ψ&

r

, β = constantψ&

rz

rx

mg

β

O

nz

l

φ&ψ&

r

, β = constantψ& , β = constantψ&

rz

rx

mg

(30 pt) 2. A missile is spinning and precessing with n = φc ψω θ3&& += = 10π rad/sec.

It precesses with nutation angle θ = 5°about a fixed direction in space.

10

1

A

C=

a) Determine the precession speed ψ& and body cone angle γ.

b) Draw the space and body cones.

(30 pt) 3. Calculate the ground track of a point on the top edge of a cylinder in free

rotation. Compute the position only for t = 1 sec and t = 2 sec.

Let No correspond to the center of mass of the cylinder. Model the cylinder’s motion

with the 3-1-3 Euler angles that we use in class (Chapter 4):

i) precession, ψ ψ, & , about +nz, then define intermediate frame C

ii) nutation, θ θ, & , about +cx, then define intermediate frame G

iii) spin, φ φ, & , about +gz, then define body reference frame B

Cylinder characteristics:

height = 3 m

radius = 1 m

mass = 1 kg

theta = 30 deg (constant)

sec

deg 5φ =& (constant)

(10 pt) 4. Show that a thin disk (a Frisbee) thrown so that its plane remains almost

parallel to the ground (cos θ ≅ 1) will wobble (precess) twice as fast as it spins. Also,

show that precession and spin will have opposite senses.

AE140

Exam #2 Crib Sheet

1. Golden Rule of Vector Differentiation: vωvv ×+= BA

dt

dB

dt

dA

2. Motion with respect to the rotating Earth:

In the Northern Hemisphere : Ω Ω Ω Ω = Ω nz = (− Ω cos λ) ex + (Ω sin λ) ez

Ω = 0.729 x 10-4 rad/sec

3. For a planar rotation through angle θ:

fx fy fz

rx cθ sθ 0

ry - sθ cθ 0

rz 0 0 1

NaQ = ΩΩΩΩ × (ΩΩΩΩ × R) + EaQ + ΩΩΩΩ × (ΩΩΩΩ × EpQ) + 2 ΩΩΩΩ × EvQ

nx

R

nz

ΩΩΩΩ

No

Eo

ny

λ

ez

ex

ey

equator

nx

R

nz

ΩΩΩΩ

No

Eo

ny

λ

ez

ex

ey

equator

fx

rx

fy

ry

θ

θ

fx

rx

fy

ry

θ

θ

If θ is a function of time: FωωωωR = θ& rz = θ& fz

dt

d sθ = θ& cθ

4. Matrix multiplication:

[3 × 3 matrix] times [3 × 1 matrix] = [3 × 1 matrix]

333231

232221

131211

aaa

aaa

aaa

31

21

11

b

b

b

=

)ba ba b(a

)ba ba b(a

)ba ba b(a

31 33 21 32 11 31

31 23 21 22 11 21

31 13 21 12 11 11

++

++

++

[3 × 3 matrix] times [3 × 3 matrix] = [3 × 3 matrix]

33 3231

232221

131211

aaa

aaa

aaa

333231

232221

131211

bbb

bbb

bbb

=

)ba ba b(a)ba ba b(a )ba ba b(a

)ba ba b(a)ba ba b(a)ba ba b(a

)ba ba b(a)ba ba b(a)ba ba b(a

33 33 23 32 13 3132 33 22 32 12 3131 33 21 32 11 31

33 23 23 22 132132 23 22 22 122131 23 21 22 1121

33 13 23 12 13 1132 13 22 12 12 1131 13 21 12 11 11

++++++

++++++

++++++

5. Moments of Inertia

Ιx = ∫Vol

(y2 + z2) dm Ιy = ∫Vol

(x2 + z2) dm Ιz = ∫Vol

(x2 + y2) dm

Ιx =∑i

im (yi2 + zi

2) Ιy =∑i

im (xi2 + zi

2) Ιz =∑i

im (xi2 + yi

2)

6. Products of Inertia

Ιxy = − ∫Vol

xy dm Ιxz = − ∫Vol

xz dm Ιyz = − ∫Vol

yz dm

Ιxy = −∑i

im xiyi Ιxz = −∑i

im xizi Ιyz = −∑i

im yizi

7. Angular Momentum Vector

N

HB/Bo = Ixωx + Ixy ωy + Ixz ωz bx

+ Ixy ωx + Iy ωy + Iyz ωz by

+ Ixz ωx Iyz ωy + Iz ωz bz

8. Newton’s Law of Rotational Motion:

MBo = N

dt

d NH

B/Bo =

B

dt

d NH

B/Bo +

NωωωωB × N

HB/Bo

9. To transform mass properties:

[ ]I' =

z'z'y'z'x'

z'y'y'y'x'

z'x'y'x'x'

III

III

III

[ l B-B’ ] =

zz'zy'zx'

yz'yy'yx'

xz'xy'xx'

lll

lll

lll

[ I ] = [ l B-B’ ] [ I’ ] [l B-B’ ]T

After multiplying the three matrices together, you will get: [ ]I =

zyzxz

yzyxy

xzxyx

III

III

III

10. For constant precession and spin:

ψ& = θ cos C)(A

φC

−

& tan θ =

C

A tan γ

11. General Gyro Equations

(a) l W sθ = A θ&& + C n ψ& sθ − A sθ cθ ψ&2

(b) 0 = Aψ&& sθ + 2 A ψ& θ& cθ − C n θ&

(c) φ& + ψ& cθ = n = constant

12. Mass Properties of a Body of Revolution

r

axial

transverse

h

( )22

transverse

2

axial

312

mI

mr2

1I

hr +=

=

Cylinder

r

axial

transverse

h

( )22

transverse

2

axial

312

mI

mr2

1I

hr +=

=

Cylinder

raxial

transverse

Thin Disk

2

transverse

2

axial

mr4

1I

mr2

1I

=

=r

axial

transverse

Thin Disk

2

transverse

2

axial

mr4

1I

mr2

1I

=

=