# Essay about AE140 S14 Problem 6

Submitted By Jared-Sagaga
Words: 379
Pages: 2

AE140: Rigid Body Dynamics
Problem 6.5.2 Solution

bz
A frisbee is flying with its plane parallel to the ground. It spins at 120 rpm.

ω1 = 120rpm r I axial

1 2
= mr
2

by

bx

For a thin disk:
I transverse

1 2
= mr
4

r = 20 cm mass = mf

vacorn = 10 m/sec massacorn = 0.1 mf

bz

by bx Angular momentum in state 1:

×
× b z = −4π bz min rev 60sec sec 1
2
Ix = Iy = mf (20cm) = 100m f cm2
4
1
2
Iz = mf (20cm) = 200m f cm2
2
H 1 = Iz ω1 b z = (200 mf )cm (- 4 π )
= −800 π mf bz sec sec Angular momentum is not conserved.

H 2 = H 1 + r × Imp 1→2

Due to the impulse of the acorn strike, angular momentum is added to the system.

where : Imp 1→2 = (mass acorn )(velocity acorn ) cm 

= (0.1m f ) − 1000 bz  sec 

cm 

r × Imp 1→2 = (- 20cm b y )× (0.1m f ) − 1000 bz  sec 

cm2
= 2000m f bx sec
H 2 = H 1 + r × Imp 1→2

cm2
= {− 800 π mf b z + 2000m f b x } sec Solve for nutation angle, θ, using components of angular momentum

2000m f θ = tan
= 38.5°
800 πmf
−1

Angular velocity after the acorn strike:

ω2 = ωx 2 b x + ωz2 b z
H2
ω2 =
=
I

2 cm (2000 mf )

sec b + x 2
(100 mf )cm

= {20 b x − 4 π b z }

2 cm (− 800 π mf )

sec b z 2
(200 mf )cm

sec

Solve for body cone angle, γ, using components of angular velocity

γ = tan

−1

ωx 2 ωz2 20
= tan
= 58°

−1

Law of Sines

φ ψ sin (180° − γ ) sin θ sin (γ − θ )
=
= ψ ω2 φ 38.5°

θ=38.5°

ω2
19.5°

γ=58°

φ

180 – γ
= 122°