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:
rev 2π rad min rad ω1 = - 120
×
× 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
2 rad cm rad 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°
4π
−1
Law of Sines
φ ψ sin (180° − γ ) sin θ sin (γ − θ )
=
= ψ ω2 φ 38.5°
