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:

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°

θ=38.5°

ω2

19.5°

γ=58°

φ

180 – γ

= 122°

φ = 12.4 rad

sec ψ = 32.7 rad

sec