Download    

Let the axes be (x,y,z), where z is the one perpendicular to the plane of the body at hand,
and let the moment of inertia for rotation around the z axis be Iz. Then-

I z   r2 dm    x 2  y 2  dm   x 2 dm   y 2 dm    x 2  02  dm    y 2  02  dm
Since on the plane the z coordinate is zero, this corresponds with-
Iz  Ix  I y


 
   
  
   

       

  
1
In a galaxy far far away there is a small, strange planet shaped like a cone. This planet has mass M , base radius R,
opening angle α and it rotates about its axis with an angular velocity ω.
A spherical alien of mass m << M and radius h lands on the planet, a distance ρ >> h from its axis.
a) At what velocity vector should the alien walk to avoid feeling dizzy, i.e. to feel no fictitious forces?
b) Why isn’t the answer simply walking precisely against the rotation, i.e. remaining stationary in an inertial frame?
c) The cone’s rotation gradually accelerates, ω = ω(t). How should the alien walk now to avoid dizziness?
d) Calculate the moment of inertia for the rotation of the planet about its axis (neglect the alien’s effect on it).
e) The alien suddenly realizes that the acceleration is its own fault! Show that it can bring the cone’s angular velocity
from ω to zero by moving at a constant inertial frame velocity v for (3M/10m)(Rω/v)2 radians in the direction of the
cone’s rotation. The alien moves by exerting equal force forward and also downwards, just to keep it barely in touch
with the surface. Neglect dissipative losses and the alien’s dizziness, and note that some external force is needed, e.g.
rocket propellant.
Solution
a+b+c) Adopt cylindrical coordinates {z, ρ, φ} with the origin at the cylinder’s tip and ω
~ = ω~ẑ.
The location of the alien is given by ~r = r cos(α/2)ẑ + ρρ̂, where ρ ≡ r sin(α/2). The fictitious forces ared~
ω
F~ = −m~
ω × (~
ω × ~r) − 2m~
ω × ~v −
× ~r
dt
(1)
ω̇
ρ and vφ = − ω2 ρ.
giving vρ = − 2ω
Hence, when the cone accelerates, the alien should walk faster in the φ direction, in proportion to ω(t), but also walk
towards the cone’s point, such that the centrifugal force doesn’t grow too fast.
The factors of 1/2 (which appear also when walking on rotating disks and spheres) may seem non-intuitive. Consider
vφ : it implies vφ,in = +ωr/2 in the inertial frame. The centripetal acceleration of the alien in the rotating frame,
−(vφ /r)2 r = −ω 2 r/4, precisely equals its centripetal acceleration in the inertial frame, i.e. no fictitious forces. Can
you similarly explain the factor 1/2 in vr ?
R 2
3M
2
d)We need to calculate I = ~r⊥
dm. Let dm = ρM dV = πR
r⊥
= ρ2 . Now2 H (NOTE THE DIFFERENT ρ!!) and ~
Z
3M
I=
πR2 H
6πM
ρ ρdρdzdφ =
πR2 H
2
Z
H
Z
z tan(α/2)
ρ3 dρ =
dz
0
0
3
M R2
10
(2)
e) To stop the rotation, the alien must do work by exerting a torque on the cone. The total angular momentum in
the system is conserved, so the alien must somehow get rid of angular momentum, for example by firing a rocket or
rotating its gyroscopes.
(c)
As the alien exerts a force Fφ
on the cone, the cone exerts an equal but opposite force on the alien. For the alien to
(ext)
maintain its angular velocity vφ =const., it must exert an additional force Fφ
(tot)
(c)
= Fφ , so in total the alien exerts
(c)
an azimuthal force Fφ
= 2Fφ .
By assumption, the alien exerts a minimal force downward, only to keep it attached to the surface, so the normal
force is negligibly small, and Fρ = −mvφ2 /ρ is responsible for the centripetal acceleration. Note that for friction, this
(tot)
is possible if µ 1, e.g. if the alien pushes strongly against irregularities in the surface. By assumption Fφ
so the work on the cone isZ
W =
F(c) · dr =
Z
φ0 +∆φ
φ0
(c)
Fφ ρdφ =
Z
φ0 +∆φ
φ0
−Fρ
ρdφ =
2
Z
φ0 +∆φ
φ0
mvφ2
mv 2
ρdφ =
∆φ ,
2ρ
2
= −Fρ ,
(3)
where ∆φ is the angular distance travelled. The cone’s rotational energy is initially Iω 2 /2, where the moment of inertia
along the axis is easily computed or looked up to be I = (3/10)M R2 . Neglecting energy losses, mv 2 ∆φ/2 = Iω 2 /2
can be solved for ∆φ.
Gyroscopic Motion:
For solution see Kleppner & Kolenkow, Sec. 7.3 and Example 7.7