MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
Physics 8.01
Problem Solving Session 11 Three Dimensional Rotation and Gyroscopes
Solutions
W14D3-1 Rotating Skew Rod Solution
Consider a simple rigid body consisting of two particles of mass m separated by a rod of length
2l and negligible mass. The midpoint of the rod is attached to a vertical axis that rotates with
angular velocity    k̂ about the z -axis. The rod is skewed from the vertical at an angle  .
Set time t  0 when the rod is in the position shown in figure below left. At t   /  the rod has
rotated to the position shown in the figure below right.
a) Find the direction and magnitude of the angular momentum about the center of mass at
t  0.
b) Find the direction and magnitude of the torque about the center of mass at time t  0 .
Solution:
a)
We use Lcm  r  p  r  mv for the angular momentum about the center of mass for each
particle:
Lcm  r1  p1  r2  p2  r1  mv1  r2  mv2
Since each particle travels at an angular speed  in a circular orbit of radius cos  , the speed
of each particle is given by v   l cos  . We choose a coordinate system shown in the figure
below
r
r
For particle 1: r1  l cos φ
i  l sin  kφ and v1  l cos φ
j . Thus
r
r
r
φ  (ml cos  φ
L cm,1  r1  mv1  (l cos  φ
i  l sin k)
j) .
After calculating the cross products, we have that the angular momentum about the center of
mass for particle 1 is
r
L cm,1  ml 2 cos  (cos  kφ  sin  φ
i) .
Note that for particle 2, r2  r1 and v 2   v1 , so
Lcm,2  r2  mv 2  r1  mv1  Lcm,1 .
Thus the angular momentum about the center of mass at time t  0 is given by
r
Lcm (0)  2ml 2 cos(cos kφ  sin  φ
i) .
The magnitude of the angular momentum about the center of mass is given by
Lcm  2ml 2 cos(cos2   sin2  )  2ml 2 cos .
b) At t  0 , the rod is rotating in the x  y plane. The figure below shows the orientation of the
rod as seen from above.
The z -component of the angular momentum about the center of mass is constant and the x component of the angular momentum about the center of mass is changing in time as the rod
rotates and is given by
r
L cm, x (0)  2ml 2 cos  sin  φ
i
The time derivative of the angular momentum about the center of mass is perpendicular the
angular momentum, points in the positive y -direction, and has a magnitude that is equal to
r
L cm,x (0)  . Therefore
r
r
dL cm
(0)  L cm,x (0)  φ
j  2ml 2 2 cos  sin  φ
j.
dt
The torque about the center of mass is given by
Therefore at t  0 , we that
r
dL cm
r
 cm 
dt
r
 cm  2ml 2 2 cos sin  φ
j.
W14D3-2 Table Problem Suspended Gyroscope Solution
A gyroscope wheel is at one end of an
axle of length d . The other end of the
axle is suspended from a string of length
s . The wheel is set into motion so that it
executes uniform precession in the
horizontal plane. The string makes a
fixed angle  with the vertical. The
wheel has mass M and moment of
inertia about its center of mass I cm . Its
spin angular speed is  . Neglect the
mass of the shaft and the mass of the
r
string. Assume    . What is the
direction and magnitude
precession angular velocity?
of
the
Solution: We shall choose as our system the wheel and axle but not the string. The force
diagram and coordinate system are shown in figure below.
The torque on the wheel with respect to the center of mass of the wheel is given by
 cm  rcm,T  T ,
(1)
In the coordinate system shown in figure below, rcm,T  d î and T  T sin  î  T cos  ĵ . So
the torque in Eq. (1) becomes
 cm  d î  (T sin  î  T cos  ĵ)  dT cos  ( î  ĵ)  d T cos  k̂ .
(2)
The direction of the torque is into the plane of figure (negative z-direction). This means that the
direction of the change of the angular momentum of the spinning wheel is also into the plane the
figure. Since the angular momentum is pointing outward (in the positive x-direction at the instant
shown in the figure below on the lerft), the change in the angular momentum points in the
negative z-direction as seen in the overhead view in the figure below on the right.
The wheel will precess counterclockwise about the vertical axis. Denote the wheel’s precessional
angular frequency by   d / dt . Note we choose the angle  to be increasing in the
counterclockwise direction and so   0 .
The spin angular momentum is directed along the axis of rotation and is equal in magnitude to
the product of the moment of inertia about the center of mass and the spin angular velocity. At
the instant depicted in the figure above, the angular momentum about the center of mass is
Lcm  I cm s î  Lcm,z k̂ .
(3)
r
dL cm
φ .
 I cm s  (k)
dt
(4)
The time derivative is then
Because
 cm 
dL cm
,
dt
(5)
substitute Eq. (2) into the left hand side of Eq. (5), and substitute Eq. (4) into the right hand side
of Eq. (5) yielding,
dT cos  k̂  I cm s  ( k̂) .
(6)
In Eq. (6), we have still not determined the tension of the string. We can use Newton’s Second
Law applied to the center of mass
F  macm .
(7)
The weight of the wheel must be balanced by the vertical component of the force supplied by the
string,
T cos   mg  0 .
(8)
Therefore the tension is
T
mg
.
cos 
(9)
We can substitute Eq. (9) into Eq. (6) and solve for the precessional frequency,

d mg
.
I cm s
(10)
Note that we are assuming the precession is uniform. This is our gyroscopic approximation and
requires us to ignore any “nutational” effects as well as requiring that  s   .
W14D3-1 Grain Mill Solution
In a mill, grain is ground by a massive wheel
that rolls without slipping in a circle on a
flat horizontal mill stone driven by a vertical
shaft. The rolling wheel has mass M , radius
b and is constrained to roll in a horizontal
circle of radius R at angular speed  . The
wheel pushes down on the lower mill stone
with a force equal to twice its weight
(normal force). The mass of the axle of the
wheel can be neglected. Express your
answers to the following questions in terms
of R , b , M ,  , and g as needed. The
goal of this problem is to find  .
a) What is the relation between the angular speed  of the wheel about its axle and the
angular speed  about the vertical axis?
b) Find the time derivative
of the angular momentum about the joint (about the point P in
r
the figure above) dL P / dt .
c) What is the torque about the joint (about the point P in the figure above?
d) What is the value of  ?
Solution: The figure below shows the pivot point along with some convenient coordinate axes.
For rolling without slipping, the speed of the center of mass of the wheel is related to the angular
spin speed by
vcm  b .
(11)
Also the speed of the center of mass is related to the angular speed about the vertical axis
associated with the circular motion of the center of mass by
vcm  R .
(12)
Therefore equating Eqs. (11) and (12) we have that
  R / b .
(13)
b) Assuming a uniform millwheel, I cm  (1/ 2) Mb 2 , the magnitude of the horizontal component
of the angular momentum about the center of mass is
Lcm,h  I cm 
1
1
Mb 2   MRb .
2
2
(14)
r
The horizontal component of L cm is directed inward, and in vector form Lcm  Lcm,h (rφ) in the
above coordinate system.
c) The axle exerts both a force and torque on the wheel, and this force and torque would be quite
complicated. That’s why we consider the forces and torques on the axle/wheel combination. The
normal force of the wheel on the ground is equal in magnitude to NWG  2mg so the third-law
counterpart, the normal force of the ground on the wheel has the same magnitude NGW  2mg .
The joint (or hinge) at point P therefore must exert a force FH,A on the end of the axle that has
two components forces an inward force F2 to maintain the circular motion and a downward
force F1 to reflect that the upward normal force is larger in magnitude than the weight.
d) About point P , FH,A exerts no torque. The normal force exerts a torque of magnitude
r
NGW R  2mgR , directed out of the page, or, in vector form,  P, N  2mgR φ . The weight exerts
r
a toque of magnitude MgR , directed into the page, or, in vector form,  P,mg  MgR φ. So the
torque about P is
r
r
r
 P   P, N   P,mg  2mgRφ  MgR φ   MgR φ.
(15)
As the wheel rolls, the horizontal component of the angular momentum about the center of mass
will rotate, and the inward-directed vector will change in the negative φ -direction.
Mathematically,
r
dLcm,h
dt
r
φ  1  MRb( )
φ,
 Lcm,h ( )
2
(16)
where we used Eq. (14) for the magnitude of the horizontal component of the angular momentum
about the center of mass. This is consistent with the torque about P pointing out of the page in
the above figure. We can now apply the torque condition that
P 
dLP
dt
(17)
that becomes using Eqs. (15) and (16)
φ  1 2 MRb( )
φ
MgR( )
2
(18)
We can now solve Eq. (18) for the angular speed about the vertical axis

2g
.
b
(19)