MATH 114 - EXTRA CREDIT PROBLEM # 5 Precession of spinning

MATH 114 - EXTRA CREDIT PROBLEM # 5
Abstract. You can turn in this problem anytime before the final exam. Work on extra credit
problems will be used to decide cases in which a course grade is in between two letter grades.
Precession of spinning wheels I
This problem is about what happens when you first let go of one end of the axle of a spinning
wheel when the other end is suspended by a rope. The problem will show that the axle starts to
rotate about the end attached to the rope. In the next extra credit problem, the motion of the
wheel once it has started to precess will be analyzed.
Suppose that a wheel starts off by spinning parallel to the y − z plane in R3 with the center of
the axle at the point (a, 0, 0) with a > 0. Suppose that the radius of the wheel is R and that it is
spinning at angular velocity ω radians per second in a counterclockwise direction as you look down
at it from a point on the x-axis with large x coordinate.We take the axle to be fixed and lying along
the x-axis for the moment (so we are not letting the axis go yet).
a. Show that if we hold the axis of the wheel fixed along the x-axis, then
r(t, θ) = (a, 0, 0) + R · (0, cos(ωt + θ), sin(ωt + θ))
describes the position of the part of the rim at time t which at time t = 0 is at position
r(0, θ). (Notice the rate of change of ωt with t is just the constant ω; this is the meaning
1
2
MATH 114 - EXTRA CREDIT PROBLEM # 5
of an angular velocity of ω radians per second.) Find the velocity vector
v(t, θ) =
d
r(t, θ)
dt
of this part of the rim at time t by differentiating the components of r(t, θ) with respect to
t (and keeping θ fixed).
b. Suppose ∆θ is some small positive number. If the wheel has a constant density ρ, the mass
of the small piece of the wheel between positions r(t, θ) and r(t, θ + dθ) is approximately
m = ρR∆θ since the length of this piece is approximately R∆θ.
For small ∆θ, the angular momentum of this piece is approximately
(0.1)
A(t, θ, ∆θ) = m (r(t, θ) × v(t, θ)) = (r(t, θ) × v(t, θ)) ρR∆θ
Use this and part (a) to compute the angular momentum A(t, θ, ∆) of this piece of the rim.
c. Subdivide the wheel into small pieces each of which have length R∆θ. Add up the angular
momentum vectors found in part (b) for each of these small pieces, and let ∆θ → 0. This
leads to an integral which gives the total angular momentum vector of the wheel:
Z 2π
(0.2)
A(t) =
(r(t, θ) × v(t, θ)) ρRdθ
θ=0
Here you should integrate each component of the vector in the integrand to get the corresponding component of A(t). Use the formula for the components of the integrand you
found in part (b) to find A(t) explicitly. Your answer should show that A(t) points outward
along the x-axis, which is the axis about which the wheel is spinning. This answer is correct
only if we assume that the axis of the wheel is pointing outward along the x-axis at all
times.
d. Suppose now that we suspend the wheel from a rope which runs down along the z axis to
the origin. At time t = 0 we let the axis of the wheel go and allow the (downward) force of
gravity to act.
Let Ã(t) be the (new) angular momentum vector of the wheel at time t. Then at t = 0
we will have Ã(0) = A(0), but as t increases there will be a difference between Ã(t) and
MATH 114 - EXTRA CREDIT PROBLEM # 5
3
A(t) since we will let the axis of the wheel move. By extra credit problem #1,
d
Ã(t) = τ (t)
dt
where τ (t) is the torque acting on the wheel. Thus for a small time interval ∆t, we should
expect
Ã(∆t) ∼
= Ã(0) + τ (0)∆t = A(0) + τ (0)∆t
so that τ (0) gives the direction in which Ã(t) moves when t = 0.
The forces acting on the wheel are then an upward force acting along the rope on the
axle and the downward force of gravity at each point on the rim. Show that the torque
acting on the system as a result of the upward force along the rope is 0 since this force acts
at the origin, which has distance 0 from itself.
e. At time t = 0, the downward force of gravity acting on the piece of the rim discussed in
part (b) is the approximated by the vector
F (0, θ, ∆θ) = (0, 0, −g)ρR∆θ
where g is the gravitation constant and ρR∆θ is approximately the mass of this piece of
the rim. Find a formula for the torque
r(0, θ) × F (0, θ, ∆θ) = r(0, θ) × (0, 0, −g)ρR∆θ
acting on this part of the rim at time t = 0. Adding up over these pieces and letting ∆θ → 0
gives the formula
Z
2π
r(0, θ) × (0, 0, −g)ρRdθ
τ (0) =
θ=0
for the torque at time t = 0. Show that this torque is a vector perpendicular to A(0) = Ã(0)
which is parallel to the y-axis. This means that Ã(t) at t = 0 will move in the direction
of the torque τ (0), leading to precession of the wheel. Will the axis of the wheel move
clockwise or counterclockwise about the z-axis when viewed from above (i.e. from a point
corresponding to large values of z along the z-axis)?