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Wobbling Disk Problem A thin, uniform disk of mass m and radius a is attached to an axle which passes through its center; a normal to the disk makes an angle θ with the axle, as shown in the figure. The axle, which can be considered to be massless, is supported a distance d from the center of the disk by bearings which apply a force F to the axle. The axle rotates with an angular frequency ω. What is the force required to maintain this uniform motion? Ignore the weight of the disk (i.e., don’t worry about the gravitational force acting on the disk).
y
z
x
θ
3
z
x
Side view of disk; 2 and y are out of the page
1
F
F
d d
(^3) ω
2
1
θ
Solution. Let’s first define an appropriate set of fixed and body coordinates, and find the relationship between them. For the fixed coordinates, chose (x, y, z) as shown in the figure, with z along the axle and x and y in the plane perpendicular to the axle (let the x − y plane pass through the center of the disk). In terms of these coordinates, the angular velocity of the disk is ω = ω ez. (1)
Let the body coordinates (attached to the disk) be (1, 2 , 3), with the origin at the center of the disk. Let e 3 be normal to the disk with e 1 and e 2 in the plane of the disk; e 2 lies along a line defined by the intersection of the disk with x − y plane, and e 1 and e 3 lie in a plane which is perpendicular to e 2 and which rotates counterclockwise about the z axis with an angular frequency ω. With these conventions, e 2 always rotates in the x − y plane with no
projection along ez. The figure below shows the projection of the unit vectors (e 1 , e 2 , e 3 ) onto the x − y plane after a time t; e 2 and e 3 make an angle ωt with the x-axis, and e 2 makes an angle ωt with the y-axis. We are now in a position to write down the relationship
x
y
z-axis is out of the page
ω t
e
e
2
3
e (^) 1
between the two coordinate systems:
e 1 = cos θ(cos ωt ex + sin ωt ey) − sin θ ez , (2)
e 2 = − sin ωt ex + cos ωt ey, (3) e 3 = sin θ(cos ωt ex + sin ωt ey) + cos θ ez. (4)
It’s easy to verify that the body coordinates are orthonormal. We also have
ex = cos θ cos ωt e 1 − sin ωt e 2 + sin θ cos ωt e 3 , (5)
ey = cos θ sin ωt e 1 + cos ωt e 2 + sin θ sin ωt e 3 , (6) ez = − sin θ e 1 + cos θ e 3. (7)
Therefore, from Eq. (1), we see that the angular velocity in the body frame is
ω = ω(− sin θ e 1 + cos θ e 3 ). (8)
Notice that the angular velocity is also constant in the body frame. This is what we should expect, since the body frame is rotating with an angular velocity ω with respect to the fixed frame, so that the time derivative of ω is
dω dt
δω δt
δω δt
so the torque has a constant magnitude and direction in the body frame. Using Eq. (3) to relate e 2 to the unit vectors in the fixed frame, we see that our two calculations (in the fixed and body frames) agree. The torque is supplied by the reaction of the bearings on the ends of the axle. The forces are in the x − y plane, and are perpendicular to N; the magnitude of the torque which they produce is 2F d, and this must equal (1/4)ma^2 ω^2 sin θ cos θ, so we find
F =
ma^2 ω^2 8 d
sin θ cos θ. (18)
The force vanishes when θ = 0 (spinning the disk about the symmetry axis) and θ = π/ 2 (spinning the disk about a diameter). In these situations the angular momentum is either parallel or antiparallel to the angular velocity, and we have a state of dynamic balance.
