Physics 410 Second Hour Exam -- Solution April 21, 2006



1. A rigid body is made of two hollow spheres each of mass M and radius a rigidly attached to the ends of a massless rod of length 2(b-a) (so that the spheres' centers are a distance b from the origin O). The body is rotating with constant angular velocity wz about a massless axis OZ fixed in space such that the rod makes an angle a with this axis. O is the mid point of the rod.
The moment of inertia of a hollow sphere about its center is 2/3Ma2.

a. Calculate the angular momentum L of the rigid body about O, using a set of principal axes, 1, 2, 3, where OZ is in the 1, 3 plane (the plane of the picture). Draw L in the picture in approximately the correct direction (assume a << b, for the drawing only).

We have to express ω and I in principal axes. By projection

ω1 = – ωz sinα                     ω3 = ωz cosα
By parallel axes,
I11 = I22 = 4/3 Ma² + 2Mb²                     I33 = 4/3 Ma²
Therefore:
L1 = – (4/3 Ma² + 2Mb²) ωz sinα                     L3 = 4/3 Ma² ωz cosα
If a is negligible, L points in the negative 1-direction.
Another way to do this is to break the total momentum up into orbital and spin. Orbital is just that of the CM for each of the masses, r × Mv = r × M(ω × r). This is perpendicular to r, in fact in the –1 - direction. The spin is ωz in the Z-direction, and since the sphere is completely symmetric, the spin angular momentum is in that same direction, 2/3 Ma²ωz.

Answer either b.(quantitative) or c.(qualitative):

b. Calculate the vector dL/dt at the moment shown in the picture, hence find the magnitude of the torque that the bearings must exert on the rigid body. Check that your answer is reasonable for the case a = 0 and a = p/2.

We have

dL/dt = ω × L = (ω3L1 – ω1L3)j = – 2Mb² ωz² sinα cosα
so the magnitude of the torque is Γ = |dL/dt| = 2Mb² ωz² sinα cosα. When α = 0, the spin axis is the principal 3-axis, when α = π/2, the spin is about the principal 1-axis. In either case L is parallel to ω, so it does not change in time, and the expression for Γ does vanish for those values. (To check the direction of the torque, note that the –j direction means counterclockwise, which is needed to provide the necessary centripetal force on the masses -- or, from the point of view of the rotating body, to balance the centrifugal force on the masses.)

c. What will happen if the bearings are suddenly removed at the moment shown in the picture? Ordinary gravity (g) is acting. Specify both the motion of the center of mass, and about the center of mass. In particular, will w suddenly change when the bearings are removed? Will it change later on? Will its magnitude change?

c. The bearings provide torque and of course also prevent the body from falling. So, without bearings, the body's center of mass (at O) would fall straight down, ZCM = -½gt². Without torque, L would no longer change, it would remain constant in magnitude and direction. Since the position of the body cannot suddenly jump, ω must also initially be the same (along the Z-axis). But because the body is not spinning about a principal axis and it has angular momentum, the angular position of the body (and of the principal axes) must change later on. To keep L constant, then, ω must change. But the kinetic energy of rotation ½ω·L is also constant. Therefore (by further argument that the endpoint of ω must lie on an ellipsoid as well as on a plane, which argument you are not required to give) the magnitude of ω will not change.
The simple case a = 0 (point particles) looks complicated and possibly confusing from this point of views: The velocities of the two point particles will stay the same immediately after the bearings are removed, but those velicites can be described as a rotation about L, so the particles will later continue rotating about L. But that means that ω does change at the moment the bearings are removed. The answer is that for a rigid body made of two point particles, the component of ω along the 3-axis is not defined, it can be assumed to be zero or non-zero, it has no effect on L.



2. A binary star system has masses m1 and m2 moving under each other's gravitational attraction in such a way that the distance a between them stays constant. Find the following quantities in terms of m1, m2, a, G:

a. the period of the system

You can do this by considering the centripetal force on one body, say the first, as if moves about the common center of mass, from which it has a distance r = (m2/M)a where M = m1 + m2:

m1rω² = m1m2aω²/M = Gm1m2/a²   hence   ω² = GM/a³
But ω = 2π/τ so the period τ = 2π a3/2(GM)-½.
You can also do it as an equivalent 1-body problem, with a mass μ = m1m2/M moving on a circle of radius a under the force Gm1m2/a².

b. the total energy of the system QIt's easiest to appeal to the virial theorem, E = ½U = –Gm1m2/2a. More explicitly, E = T + U = ½μ(aω)² – Gm1m2/a.