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 w_{z} 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}/_{3}Ma^{2}.

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

ω_{1} = – ω_{z} sinα ω_{3} = ω_{z} cosα
By parallel axes,

I_{11} = I_{22} = ^{4}/_{3} Ma² + 2Mb²
I_{33} = ^{4}/_{3} Ma²Therefore:

L_{1} = – (^{4}/_{3} Ma² + 2Mb²) ω_{z} sinα
L_{3} = ^{4}/_{3} Ma² ω_{z} 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** × M**v** = **r** × M(**ω** × **r**). This is perpendicular to **r**, in fact in the –1 - direction. The spin is ω_{z} in the Z-direction, and since the sphere is completely symmetric, the spin angular momentum is in that same direction, ^{2}/_{3} Ma²ω_{z}.

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

b. Calculate the **vector** d**L**/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

d**L**/dt = **ω** × **L** = (ω_{3}L_{1} – ω_{1}L_{3})**j**
= – 2Mb² ω_{z}² sinα cosα
so the magnitude of the torque is Γ = |d**L**/dt| = 2Mb² ω_{z}² 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, Z_{CM} = -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 ½**ω**·**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 m_{1} and m_{2} 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 m_{1}, m_{2}, 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 = (m_{2}/M)a where M = m_{1} + m_{2}:

m_{1}rω² = m_{1}m_{2}aω²/M = Gm_{1}m_{2}/a² hence ω² = GM/a³
But ω = 2π/τ so the period τ = 2π a^{3/2}(GM)^{-½}.

You can also do it as an equivalent 1-body problem, with a mass μ = m_{1}m_{2}/M moving on a circle of radius a under the force Gm_{1}m_{2}/a².

b. the **total energy** of the system
QIt's easiest to appeal to the virial theorem, E = ½U = –Gm_{1}m_{2}/2a. More explicitly, E = T + U = ½μ(aω)² – Gm_{1}m_{2}/a.