## Conservation of Angular Momentum

We have previously seen that, for a system of particles, conservation of moment of momentum dictated that

in which Ho and Hc refer to the moment of momentum with respect to fixed point O and center of mass, respectively. The symbols Mo and Mc represent, respectively, the resultant moment of external forces about point O and the center of mass.

When a rigid object undergoes planar motion, we have seen that its moment of momentum (angular momentum) can be expressed by the following simple equations:

When one substitutes these expressions into Eqn. 4.28 depicting the conservation of angular momentum, we arrive at the following equation:

As we noted before, a is the angular acceleration of the rigid body B and Io, the mass moment of inertia with respect to point O that is fixed in E, is related to Ic by the following equation:

Io = mr2 + Ic in which r is the distance between center of mass of the object and point O.

Equation 4.30 is the fundamental equation governing the rate of rotation of a rigid object. It is similar in structure to the equation governing the motion of center of mass, recapitulated below:

where m is the mass of the rigid body, ac is the acceleration of the center of mass, and F is the resultant external force acting on the rigid body. It is clear from Eqns. 4.30 and 4.31 that mass is the measure of an object's resistance to uniform acceleration, and mass moment of inertia is the measure of a rigid body's resistance to changes in the rate of rotation. The resultant external force determines the path of the center of mass during motion, whereas the resultant external moment is responsible for a change in the angular velocity of a rigid body undergoing planar motion. Equations 4.30 and 4.31 comprise a complete set of equations of motion for a rigid object. In the following, we present a number of examples that illustrate the use of these equations.

Example 4.4. Swinging of a Disk Around a Fixed Axis. A disk of radius r and mass m is hinged at point O. It is otherwise free to swing under the application of an external force (Fig. 4.9). Let \$ be the angle that the radial line of the disk passing through point O makes with the vertical axis. Derive the equations of motion of the disk.

Solution: The free-body diagram shown in Fig 4.9 illustrates the forces acting on the disk. These forces are the gravitational force —mg e2 acting at the center of mass of the disk and the reaction force (F1 e1 + F2 e2) exerted by the pin at point O on the disk B. The only moment acting on the disk is the moment of the gravitational force, and through vector multiplication it can be shown to be given by the following equation:

The moment of inertia of the disk with respect to its center of mass is Ic = mr2/2 (see Appendix 2). Substituting this value into Eqn. 4.27 we find Io = 3mr2/2. Conservation of angular momentum with respect to point O yields:

In the case of small oscillations, sin \$ can be replaced by \$. The small angle swinging solution for Eqn. 4.33 is then equal to

\$ = A sin [(2g/3r)1/2 t] + B cos [(2g/3r)1/2 t] (4.34)

where A and B are arbitrary constants to be determined by the initial conditions. For the initial conditions \$ = v/6 and (d\$/dt) = 0 at t = 0, we find that A = 0 and B = ^/6. This is equivalent to the frequency of a pen- Figure 4.9a,b. A pendulum composed of a disk with uniform distribution of mass (a). Free-body diagram of the disk showing external forces acting on it is also illustrated (b).

F2e2

-mge2

dulum (weightless rod and bob with mass m) with length equal to rg = 3r/2. Reaction forces acting on the disk at the pin joint O can be computed by using the law of motion for the center of mass, as was illustrated for the swinging of a rod in the previous example.

Example 4.5. Rotation of an Airborne Disk. Let us consider the rotation of a disk as it moves in air. It is given an initial angular velocity of w e3 before it becomes airborne. How does its angular velocity change as it traverses a parabolic path in air?

Solution: Because the resultant gravitational force acts on the center of mass, it creates no moment with respect to the center of mass. The angular momentum with respect to the center of mass must remain constant. According to Eqn. 4.30b, the rotational velocity of an object moving in air is constant and this constant is equal to the initial angular velocity at the time the object becomes airborne.

Example 4.6. An Airborne Rod Breaks into Two. A rod of length L and mass m rotating counterclockwise in air with angular velocity w e3 sud denly breaks into two parts. Determine the angular velocity of the two pieces of rod immediately after the breakup. Assume the rods to be of equal length (L/2).

Solution: Before the rod breaks into two, its angular momentum with respect to its center of mass is given by the following equation:

in which the term on the right in parentheses is the mass moment of inertia of the rod with respect to its center of mass. Because the only external force that acts on the object before, during, and after the break is the force of gravity and this force creates no moment with respect to the center of mass, angular momentum Hc must be conserved. Immediately after the breakup, the original rod is replaced by a system composed of two rigid bodies. Using Eqn. 4.22 for each of the two rods and summing the two equations, the moment of momentum about point C immediately after the break can be shown to be

in which of e3 is the angular velocity of the two rods after the break, and the term in the brackets represents the moment of inertia of one-half of the original rod with respect to its end that coincides with the center of mass of the system. Equating these equations we find that of = o

Thus, immediately after the breakup, the two rods rotate with the same angular velocity as before the breakup. 