If the system is an isolated one, no net torque acts on the object. Thus the angular momentum of the object must remain constant. Since L = Iσ , if I is doubled, σ must be halved. Thus the final angular velocity is equal to one half its original value.

We solve this problem using the principle of conservation on angular momentum. Initially the angular momentum of the system is entirely from the rotating disk: L _o = Iσ = 10I , where I is the moment of inertia of the rotating disk. When the second disk is added, it has the same moment of inertia as the first one. Thus I _f = 2I . With this information we can use conservation of angular momentum:

Thus the two disks have a final angular velocity of 5 rad/s, exactly half the initial velocity of the single disk. Notice that we got this answer without knowing either the mass of the disks or the moment of inertia of the disks.

Comets travel in wide elliptical paths, approaching the sun almost head on, then quickly rotating around the sun, and travelling back into space, as shown in the figure below:

To compute angular momentum, we can take the sun as our origin. As the comet approaches the sun, its radius, and thus its moment of inertia, decreases. To conserve angular momentum, then, the angular velocity of the comet must increase. In this way, the velocity of the comet increases as it approaches the sun.

As the string winds around the peg, the radius of rotation of the particle decreases, causing a decrease in the moment of inertia of the particle. The tension in the string acts in the radial direction, and thus does not exert a net force on the particle. Thus momentum is conserved and, as the moment of inertia of the particle decreases, its speed increases. Recall that v = σr . Thus the initial angular velocity of the particle is σ _o = v/r = 3 rad/s. In addition, the initial moment of inertia of the particle is I _o = mR ² = 4m . We want to find r , the radius of the string when the particle has a speed of 20 m/s. At this point, the angular velocity of the particle is σ _f = v/r = 20/r and the moment of inertia is I _f = mr ² . We have the initial and final conditions of the problem, and need only apply the conservation of angular momentum to find our value for r :

.4 meters of the string has wound around the peg when the velocity of the particle is 20 m/s.

Just as we used conservation of linear momentum to solve linear collisions, we use conservation of angular momentum to solve angular collisions. Firstly, we define the positive direction as the counterclockwise direction. Thus the total momentum of the system is simply the sum of the individual angular momenta of the particles:

Since the two particles move in opposite directions,

After they collide, the mass of the two particles together is 3 kg, and thus the large particle has a moment of inertia of 3r ² , and a final angular velocity of v _f/r . Thus L _f = (3r ²)(v _f/r) = 3rv _f . Since no net external force acts on the system, we can use the conservation of angular momentum to find v _f :

Thus the final particle has a velocity a third of the initial velocity of each particle, and it moves in the counterclockwise direction.