A spinning object has angular momentum, represented by l or L.
In a rotational situation, impulse is the product of a torque and the time interval over which the torque acts.
Work (a torque acting over an angle) produces a change in kinetic energy. An impulse (a torque acting over a time interval) produces a change in angular momentum.
τ = Iα
τ = I dω/dt = dL/dt
Expressed as an integral, this becomes:
ΔL = ∫ τ dt
The impulse is the area under the torque vs. time graph.
If the torque is constant: ΔL = τ Δt
The Law of Conservation of Angular Momentum states that, when no external torques act on a system, the angular momentum of the system is conserved.
Always remember that angular momentum is a vector when applying this law.
A spinning figure skater is an excellent example of angular momentum conservation. The skater starts spinning with her arms outstretched, and has a a rotational inertia of Ii and an initial angular velocity of ωi. When she moves her arms close to her body, she spins faster. Her moment of inertia decreases, so her angular velocity must increase to keep the angular momentum constant.
Conserving angular momentum:
Li = Lf
Ii ωi = If ωf
In this process, what happens to the skater's kinetic energy?
In this case the kinetic energy actually increases.
Ki = ½ Ii ωi2 = ½ (Ii ωi) ωi
Kf = ½ If ωf2 = ½ (If ωf) ωf
The figure skater does work on her arms and hands to speed them up as she brings them closer to her body - that's where the extra energy comes from.
A person standing on a turntable while holding a bicycle wheel is an excellent place to observe angular momentum conservation in action. The person is initially not rotating on the turntable, and the bicycle wheel is rotating about a horizontal axis.
The initial angular momentum about a vertical axis is zero.
If the person re-positions the bicycle wheel so its rotation axis is vertical, the wheel exerts a torque on the person during the re-positioning that makes the person spin in the opposite direction as the wheel. The angular momenta cancel, so that L = 0 at all times about a vertical axis.
As long as the rotating platform is well-balanced and there are no net external torques acting, re-positioning the bike wheel so its rotation axis is horizontal again should stop the person's rotation. Flipping the bike wheel over so that its rotation axis is again vertical but reversed will make the person spin in the opposite direction.
Sarah, with mass m and velocity v, runs toward a playground merry-go-round and jumps on at its edge. Sarah and the merry-go-round (mass M, radius R, and moment of inertia I = cMR2) then spin together with a constant angular velocity ωf. If Sarah's initial velocity is tangent to the circular merry-go-round, what is ωf?
We can apply angular momentum conservation here. Sarah's angular momentum before the collision equals the angular momentum of the system after the collision.
It's not obvious that Sarah has an initial angular momentum, although she clearly has a linear momentum. That linear momentum can be transformed to an angular momentum the same way forces produce torques.
For any linear momentum p, you can treat that as an angular momentum L about a particular axis of rotation using:
L = r × p
The magnitude of the angular momentum is L = r p sin(θ), where θ is the angle between r and p.
In this example Sarah's linear momentum mv can be transformed to an initial angular momentum Li = Rmv.
Applying angular momentum conservation:
Total angular momentum before = total angular momentum after
Rmv + 0 = Itotal ωf
Itotal = The moment of inertia of the merry-go-round plus Sarah's moment of inertia.
Itotal = cMR2 + mR2
Therefore: ωf = Rmv / Itotal = mv / (cMR + mR)
Numerical example
Let's say m = 25 kg; v=4 m/s; M = 50 kg; R = 2 m; c = ½
ωf = 100 / (50+50) = 1 rad/s.