## Angular momentum

Assume a particle has angular velocity ω about a pivot point.  We define the angular momentum L of the particle about the point as L = r × p, where r is the displacement vector of the particle from the pivot point and p is its momentum.  The direction of L is perpendicular to both r and p
Let r and p lie in the x-y plane, as shown in the figure on the right.
Then L = r p sinθ k.  For a particle moving in a circular path sinθ = 1,
L = r p k = r m v k = mr2ω k = Iω,
since ω = ωk and the moment of inertial of te particle about the pivot point is mr2.

L = Iω.

The change in angular momentum of the particle is

dL/dt = d(r × p)/dt = r × dp/dt + dr/dt × p.

The last term on the right is proportional to v × vand therefore is zero.  We have

dL/dt = r × dp/dt = r × F = τ.

τ = dL/dt is the rotational analog of Newton's second law,  F = dp/dt.

The angular momentum of a rigid object rotating about an axis is the sum of the angular momenta of all its parts.  It is a measure of an object's rotational motion about this axis.  The angular momentum L about one of the object's symmetry axis is the product of the object's moment of inertia I times its angular velocity ω about this symmetry axis.

#### Problem:

A light rod 1 m in length rotates in the xy plane about a pivot through the rod's center.  Two particles of mass 4 kg and 3 kg are connected to its ends.  Determine the angular momentum of the system at the instant the speed of each particle is 5 m/s.

Solution:
We assume that the mass and moment of inertia of the rod can be neglected.
Let the z-axis pass through the center of the rod and point out of the page.
The moment of inertia of the system about the z-axis is
3 kg*(0.5 m)2 + 4 kg*(0.5 m)2 = 1.75 kgm2.
The angular velocity of the system is ω = (v/r)k = ((5 m/s)/(0.5 m))k = (10/s)k.
Here k is a unit vector or direction indicator pointing in the z-direction (out of the page).
The angular momentum of the system is L = Iω = (17.5 kgm2/s)k.

Angular momentum is a vector.  For a single particle, its direction is the direction of the angular velocity (given by the right hand rule).  The angular momentum of an object is changed by giving it an angular impulse.  An angular impulse ΔL is a change in angular momentum.  You give an object an angular impulse by letting a torque act on it for a time interval Δt.

ΔL = τΔt
angular impulse = torque * time

If an object has many independently rotating parts, the total angular momentum of the object is the sum of the angular momenta of all its parts.

#### Problem:

You usually give your closet door a gentle push and it swings closed gently in 5 seconds.  But today you are in a rush and exert 3 times the normal torque on it.
(a)  If you push on it for the usual time with this increased torque, how will its angular momentum differ from the usual value?
(b)  How long will it take the closet door to swing closed after your hurried push?

Solution:
(a)  When you push on the door, you exert a torque (force times lever arm) for a certain amount of time, and the door gains angular momentum L = Iω.  If you increase the force by a factor of 3 and push for the same amount of time, the angular momentum increases by a factor of three.  The moment of inertia of the door does not change, so its angular velocity must increase by a factor of 3.

(b)  angular velocity = angular displacement / time,  time = angular displacement / angular velocity
The time it will take the door to close will decrease by a factor of three.
time = (5 s)/3 = 1.66 s .

### Conservation of angular momentum

The total angular momentum of a a single object is constant if no external torque acts on the object.  An object cannot exert a torque on itself.  The total angular momentum of two interacting objects is also constant if no external torque acts on the objects.  Newton's third law tells us the forces the objects exert on each other are equal in magnitude and opposite in direction.  The interaction forces produce torques equal in magnitude and opposite in direction.  These torques change the angular momentum of each object by the same amount, but the changes will have opposite directions.  When we sum them up to find the change in the total angular momentum, we obtain zero.

If no external torque acts on a system of interacting objects, then their total angular momentum is constant.

In the video clip shown below the total angular momentum of the system points upward.  The person is stopping a spinning wheel and the stool starts to spin.

Some of the first few frames

As the person applies a torque to the wheel, the wheel applies a torque to the person.  The magnitudes of the angular momenta of the wheel and of the person change at the same rate, but their sum remains constant.

#### Problem:

A 60 kg woman stands at the rim of a horizontal turntable having a moment of inertia of 500 kgm2, and a radius of 2 m.  The turntable is initially at rest and is free to rotate about a frictionless vertical axis through its center.  The woman then starts walking around the rim clockwise (as viewed from above the system) at a constant speed of 1.5 m/s relative to the Earth.
(a)  In what direction and with what angular speed does the turntable rotate?
(b)  How much work does the women do to set herself and the turntable in motion?

Solution:
The system consists of the woman and the turntable.  No external torques act on the system, so the total angular momentum of the system is conserved.  It is zero before the woman starts to walk, and it is zero afterwards.

(a)  When the women walk with angular velocity
ω = -(v/r) k = -((1.5 m/s)/(2 m)) k = -(0.75/s) k
her angular momentum is
L = Iω = (-60 kg * 4 m2 * 0.75/s)k = -(180 kgm2/s)k.
The angular momentum of the turntable will be
L = (180 kgm2/s)k
and its angular velocity
ω = (180 kgm2/s)k/(500 kgm2) = (0.36/s)k.
The turntable turns counterclockwise.

(b)  The kinetic energy of the women is
½Iω2 = ½ 240 kgm2*(0.75 s)2 = 67.5 J.
The kinetic energy of the turntable is
½Iω2 = ½ 500 kgm2*(0.36 s)2 = 32.4 J.
The work done by the women is W = (67.5 + 32.4)J = 99.9 J.

The total angular momentum about any axis in the universe is conserved.  The angular momentum of a single object, however, changes when a net torque acts on the object for a finite time interval.  Conversely, if no net torque acts on an object, then its angular momentum is constant.

#### Problem:

A puck of mass 80 g and radius 4 cm slides along an air table at a speed of 1.5 m/s.  It makes a glancing collision with a second puck of radius 6 cm and mass 120 g (initially at rest) such that their rims just touch.  The pucks stick together and spin after the collision.
(a) What is the angular momentum of the system relative to the center of mass?
(b) What is the angular velocity about the center of mass?

Solution:
Let the center of the 120 g mass be at the origin before the collision.

(a)  Before the collision, the y-coordinate of the CM is
(m1y1 + m2y2)/M = (0.08/0.2)0.1 m = 0.04 m.
The x-coordinate of the CM is (m1x1 + m2x2)/M = (0.08/0.2)x1 = (0.4)x1.
The velocity of the CM is
vCM = dxCM/dt = (0.4) dx1/dt  = (0.4)1.5 m/s = 0.6 m/s in the x-direction.
In the lab frame the particle moves with velocity v = 1.5 m/s i
and the CM moves with vCM = 0.6 m/s i.
With respect to the CM m1 moves with velocity v1 = v - vCM =  0.9 i m/s ,
and m2 moves with velocity v2 = 0 - vCM =  -0.6 i m/s.
The angular momentum of the system about the CM is
L
= -(m1v1(y1 - yCM) + m2v2yCM ) k = -(7.2*10-3 kg m2/s) k.

(b)  In the collision momentum and angular momentum are conserved.
The total angular momentum about the CM after the collision is Iω = (7.2*10-3 kg m2/s )k.
The moment of inertia of the system about the CM is I = I1 +  I2, where I1 = ICM1 + M1R12, and I2 = ICM2 + M2R22.
I1 = (1/2) 0.08 kg(0.04 m)2 + 0.08 kg(0.06 m)2 = 3.52*10-4 kg m2.
I2 = (1/2) 0.12k g(0.06 m)2 + 0.12 kg(0.04 m)2 = 4.08*10-4 kg m2.
I = I1 + I2 = 7.6*10-4kg m2.
ω = L/I = (9.47/s) k.
To find the moment of inertia of the composite object after the collision the parallel axis theorem was used.

### The sweet spot

When you push on an object with a force directed towards its center of mass, the object will accelerate.  But it will not start rotating about its center of mass.  You are not applying a torque.  The lever arm is zero.  No torque implies no angular acceleration.  When you push on it with a force not directed towards the center of mass, you exert a torque about the CM, because the force now has a lever arm.  This will result in linear as well as angular acceleration of the object.  The linear acceleration is a result of the force and the angular acceleration is a result of the torque.

If a ball hits a bat right at the center of mass the bat will accelerate backward without rotating.  The bat's handle will jerk backward in the batter's hand.  If the ball hits farther away from his hand, the bat will accelerate backward, but at the same time start rotating about its center of mass.  This rotation will move the handle forward, while the translation moves it backward.  If the ball hits at just the right spot, called the center of percussion, the backward and forward accelerations exactly cancel and the batter can swing the bat smoothly without feeling much of a jerk.  The center of percussion of baseball bats, tennis rackets, golf clubs, and other sporting equipment is called one of their sweet spots.