## Tuesday, August 21, 2018

### Unizor - Physics4Teens - Mechanics - Dynamics - Angular Momentum

Notes to a video lecture on http://www.unizor.com

Angular Momentum

Recall the rotational equivalent of the Newton's Second Law:

τ = I·α

where

τ = F·r - a torque, a product of tangentially applied force F and the distance from the point of application of force to the axis of rotation r;

I = m·R² - a moment of inertia, a product of inertial mass of a point-object by a square of its distance from the axis of rotation R (might be different from the distance r above);

α - angular acceleration of rotation.

It is very important to correspond straight line translational elements of motion with their rotational counterparts:

 Translation Rotation ForceF Torqueτ = F·r Linear Accelerationa Angular Accelerationα = a/r InertialMassm Momentof InertiaI = m·r² Newton'sSecond LawF=m·a RotationalEquivalentτ = I·α

Now we will consider a rotational equivalent of a familiar equation
between impulse and momentum in translational motion along a straight
line.

As is known from the previous material, there is a correspondence between these quantities: impulse of the force F exhorted during an infinitesimal time period dt equals to an increment of the momentum d(m·v), where m is mass and v - velocity of an object:

dt = d(m·v)

Let's see if the corresponding rotational characteristics of motion have similar dependency.

We know the rotational equivalent of the Newton's Second Law τ=I·α.

Multiplying it by dt, we obtain

τ·dt = I·α·dt

Since a product of angular acceleration α and infinitesimal time interval dt is an infinitesimal increment of angular speed dω, there is an equality

τ·dt = I·dω = d(I·ω)

The latter fully corresponds to an equation between impulse and momentum
of translational movement. It represents the relationship between rotational impulse τ·dt of torque τ during time interval dt and increment of rotational (angular) momentum d(I·ω) of an object with a moment of inertia I and angular speed ω.

We can continue our table of correspondence between translational motion
along a straight line and rotation along a circular trajectory.

 Translation Rotation ImpulseF·dt Rotational Impulseτ·dt Momentumd(m·v)=F·dt Rotational Momentumd(I·ω)=τ·dt

Trivial consequence of an equation

τ·dt = d(I·ω)

is that

τ = d(I·ω)/dt = (I·ω)'

From this follows that if the balance of all torques acting on an object is zero, the rotational momentum remains constant.

This is the Law of Conservation of the rotational (angular) momentum.