*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

**- a**

*τ = F·r**torque*, a product of tangentially applied force

**and the distance from the point of application of force to the axis of rotation**

*F***;**

*r***- a**

*I = m·R²**moment of inertia*, a product of inertial mass of a point-object by a square of its distance from the axis of rotation

**(might be different from the distance**

*R***above);**

*r***-**

*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 |

InertialMass m | Momentof Inertia I = m·r² |

Newton'sSecond Law F=m·a | RotationalEquivalent τ = 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

**exhorted during an infinitesimal time period**

*F**d*equals to an increment of the momentum

**t***d*, where

**(m·v)****is mass and**

*m***- velocity of an object:**

*v*

*F·**d*

**t =**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

*d*, we obtain

**t**

*τ·**d*

**t = I·α·**d**t**Since a product of angular acceleration

**and infinitesimal time interval**

*α**d*is an infinitesimal increment of angular speed

**t***d*, there is an equality

**ω**

*τ·**d*

**t = I·**d**ω =**d**(I·ω)**The latter fully corresponds to an equation between impulse and momentum

of translational movement. It represents the relationship between

*rotational impulse*

*of*

**τ·**d**t***torque*

**during time interval**

*τ**d*and increment of

**t***rotational (angular) momentum*

*d*of an object with a

**(I·ω)***moment of inertia*

**and**

*I**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·**d*

**t =**d**(I·ω)**is that

*τ =**d*

**(I·ω)/**d**t = (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**.

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