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

__Conservation of__

Angular Momentum

Angular Momentum

We ended up the previous lecture deriving the Law of Conservation of

Angular Momentum. In this lecture we will continue discussing this

particular conservation law.

We will consider a thought experiment suggested by Arkady Kokish in our private discussion of this topic.

The results of this thought experiment will confirm the Angular Momentum Conservation Law.

Consider a point-object of mass

**with no external forces (like gravity) present and two threads, short of length**

*M***and long of length**

*r***tied to it.**

*R*A person holds both threads and starts spinning an object with angular velocity

**. It will rotate along a circular trajectory of a radius**

*ω*_{1}**(the length of a shorter thread).**

*r*Let's analyze what happens if this person lets a shorter thread go, while holding the long one.

On the picture above the short thread is let go when an object was at

the top. Then, since nothing holds it on a circular trajectory, it will

fly along a tangential line maintaining the same linear speed it had

when it was rotating, that is

**.**

*r·ω*_{1}But our object will not fly too far. The longer thread will stop it, when it will be on a distance

**from a center of rotation.**

*R*Here is a picture of this particular moment, when the long thread is fully extended.

At this moment the linear speed of an object flying (on the picture) left to right equals to

**.**

*V=r·ω*_{1}Let's represent this vector as a sum of two vectors:

**perpendicular to a line representing the fully extended long thread (tangential to a future circular trajectory of a radius**

*V*_{t}**) and**

*R***going along that thread line (radial).**

*V*_{r}Assuming an angle between a short thread at a moment it was let go and a

long one at the moment it is fully extended, stopping tangential

movement of an object, is

**, it is obvious that**

*φ***.**

*r=R·cos(φ)*At the moment when the long thread is fully extended the tension of the

thread instantly (actually, during a very short interval of time Δ

**) nullifies the radial speed**

*t***. To be exact, its moment of inertia**

*V*_{r}**will be equal to an impulse of the tension force**

*M·V*_{r}**during interval of time Δ**

*T***:**

*t***Δ**

*T·*

*t = M·V*_{r}This tension force decelerates the radial speed to zero during the interval of time Δ

**. The shorter this interval - the more ideal our experiment is.**

*t*In an ideal case of non-stretchable thread that can withstand infinite tension the interval Δ

**is infinitely small, and deceleration is infinitely large in magnitude and short in time.**

*t*In this case the radial speed

**will be brought down to zero instantaneously.**

*V*_{r}At the same time the magnitude of tangential vector of speed

**is not affected by the tension since this vector is perpendicular to a tension force.**

*V*_{t}We see that the angle between vectors

**and**

*V=r·ω*_{1}**is the same as between threads**

*V*_{t}**.**

*φ*Therefore,

*V*_{t}= r·ω_{1}·cos(φ)But

*cos(φ) = r/R*from which follows

*V*_{t}= r·ω_{1}·r/RTangential linear speed

**is related to angular speed of rotation**

*V*_{t}**of our object on a new radius**

*ω*_{2}**as**

*R***.**

*V*_{t}=R·ω_{2}Hence,

*R·ω*_{2}= r·ω_{1}·r/R

*R²·ω*_{2}= r²·ω_{1}

*M·R²·ω*_{2}= M·r²·ω_{1}Expression

**represents the moment of inertia**

*M·r²***of an object of mass**

*I*_{1}**rotating on a radius**

*M***.**

*r*Analogously, expression

**represents the moment of inertia**

*M·R²***of the same object of mass**

*I*_{2}**rotating on a radius**

*M***.**

*R*Therefore, we have derived the Law of Conservation of the Angular Momentum:

*I*_{1}·ω_{1}= I_{2}·ω_{2}
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