W9.1 Angular momentum Banana Skins

By far the most common misconception about angular momentum is ......

WRONG!
Only a particle moving in a circle has angular momentum WRONG!

Not true - review Key Point 5.16; there is no mention of rotation here!

With that in mind, let’s try a little problem…

Graphic - Setup for Angular Momentum Bananaskin problem

The figure shows a particle moving in a straight line at constant speed, past some point O. Describe the behaviour of the particle’s angular momentum about O.

Solution

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Solution

It is the same at all points along the line : 12 kgm2s-1, in a direction into the page. It is tempting, but incorrect, to think it is a maximum when the radius and velocity vectors are perpendicular.

Let’s consider a general case:

Graphic - Solution setup of Angular Momentum problem

At the point P, the distance away from the origin is given by $\frac{2}{sin\phi}$ The (vector) equation for the total angular momentum, $\vec{L}$, at any point is given by

\begin{displaymath} \vec{L}=m(\vec{r} \times \vec{v})\end{displaymath}

This vector is directed into the page and has magnitude given by L, where

\begin{displaymath} L = mvr~sin \phi = 2 \times \frac{2}{sin\phi} \times 3 \times sin \phi = 12 kgm^2s^{-1} \end{displaymath}

So, in the case where $\vec{v}$ and $\vec{r}$ are perpendicular, their cross product is a maximum (sin~φ=1) but this is balanced by $\vec{r}$ being its minimum value.

You may like to think about this when you are tackling the roundabout question (Q6.15)