Let us set up a versor system with the origin in the center of the circle described.

I have 2 solutions: a more complicated one (the first) and a simpler one (the second).

The first solution is more complex, but will work pretty much 100% of the time. The second one is slicker though

.

**1st solution:**
Writing

**v** and

**r** with respect to theta we have:

**v**(theta) = - v sin(theta)

**i** + v cos(theta)

** j** **r**(theta) =

**PO **+ ( R cos(theta)

**i** + R sin(theta)

**j** )

, which gives us:

**r**(theta) = R ( 1 + cos(theta) )

**i **+ R sin(theta)

**j**
(we have set

**PO** = R

**i** )

Setting

**L** =

**r** x m

**v** (we will have to use the determinant form of the vectorial product),

we will have:

**L**(theta) = - mvr sin^2(theta)

**k **- mvr cos(theta) ( 1 + cos(theta) )

**k**
Or written in scalar form:

L(theta) = mvr ( sin ^2(theta) + cos(theta) + cos^2(theta) )

L(theta) = mvr ( 1 + cos(theta) )

***

Setting theta = wt = (v/R) t, we have:

L(t) = mvr ( 1 + cos(wt) ) = mvr ( 1 + cos (vt/R) )

**2nd solution:**
We know that L =

**r** x m

**v**. Well, we can write r as follows:

**r** =

**PO** +

**r0** , where

**r0** is the position vector of the point at any time.

Well, then

**L** =

**PO** x m

**v** +

**r0** x m

**v**
As r0 is perpendicular to v at any time, and PO makes at all times the angle 90-theta with v,

writing the equation in scalar form gives us

L(theta) = mv(PO)sin(90-theta) + mvr0

But r0=PO=R, and sin(90-theta) = cos(theta).

Then L(theta) = mvr cos(theta) + mvr = mvr ( 1 + cos(theta) ).

From now on we use the same method from solution #1. (see mark ***)