T find the angular momentum

Oct 2013
48
0
Can you suggest me please how to set my equations in order to get the angular momentum in this problem?


A particle of mass m moves in a circle of radius R at a constant speed v, as shown below. The motion begins at point Q at time t = 0. Determine the angular momentum of the particle about the axis perpendicular to the page through point P as a function of time. (Use any variable or symbol stated above along with the following as necessary: t.)




I have the answer, but I have no idea how to get it,

Since the angular momentum is m times the cross product of radius and velocity, I tried to set the y and x coordinates of these two vectors but I think I miss something in the velocity vector
 

Attachments

Last edited:
Apr 2014
4
0
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 mv (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 mv. 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 mv + r0 x mv
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 ***)
 
Last edited: