Angular momentum as a function of time question

goli12
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Homework Statement


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.
http://www.webassign.net/serpse8/11-p-017.gif

Homework Equations


L=Iω=rp=m(r x v)
p=mv
v=rω

The Attempt at a Solution



What I did first was finding the angular momentum if it rotated around the origin O. This was straightforward and hence i got the answer of L=m(v x r). However, what got me is finding the angular momentum about the point p. So using the given information, i found the moment arm during t=0 to be 2R, 2x the distance of the radius and with v perpendicular to the moment arm, the angular momentum about point p at t=0 is 2mvr.

However, i realize that the distance between the mass and point p is changing as it moves around in a circle and got completely lost :(

Can you guys please help me with this question with hints please would be really grateful as this question is really getting to me... Cheers
 
Last edited:
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Consider the mass to be at a general angle with Q. What will its angular momentum about point p at this instant (in terms of the general angle) ? What is the relation between time and the general angle ? Answer requires some geometry.
 

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