Physics Help Forum T find the angular momentum

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 Apr 4th 2014, 08:13 AM #1 Member     Join Date: Oct 2013 Posts: 48 T find the angular momentum 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 Attached Thumbnails   Last edited by dokrbb; Apr 4th 2014 at 08:24 AM.
 Apr 15th 2014, 12:08 PM #2 Junior Member   Join Date: Apr 2014 Posts: 4 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 by lbicsi; Apr 15th 2014 at 12:12 PM.

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# the angular momentum of a particle of mass m and radius r

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