Re: Physics
posted by COFFEE on .
Posted by COFFEE on Friday, March 30, 2007 at 4:25am.
A 1.1 kg particlelike object moves in a plane with velocity components Vx = 30 m/s and Vy = 90 m/s as it passes through the point with (x, y) coordinates of (3.0, 4.0) m.
(a) What is its angular momentum relative to the origin at this moment?
_____ kg*m^2/s k
(b) What is its angular momentum relative to the point (2.0, 2.0) m at this same moment?
_____ kg*m^2/s k

i drew a points on a flat xy plane.. but im not sure if i drew it correctly. i have the Vx component towards the positive X direction and the Vy towards the negative Y direction.
angular momentum = r x p
r x p = m (r x v)
......= 1.1 (r x 30)
......= 1.1r x 30
33 = 1.1r
30 = r
.. im stuck.. help!
For Further Reading
* Physics  bobpursley, Friday, March 30, 2007 at 7:52am
1. Angular momentum= R x mV Where mV is the linear momentum, x is the cross product, and R is the position vector.
Angular momenum= m(3i4j)x(30i+90j)
= m270k + m 120k
= m*390 k
check my thinking.
==================
thank you, i understand that part.
..but if the origin was changed, in this case in part b: (2.0, 2.0) m
how would the 'Angular momentum= R x mV' equation change? is the linear momentum the one affected?
R changes: New R is RRo where Ro is the new point for the origin.
You now have M=(RRo)xV
thanks! got it :)

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