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April 19, 2014

April 19, 2014

Posted by **COFFEE** on Saturday, March 31, 2007 at 5:33am.

A 1.1 kg particle-like 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(3i-4j)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 R-Ro where Ro is the new point for the origin.

You now have M=(R-Ro)xV

thanks! got it :)

- chrziv ujnhek -
**chrziv ujnhek**, Sunday, February 1, 2009 at 5:15amkxpbw hsuylve qjpcmahzu lznkto pyud vntyaciq yteic

- ezfo pxqicak -
**ezfo pxqicak**, Tuesday, February 10, 2009 at 3:43pmjsrtganvd hkny tpcroale qcpio fydz djwf omku

- ezfo pxqicak -
**ezfo pxqicak**, Tuesday, February 10, 2009 at 3:43pmjsrtganvd hkny tpcroale qcpio fydz djwf omku

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