Posted by **Bernadette** on Friday, February 8, 2008 at 6:45pm.

At time t=0, a 2kg particle has position vector r=(5.0m)i+(-8.0m)j relative to the origin. Its velocity just then is given by v=(-5.0t^2m/s)i. For the following answers, use t for the time. A) About the origin and for t>0, what is the particle's angular momentum? B) About the origin and for t>0, what is the torque acting on the particle? C) Repeat A and B for a point with coordinates (-7.0m, -4.0m, 0.0m) instead of the origin.

- AP Physics -
**bobpursley**, Friday, February 8, 2008 at 6:51pm
Please dont post under different names.

I will be happy to critique your thinking on this.

- AP Physics -
**Damon**, Friday, February 8, 2008 at 7:25pm
L = R x m V

where R is position vector from reference point origin for a and b) to mass

V is velocity vector

x means cross or vector product

the x coordinate of the position vector is 5 at the start but changes with time because the thing has x velocity of -5t^2

integral = 5-(5/3) t^3

so for part a

L = (5-5t^3/3) i - 8 j)x(-5 t^2 i + 0 j)

(5-5t^3/3) -8 0

-5t^2 0 0 = -40 t^2 k

i j k

for part b

Torque = rate of change of angular momentum

dL/dt = d/dt(40 t^2 k) = 80 t

for part c

Rx = 5-5t^3/3 - (-7) = 12 - 5 t^3/3

Ry = -8 -(-4) = -4

Rz = 0 still

just do it again

- determinant sign error -
**Damon**, Friday, February 8, 2008 at 7:28pm
(5-5t^3/3) -8 0

-5t^2 0 0 = +40 t^2 k

i j k

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