AP Physics
posted by Bernadette on .
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.

Please don't post under different names.
I will be happy to critique your thinking on this. 
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 = (55t^3/3) i  8 j)x(5 t^2 i + 0 j)
(55t^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 = 55t^3/3  (7) = 12  5 t^3/3
Ry = 8 (4) = 4
Rz = 0 still
just do it again 
(55t^3/3) 8 0
5t^2 0 0 = +40 t^2 k
i j k