Newton's second law for angular form?

Question

A 3.0 kg toy car on the x axis has velocity v = -2.0 t^3 m/s along that axis. About the origin and for t>0, what are a) the car's angular momentum and b) the torque acting on the car? c) Repeat a) and b) for a point with coordinates (2.0m, 5.0m, 0) instead of the origin. d) repeat a) and b) for a point with the coordinates (2.0m, -5.0m, 0) instead of origin

Answer 1

The angular momentum of an object is given by:

L=m r cross v  (where r and v are vectors)

if you dont know the cross product, use r v sin(t) where sin (t) is the sin of the angle between the the r and v vectors

in part a) the angular momentum is zero since r and v are in the same direction; since L is always zero, dL/dt=0 and there is no torque

c) here there will be an angular momentum since the r vector does not lie along the x axis...so when the car is at a point (x,0) on the x axis, its r vector is simply (x-2) i -5j where i and j are unit vectors in the x and y directions; the cross product of r cross v becomes:

[(x-2)i -5j]cross[-2t^3 i] = 10t^3 (-k) = -10t^3 k

where k is a unit vector in the z direction

(i cross i=0; j cross i = -k)

so the angular momentum points down

torque is simply dL/dt, so differentiate L to get torque