Let a particle of mass m travel along a differentiable path x in a Newtonian vector field F (i.e., one that satisfies Newton’s second law F = ma, where a is the acceleration of x). We define the angular momentum l(t) of the particle to be the cross product of the position vector and the linear momentum mv, that is,
l(t) = x(t)×mv(t).
(Here v denotes the velocity of x.) The torque about the origin of the coordinate system due to the force F is the cross product of position and force:
M(t) = x(t)×F(t) = x(t)×ma(t).
(Also see §1.4 concerning the notion of torque.) Show that
Thus, we see that the rate of change of angular momentum is equal to the torque imparted to the particle by the vector field F.
We need at least 10 more requests to produce the solution.
0 / 10 have requested this problem solution
The more requests, the faster the answer.