11.8:
Angular Momentum: Single Particle
Every rotating object, like a spinning top, possesses an angular momentum L, which is mathematically expressed as the cross-product of the position vector and the linear momentum, which is mass times velocity.
The magnitude of angular momentum equals to mvrsinθ, and its SI units are kilogram meter square per second.
For example, consider a particle moving in the xy-plane and having a position vector r. The right-hand thumb gives the direction of angular momentum if the fingers are curled in the direction of rotation.
The time derivative of angular momentum can be expressed using the vector rule for the derivatives. The first term contains the cross-product of the velocity vector with itself and hence equals zero.
In the second term, mass times acceleration is the net force acting on the particle; recall that the cross product of r and F is torque. Thus, the rate of change of angular momentum is equal to the torque exerted by the net force acting on the particle.
11.8:
Angular Momentum: Single Particle
Angular momentum is directed perpendicular to the plane of the rotation, and its magnitude depends on the choice of the origin. The perpendicular vector joining the linear momentum vector of an object to the origin is called the “lever arm.” If the lever arm and linear momentum are collinear, then the magnitude of the angular momentum is zero. Therefore, in this case, the object rotates about the origin such that it lies on the rim of the circumference defined by the lever arm magnitude.
The net torque acting on rotating bodies causes the angular momentum to change, which is a rotational analog for Newton's second law of motion in terms of momentum. It is important to note that this is valid as long as both torque and angular momentum are measured to the same origin, fixed to an inertial frame of reference.
This text is adapted from Openstax, University Physics Volume 1, Section 11.2: Angular Momentum.