In the instant of Figure, two particles move in an xy plane. Particle P₁has mass 6.5kgand speed v₁ = 202m/s, and it is at distance d₁ = 105mfrom point O. Particle P₂has mass 3.1kgand speed v₂ = 3.6m/sand it is at distance d₂ = 2.8mfrom point O. (a) What is the magnitude of the net angular momentum of the two particles about O? (b) What is the direction of the net angular momentum of the two particles about O?

m₁ = 6.5 kg

v₁ = 2.2 m/s

d₁ = 1.5 m

m₂ = 3.1 kg

v₂ = 3.6 m/s

d₂ = 2.8 m

The problem deals with the calculation of angular momentum. The angular momentum of a rigid object is product of the moment of inertia and the angular velocity. It is analogous to linear momentum. The angular momentum can be found using right-hand rule. The counterclockwise rotation of the particles position gives positive angular momentum and clockwise rotation of the particles position gives negative angular momentum.

Formulae:

$l=r_{\perp }mv$

The magnitude of the angular momentum of the particle P₁ is

$\begin{array}{rcl}{l}_1& =& d_{1\perp }{m}_1{v}_1\\ & =& 1.5m\times 6.5kg\times 2.2m/s\\ & =& 21.45kg\frac{m^2}{s}\end{array}$

By using right hand rule for vector product, ${\stackrel{\rightharpoonup }{L}}_1$is negative or along –z axis.

The magnitude of the angular momentum of the particle P₂ is

$\begin{array}{rcl}{l}_2& =& d_{2\perp }{m}_2{v}_2\\ & =& 2.8m\times 3.1kg\times 3.6m/s\\ & =& 31.25kg\frac{m^2}{s}\end{array}$

By using right hand rule for vector product, ${\stackrel{\rightharpoonup }{l}}_2$is positive or along +z axis.

The two angular momentum vectors are in opposite direction. Hence, magnitude of angular momentum difference gives the magnitude of net angular momentum.

$\begin{array}{rcl}L& =& l_2-l_1\\ & =& 31.25kg\frac{m^2}{s}-21.45kg\frac{m^2}{s}\\ & =& 9.8kg\frac{m^2}{s}\end{array}$

The net angular momentum is positive. Hence, it is going along +z axis.