91Ó°ÊÓ

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:

$l=r_{âŠ\yen }mv$

The magnitude of the angular momentum of the particle P₁ is

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

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

The magnitude of the angular momentum of the particle P₂ is

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

By using right hand rule for vector product, ${\stackrel{â‡\P Ä}{l}}_2$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.