91Ó°ÊÓ

The rotating analogue of linear momentum is angular momentum (also known as moment of momentum or rotational momentum). Because it is a conserved quantity—the total angular momentum of a closed system remains constant—it is a significant quantity in physics.

Use the equation that indicates the rate of change of the particle's angular acceleration is equal to the torque acting on it.

${\stackrel{→}{\mathrm{τ}}}_A=\frac{d{\stackrel{→}{L}}_A}{dt}$

Here,$\frac{d{\stackrel{→}{L}}_A}{dt}$is the rate of change of angular momentum of particle at location A, and${\stackrel{→}{\mathrm{τ}}}_A$is the torque on the particle at location A.

The above expression can also be written as,

${\stackrel{→}{\mathrm{τ}}}_A=\frac{{\stackrel{→}{L}}_{fA}-{\stackrel{→}{L}}_{iA}}{t_f-t_i}$

Here, ${\stackrel{→}{L}}_{fA}$is the final angular momentum of the particle, ${\stackrel{→}{L}}_{iA}$is the initial angular momentum of the particle, and $t_f-t_i$is the time for which the torque is applied.

Substitute $0.1s$for $t_f-t_i$$\left(10,-12,20\right)N·m$${\stackrel{→}{\mathrm{τ}}}_A.\left(3,5,-2\right)kg·m^2/s$in ${\stackrel{→}{L}}_{iA}$

${\stackrel{→}{\mathrm{τ}}}_A=\frac{{\stackrel{→}{L}}_{fA}-{\stackrel{→}{L}}_{iA}}{t_f-t_i}$

$\left(10,-12,20\right)N·m=\frac{{\stackrel{→}{L}}_{fA}-\left(3,5,-2\right)kg·m^2/s}{0.1s}$

${\stackrel{→}{L}}_{fA}=\left(4,3.8,0\right)kg·m^2/s$

Therefore, the angular momentum of the particle at $15.1s$is $\left(4,3.8,0\right)kg·m^2/s$.