At $t=15s,$ a particle has angular momentum relative to location$A$ . A constant torque relative to location$A$ acts on the particle. At$t=15.1s.$ what is the angular momentum of the particle?

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.

${\overrightarrow{\tau }}_A=\frac{d{\overrightarrow{L}}_A}{dt}$

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

The above expression can also be written as,

${\overrightarrow{\tau }}_A=\frac{{\overrightarrow{L}}_{fA}-{\overrightarrow{L}}_{iA}}{t_f-t_i}$

Here, ${\overrightarrow{L}}_{fA}$is the final angular momentum of the particle, ${\overrightarrow{L}}_{iA}$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$$\left(10,-12,20\right)N·m$${\overrightarrow{\tau }}_A.\left(3,5,-2\right)kg·m^2/s$in ${\overrightarrow{L}}_{iA}$

${\overrightarrow{\tau }}_A=\frac{{\overrightarrow{L}}_{fA}-{\overrightarrow{L}}_{iA}}{t_f-t_i}$

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

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

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