A particle is acted on by two torques about the origin: $\overrightarrow{{\mathbf{\tau }}_{\mathbf{1}}}$has a magnitude of$\mathbf{2}\mathbf{.0}\mathbf{\text{Nm}}$and is directed in the positive direction of the$\mathit{x}\mathbf{}$axis, and$\overrightarrow{{\mathbf{\tau }}_{\mathbf{2}}}$has a magnitude of$\mathbf{4}\mathbf{.0}\mathbf{\text{\hspace{0.17em}Nm}}$and is directed in the negative direction of the yaxis. In unit-vector notation, find$\mathbf{}\overrightarrow{\mathbf{d}\mathbf{l}}\mathbf{/}\mathit{d}\mathit{t}$, where$\overrightarrow{\mathbf{l}}\mathbf{}$ is the angular momentum of the particle about the origin.

${\overrightarrow{\tau }}_1=\left(\text{2.0N.m}\right)$

${\overrightarrow{\tau }}_2=(-4.0\text{N.m})$

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 rate of angular momentum by using concept of Newton’s second law in rotational motion form.

Formulae:

$\frac{d\overrightarrow{l}}{dt}={\overrightarrow{\tau }}_{net}$

According to the Newton’s second law, in angular form, the sum of all torques acting on a particle is equal to the time rate of the change of the angular momentum of that particle.

$\begin{array}{c}\frac{d\overrightarrow{l}}{dt}={\overrightarrow{\tau }}_{net}\\ ={\overrightarrow{\tau }}_1+{\overrightarrow{\tau }}_2\\ =\left(2.0\text{N.m}\right)-\left(4.00\text{N.m}\right)\end{array}$

In magnitude, the rate of change of angular momentum of the particle is

role="math" localid="1661748704827" $\begin{array}{c}\frac{dl}{dt}=\sqrt{{\left(\text{2.0N.m}\right)}^2+{(-4.00\text{N.m})}^2}\\ =4.5\text{N.m}\end{array}$

The direction of the rate of change of angular momentum of the particle is

$\begin{array}{c}\tan \theta =\frac{y}{x}\\ =\frac{-4.0\text{N.m}}{2.0\text{N.m}}\end{array}$

$\Rightarrow \theta =-{63}^o$

The negative sign shows that the angle is measured in clockwise direction with respect to unit vector.$\widehat{i}$