Question

A particle is acted on by two torques about the origin: $\overrightarrow{{\tau }_1}$ has a magnitude of $2.0Nm$ and is directed in the positive direction of the $x$ axis, and $\overrightarrow{{\tau }_2}$ has a magnitude of $4.0Nm$ and is direction in the negative direction of the $y$ axis. In unit vector notation, find $d\overrightarrow{l}/dt$, where $\overrightarrow{l}$ is the angular momentum o the particle about the origin.

Answer 1

The rate of change of the angular momentum is
$\frac{\overrightarrow{dl}}{dt}={\tau }_1+{\tau }_2=(2.0N.m)\hat{i}-(4.0N.m)\hat{j}$
Consequently the vector $\overrightarrow{dl}/dt$ has a magnitude $\sqrt{(2.0N.m{)}^2+(-4.0N.m{)}^2}=4.5N.m$ and is an angle $\theta$ (in the $xy$ plane, or a plane parallel to it) measured from the positive $x$ axis,

where, $\theta ={\tan }^{-1}\left(\frac{-4.0N.m}{2.0N.m}\right)=-{63}^o$

the negative sign indicating that the angle is measured clockwise as viewed from above (by a person on the $+z$ axis)