Question

The angular momentum of a particle relative to a certain point $O$ varies with time as $\overrightarrow{M}=a+\overrightarrow{b}{t}^2$, where $\overrightarrow{a}$ and $\overrightarrow{b}$ are constant vectors, with $\overrightarrow{a}\perp \overrightarrow{b}$. If the force moment $\overrightarrow{N}$ relative to the point $O$ acting on the particle when the angle between the vectors $\overrightarrow{N}$ and $\overrightarrow{M}$ equals ${45}^{\circ }$ is given as $\overrightarrow{N}=x\sqrt{\frac{a}{b}}\overrightarrow{b}$. Find $x$.

Answer 1

Force moment relative to point $O$,
$\overrightarrow{N}=\frac{d\overrightarrow{M}}{dt}=2\overrightarrow{bt}$
Let the angle between $\overrightarrow{M}$ and $\overrightarrow{N}$,
$\alpha ={45}^{\circ }$ at $t=t_0$,
Then $\frac{1}{\sqrt{2}}=\frac{\overrightarrow{M}\cdot \overrightarrow{N}}{|\overrightarrow{M}||\overrightarrow{N}|}=\frac{(\overrightarrow{a}+\overrightarrow{b{t}_0^2})\cdot \left(2\overrightarrow{b{t}_0}\right)}{\sqrt{a^2+b^2{t}_0^4}2b{t}_0}$
$=\frac{2b^2{t}_0^3}{\sqrt{a^2+b^2{t}_0^4}2b{t}_0}=\frac{b{t}_0^2}{\sqrt{a^2+b^2{t}_0^4}}$
So, $2b^2{t}_0^4=a^2+b^2{t}_0^4$ or, $t_0=\sqrt{\frac{a}{b}}$ (as $t_0$ cannot be negative)
It is also obvious from the figure shown below that the angle $\alpha$ is equal to ${45}^{\circ }$ at the moment $t_0$, when $a=b{t}_0^2$, i.e. $t_0=\sqrt{\frac{a}{b}}$ and $\overrightarrow{N}=2\sqrt{\frac{a}{b}}\overrightarrow{b}$