Answer to Question #122783 in Math for Felix Coleman

Question

The position vector r of a particle at time t is
r = a cos (ωt)i + b sin (ωt)j
where a, b, ∈ R and a 6= b.
(a) Find the velocity and acceleration vectors v and a respectively.
(b) Show that
ω
2
|v|
2 + |a|
2 = ω
4
(a
2 + b
2
),
and find the times at which the velocity and the acceleration vectors are perpendicular.

Answer 1

(a)

$\bar{r}=a\cos(\omega t)\bar{i}+b\sin(\omega t)\bar{j}$

$\bar{v}=\bar{r'}=-a\omega\sin(\omega t)\bar{i}+bw\cos(\omega t)\bar{j}$

$\bar{a}=\bar{r''}=-a\omega^2\cos(\omega t)\bar{i}-bw^2\sin(\omega t)\bar{j}$

$\bar{a}=-\omega ^2\bar{r}$

(b)

$|\bar{v}|=\sqrt{(-a\omega\sin(\omega t))^2+(b\omega\cos(\omega t))^2}=$ $=\omega\sqrt{a^2+b^2}$

$|\bar{a}|=\sqrt{(-a\omega^2\cos(\omega t))^2+(b\omega^2\sin(\omega t))^2}=$ $=\omega^2\sqrt{a^2+b^2}$

$|\bar{v}|^2\omega^2=\omega^4(a^2+b^2)$

$\bar{v}\perp\bar{a}=>\bar{v}\cdot \bar{a}=0$

$-a\omega\sin(\omega t)(-a\omega^2\cos(\omega t))+bw\cos(\omega t)(-bw^2\sin(\omega t))=$

$={1\over 2}\omega^3(a^2+b^2)\sin2\omega t=0$

$2\omega t=\pi n, n=0,1,2,...$

$t={\pi n\over 2\omega}, n=0, 1,2,...$