Answer to Question #225438 in Calculus for Unknown346307

Question

Question:

A particle moves such that its vector is given by 𝑟

= cos 𝜔𝑡𝑖 + sin 𝜔𝑡𝑗 where 𝜔 is a constant. Show that:

(i)

Velocity, 𝑣

of the particle is perpendicular to 𝑟

(ii)

Acceleration, 𝑎

is directed towards the origin and has magnitude proportional to the

distance from the origin.

(iii) 𝑟

𝑋 𝑣

is a constant vector

Answer 1

The position vector is given by, $\vec{r}=cos\omega t \hat{i}+sin\omega t \hat{j}$

(a) Velocity is given by,

$\vec{v}= \frac{d}{dt}\vec{r}=\frac{d}{dt}(cos\omega t \hat{i}+sin\omega t \hat{j}) = -\omega sin\omega t \hat{i}+\omega cos\omega t \hat{j}$

If velocity is perpendicular to position then, $\vec{r}.\vec{v} = 0$

$\vec{r}.\vec{v} = (cos\omega t \hat{i}+sin\omega t \hat{j}).(-\omega sin\omega t \hat{i}+\omega cos\omega t \hat{j})$

$\vec{r}.\vec{v} = -\omega sin\omega t cos\omega t+\omega sin\omega t cos\omega t = 0$

So, velocity and position are perpendicular.

(b) $\vec{a} = \frac{d}{dt} \vec{ v } =\frac{d}{dt}( -\omega sin\omega t \hat{i}+\omega cos\omega t \hat{j}) = -\omega^2 cos\omega t\hat{i} -\omega^2 sin\omega t\hat{j} = -\omega^2 \vec{r}$

A negative sign indicates that it is acting toward the origin.

$|\vec{a}| = \omega^2|\vec{r}| \implies |\vec{a}| \propto|\vec{r}|$

(c) $\vec{r} \times \vec{v} = (cos\omega t \hat{i}+sin\omega t \hat{j}) \times ( -\omega sin\omega t \hat{i}+\omega cos\omega t \hat{j})$

$\vec{r} \times \vec{v} = (\omega cos^2\omega t+ \omega sin^2\omega t)\hat{k} = \omega \hat{k}$

Hence, it is a constant vector.