Answer to Question #230101 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 𝑣is a constant vector

Answer 1

(i)

$$v=\dfrac{dr}{dt}=-\omega\sin \omega ti+\omega\cos \omega tj$$

$$r\cdot v=\cos \omega t(-\omega\sin \omega t)+\sin \omega t(\omega \cos \omega t)=0$$

Hence $r \perp v.$

(ii)

$$a=\dfrac{d^2r}{dt^2}=\dfrac{dv}{dt}=-\omega^2\cos \omega ti-\omega^2\sin \omega tj$$

$$=-\omega^2(\cos \omega ti+\sin\omega tj)=-\omega^2r$$

Hence 𝑎 is directed towards the origin and has magnitude proportional to the  distance from the origin

$$a=(-\omega^2)r$$

(iii)

$$r\times v=\begin{vmatrix} i & j & k \\ \cos\omega t & \sin\omega t & 0\\ -\omega \sin\omega t & \omega\cos\omega t & 0\\ \end{vmatrix}$$

$$=k\begin{vmatrix} \cos\omega t & \sin\omega t \\ -\omega \sin\omega t & \omega \cos\omega t \end{vmatrix}$$

$$=(\omega\cos^2\omega t+\omega\sin^2\omega t)k$$

$$=\omega k$$

Therefore $r\times v$ is a constant vector ($\omega=const$ ).