Question

A particle moves so that its position vector ˜r at time t is = ˜r ˜a coswt +˜bsinwt, where w is a constant and ˜a and ˜b are constant vectors. Show that (a) ˜ ˜ r × r˙ is independent of t, ˜

Answer 1

(a)

$$|\vec a|=|\vec b|=1, \vec a\perp\vec b$$ $$\vec r\cdot \vec r=(\vec a\cos\omega t+\vec b\sin \omega t)\cdot(\vec a\cos\omega t+\vec b\sin \omega t)$$

$$=\cos^2\omega t+\sin^2\omega t=1$$

(b)

$$\vec v=\dfrac{d\vec r}{dt}=-\omega\vec a\sin\omega t+\omega\vec b\cos \omega t$$

$$\dfrac{d\vec v}{dt}=-\omega^2\vec a\cos\omega t-\omega^2\vec b\sin \omega t$$

$$=-\omega^2\vec r$$

The acceleration is everywhere towards the origin and proportional to $\vec r.$