Question

A particle moves so that its position vector ˜ r at time t is ˜ r = ˜ a cos wt + ˜ b sin wt, where w is a constant and ˜ a and ˜ b are constant vectors. Show that (a) ˜ r · ˙ ˜ r is independent of t, (b) the acceleration is everywhere towards the origin and proportional to ˜ r

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 $\vec r.$