Question

F(t) cRsin(wt)â+Rcos(wt)ý, where c, R, w are positive constants Find the angle between the velocity vector and the acceleration vector as a function of time. In particular, what is this angle at a time t T/2w? What does your expression reduce to when c 1? Can you What shape does the particle trace out as it moves through space? explicitly show that the particle's trajectory satisfies the defining equation for this shape?

Answer 1

given

$\vec r = c R \sin (wt)\hat x+R \cos (t w)\hat y$

velocity is

$\vec v=\frac{d\vec r }{dt}= c R w \cos (wt)\hat x-R w \sin ( wt)\hat y$

acceleration is

$\vec a=\frac{d\vec v }{dt}= -c R w^2 \sin (wt)\hat x-R w^2 \cos ( wt)\hat y=-w^2\vec r$

The dot product of velocity vector and acceleration vector is

$\vec a.\vec v =-c^2 R^2 w^3+R^2w^3\cos(wt) \sin (wt)=(1-c^2)R^2 w^3\cos(w t)\sin(w t)$

and

$|\vec a|=Rw^2\sqrt{\cos^2(wt)+c^2\sin^2(w t)}$

$|\vec v|=Rw\sqrt{\sin^2(wt)+c^2\cos^2(w t)}$

The angle between acceleration vector and velocity vector is

$\cos\theta =\frac{\vec a.\vec v}{|\vec a||\vec v|}=\frac{\left(1-c^2\right) \cos(w t)\sin ( wt)}{ \sqrt{c^2 \cos ^2( wt)+\sin ^2( wt)} \sqrt{c^2 \sin ^2( wt)+\cos ^2(wt)}}$

Hence

$\theta =\cos^{-1}\left(\frac{\left(1-c^2\right) \cos(w t)\sin ( wt)}{ \sqrt{c^2 \cos ^2( wt)+\sin ^2( wt)} \sqrt{c^2 \sin ^2( wt)+\cos ^2(wt)}} \right )$

when  

$t=2\pi/w$

$\sin(w t)=\sin(2\pi)=0$

then the angle is

$\theta =\cos^{-1}\left(0 \right )=\pi/2 = 90^o$

When c = 1

$\vec a.\vec v = 0$

Hence

$\theta =\cos^{-1}\left(0 \right )=\pi/2 = 90^o$

b) The shape of the trajectory is ellipse

given

$x=cR\sin( w t)$

and

$y=R\cos( w t)$

Hence

$(x/c)^2+y^2=R^2(\sin^2(wt)+\cos^2(wt))=R^2$

Hence

$\frac{x^2}{c^2 R^2}+\frac{y^2}{R^2}=1$

This the equation of ellipse. When c = 1 , the trajectory becomes a circle.