Recently I am self-learning "Introduction to classical mechanics" by David Morin. There is a problem in chapter as stated below:
Walking east on a turntable
A person walks at constant speed $v$ eastward with respect to a turntable that rotates counterclockwise at constant frequency $ω$. Find the general expression for the person’s coordinates with respect to the ground (with the $x$-direction taken to be eastward).
Actually, I can fully understand the method used by the book but I tried to use another method to get the x-coordinate and somehow I cannot get it correct. Could anyone explain why the two methods are not coherent please?
Solution by the book
$$\dot x=v-u\,sin\,\theta \quad... (1) $$ $$\dot y=u\,cos\,\theta\quad...(2)$$ $$u=r\omega, \,y=r\,sin\,\theta, \, x=r\,cos\,\theta\quad...(3)$$
By substituting equations (3) into (1) and (2), we get
$$\dot x=v-y\omega\quad...(4)$$ $$\dot y=x\omega\quad...(5)$$ By differentiating (4), we get $$\ddot x=-\dot y\omega\quad...(6)$$ By substituting (5) into (6), we get $$\ddot x=-\omega^{2} x\quad...(7)$$ Finally, the solution of (7) is $$x=x_mcos(\omega t+\phi)$$ To cater for easy comparison, I choose an initial condition "At t=0, x=0", so $$x=\frac{v}{\omega}\,cos(\omega t-\frac{\pi}{2})$$ $$x=\frac{v}{\omega}\,sin\,\omega t$$
My method
According to my knowledge, radial velocity is only contributed by component of v in alignment with the radial direction, so $$\dot r=v\,cos\,\theta$$ $$\theta=\omega t$$ By substituting and integrating both sides of the equations, $$r=\int v\,cos\,\omega t \,dt$$ $$r=\frac{v}{\omega}\,sin\,\omega t$$ Finally, $$x=r\,cos\,\omega t=\frac{v}{\omega}\,sin\,\omega t\,cos\,\omega t$$
x-coordinate by the book method: $x=\frac{v}{\omega}\,sin\,\omega t$
x-coordinate by my method: $x=r\,cos\,\omega t=\frac{v}{\omega}\,sin\,\omega t\,cos\,\omega t$
Here, we can see that my answer has an extra cosine function inside. Can anyone explain to me what I did wrong in this question? Thank you so much!
Edit 1 (15th August) I have made a mistake in assuming $\theta=\omega t\; (or\, \dot \theta=\omega)$. The correct expressions are
$$\dot \theta=\omega -\frac{v}{r}sin\,\theta$$ $$\dot r=v\,cos\theta$$
May I know is there anyway that I could use to solve r(t) and $\theta (t)$ without using cartesian coordinates?
