Locating a point in circular orbit on the Cartesian plane after some $t$ seconds

Question

The second hand of an analog clock has angular velocity $\omega=\pi/30$ rad.s^-1. The blue body in the image below mimics the hand's clockwise motion on the Cartesian plane with the center of revolution at $(0,0)$, the radius $r$ being, say, $2$ units, and initial position $(0,2)$. How can we determine the body's coordinate location $(x,y)$ after t seconds?
From here, I was under the impression that we can calculate it as follows:
$x=r*\cos(\omega*t)$
$y=r*\sin(\omega*t)$
Taking t to be $30$ seconds, this gives us:
$x=2*\cos(\pi/30*30)=1.997$
$y=2*\sin(\pi/30*30)=0.110$
Problem is, in reality, after $30$ seconds, the point should be at $(0,-2)$. Why did the formulae give conflicting results?

Answer 1

The formula you have written is for when you taking angle $\omega t$ from horizontal $x$-axis but in your case, the point is starting from the y-axis thus it's necessary to take the angle $\omega t$ from the positive y-axis. This will turn your formula to $$y=r\cos(\omega t)$$ $$x=r\sin(\omega t)$$ Now you can proceed from here. :)

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We are given $\omega=\pi/30$ rad-sec$^{-1}$ After $30$ sec $$y=2\cos\left(\frac{\pi}{30}\cdot 30\right)=2\cos(\pi)=-2$$ and $$x=2\sin\left(\frac{\pi}{30}\cdot 30\right)=2\sin(\pi)=0$$