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?

Young Kindaichi · Accepted Answer

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
$$y=rcos(omega t)$$
$$x=rsin(omega t)$$
Now you can proceed from here. :)

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

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

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