Let's call the angle between OM and ON as $\theta$. Since OM and OP are rotating in opposite directions and OM is rotating at a faster rate, the angle between OM and OP, which we will call $\phi$, is decreasing at a constant rate of $30^\circ - 6^\circ = 24^\circ$ per second.
We want to find the smallest value of t such that $\theta$ is an integer multiple of $360^\circ$ and $\theta - \phi$ is an integer multiple of $360^\circ$. This means that OM and OP will coincide again when both angles have made a whole number of revolutions.
We know that $\theta$ increases at a rate of $30^\circ$ per second, so after t seconds, $\theta = 30t^\circ$. Since $\theta$ must be an integer multiple of $360^\circ$, we can write this as $30t = 360n$ for some integer n. Simplifying, we find $t = 12n$.
Similarly, we know that $\phi$ decreases at a rate of $24^\circ$ per second, so after t seconds, $\phi = -24t^\circ$. Since $\theta - \phi$ must be an integer multiple of $360^\circ$, we can write this as $30t - (-24t) = 360m$ for some integer m. Simplifying, we find $54t = 360m$. Dividing both sides by 6, we have $9t = 60m$, which means $t = \frac{60}{9}m = \frac{20}{3}m$.
Since both t and m must be integers, we can see that the smallest possible value for t is 20, which gives us $t = \frac{20}{3}m = \frac{20}{3} \cdot 6 = 40$.
At t=30 minutes, or t=1800 seconds, we can calculate how many more revolutions ray OP has completed than ray OM by finding the difference in their angles. OM has completed an angle of $30^\circ \cdot 1800 = 54000^\circ$. OP has completed an angle of $6^\circ \cdot 1800 = 10800^\circ$. The difference is $54000^\circ - 10800^\circ = 43200^\circ$. Since there are 360 degrees in a revolution, this means that OP has completed $\frac{43200}{360} = \boxed{120}$ more revolutions than OM.
Rays OM and ON and OP coincide at time t=0. At the same instant in time, OM and OP begin to rotate in the plane about point O, but in opposite directions. Ray ON remains fixed. Ray OM rotates at a constant rate of 30˚ per second with respect to ON, while OP rotates at a constant rate of 6˚ per second. What is the smallest possible value of t, in seconds, in which rays OM and OP will coincide again? When t=30 mins, how many more revolutions has ray OP completed than ray OM?
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