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?
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.
To find the smallest possible value of t in which rays OM and OP will coincide again, we need to determine when their angles from ON are equal again.
Given that OM rotates at a rate of 30 degrees per second with respect to ON, and OP rotates at a rate of 6 degrees per second, we can set up the following equation to represent their angular positions at time t:
30t = -6t + 360n
where n is the number of complete revolutions made by OP.
Simplifying the equation, we have:
36t = 360n
To find the smallest possible value of t, we need to find the smallest positive integer value for n that will satisfy this equation.
Since 36 and 360 share a common factor of 36, we can divide both sides of the equation by 36 to simplify further:
t = 10n
Therefore, the smallest possible value of t in which rays OM and OP will coincide again is when n = 1:
t = 10 * 1 = 10 seconds
So, the smallest possible value of t is 10 seconds.
To find the number of revolutions completed by ray OP when t = 30 minutes, which is equal to 30 * 60 = 1800 seconds, we can substitute this value into the equation:
36t = 360n
36 * 1800 = 360n
64800 = 360n
Dividing both sides by 360, we find:
n = 64800 / 360
n = 180
Therefore, ray OP has completed 180 complete revolutions when t = 30 minutes.
To find how many more revolutions ray OP has completed than ray OM when t = 30 minutes, we need to determine the number of revolutions made by ray OM at this time. We know that ray OM rotates at a rate of 30 degrees per second, so the number of revolutions made by ray OM in 1800 seconds is:
1800 / (360 / 30) = 1800 / 6 = 300
Therefore, ray OP has completed 180 more revolutions than ray OM when t = 30 minutes.
To find the smallest possible value of t in which rays OM and OP will coincide again, we need to determine the time it takes for the two rays to align with each other.
Since OM rotates at a constant rate of 30˚ per second with respect to ON, we can express its angle in relation to ON as θ1 = 30t, where θ1 is the angle made by OM with respect to ON at time t.
Similarly, OP rotates at a constant rate of 6˚ per second. Therefore, its angle in relation to ON can be expressed as θ2 = -6t, where θ2 is the angle made by OP with respect to ON at time t.
To find when rays OM and OP coincide, we need to set their respective angles equal to each other: θ1 = θ2.
Substituting the expressions for θ1 and θ2, we get: 30t = -6t.
Solving for t, we find: 30t + 6t = 0 → 36t = 0.
Therefore, the smallest possible value of t in which rays OM and OP will coincide again is t = 0 seconds.
Now, to determine how many more revolutions ray OP has completed than ray OM when t = 30 minutes, we need to calculate their respective revolutions at that time.
Since OM rotates at a rate of 30˚ per second, the number of revolutions made by OM at time t is given by: Rev_OM = θ1 / 360°.
Substituting the expression for θ1, we get: Rev_OM = (30t) / 360°.
At t = 30 minutes = 30 * 60 seconds = 1800 seconds, we can calculate Rev_OM as: Rev_OM = (30 * 1800) / 360° = 150 revolutions.
On the other hand, OP rotates at a rate of 6˚ per second. Therefore, the number of revolutions made by OP at time t is given by: Rev_OP = θ2 / 360°.
Substituting the expression for θ2, we get: Rev_OP = (-6t) / 360°.
At t = 30 minutes = 1800 seconds, we can calculate Rev_OP as: Rev_OP = (-6 * 1800) / 360° = -30 revolutions.
Hence, when t = 30 minutes, ray OP has completed 30 more revolutions than ray OM.