1. Nov 4, 2013

### aleksbooker

1. The problem statement, all variables and given/known data

A rotating beacon is located 2 miles out in the water. Let A be the point on the shore that is closest to the beacon. As the beacon rotates at 10 rev/min, the beam of light sweeps down the shore once each time it revolves. Assume that the shore is straight. How fast is the point where the beam hits the shore moving at an instant when the beam is lighting up a point 2 miles along the shore from point A?

2. Relevant equations

This is a self-study question that I took from the Ohio State University Coursera course (7.08 A Beacon Problem, if you want to see it for yourself). When I first saw it, it looked a lot like a physics question, so I tried to solve it using physics - turn rev/min into angular velocity, use that to calculate tangential velocity, then find x-component of that tangential velocity $\vec{v}_{tx}$, which should be the speed at the which the beam of light is moving along the shore at that instant.

My (physics-derived) answer, 40π, is exactly half of the correct (calculus-derived) answer, 80π. That makes me think that somewhere along the way, I must have made some mistake or misplaced a 2, but I can't tell where. Why is my physics answer different from the correct answer (calculated via related rates)?

3. The attempt at a solution

Let's say that the beam of light hits the shore at point B, 2 miles to the right of point A. Since the beacon light has uniform circular motion, we should be able to calculate $v_t$ like so:

$\omega = \frac{10rev}{min} \cdot \frac{2 \pi rad}{rev} = \frac{20 \pi rad}{min} \\ r = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2 \sqrt{2} mi \\ v_t = \omega \cdot r \\ v_t = \frac{20 \pi \, rad}{min} \cdot 2\sqrt{2}mi = 40 \pi \sqrt{2} \frac{mi}{min} \\$

I'm assuming that $\vec{v_t}$ is perpendicular to the beam of light, and that $\vec{v_t}_x$ runs along the x-axis. I'm also assuming that the angle between $\vec{v_t}$ and $\vec{(v_t)}_x$ is $45^\circ$, since $\vec{v_t}$ is always parallel to the direction of motion and the complementary angle must be 45 when $\theta = 45^\circ$ (the triangle totals 180, and since both the x and y sides are 2, their angles must be 45 each), then...

$\cos{45} = \frac{\vec{(v_t)}_x}{\vec{v_t}} \\ \vec{(v_t)}_x = \vec{v_t} \cdot 20 \pi \cos{45} = 40 \pi \\ \vec{(v_t)}_x = 40 \pi$

But...

The calculation via related rates gives this solution...

$\frac{d\theta}{dt} = 2\pi \cdot 10 = 20\pi \\ \tan{\theta} = \frac{x}{2} = \frac{1}{2}x \\ (\tan{\theta})' = (\frac{1}{2}x)' \\ \sec^2{\theta} \cdot \frac{d\theta}{dt} = \frac{1}{2} \cdot \frac{dx}{dt} \\ \frac{dx}{dt} = 2 \cdot \sec^2{\theta} \cdot 20\pi \\ = 2 \cdot (\frac{1}{\cos{45}})^2 \cdot 20\pi \\ = 2 \cdot (\frac{1}{\frac{\sqrt{2}}{2}})^2 \cdot 20\pi \\ = 2 \cdot \frac{1}{\frac{2}{4}} \cdot 20\pi \\ = 2 \cdot 2 \cdot 20\pi \\ = 80\pi$

I totally accept the calculus solution and explanation. I just don't understand why I couldn't get the same answer through physics.

File size:
57.8 KB
Views:
34
2. Nov 4, 2013

### tiny-tim

hi aleksbooker! welcome to pf!

you multiplied by cos45° = 1/√2, instead of dividing by it (so your answer was exactly half)

(your reality-mistake was that the "tangential" speed is a component of the actual speed, but you were treating it as the other way round)

3. Nov 4, 2013

### aleksbooker

Thanks, tiny-tim! That was such a frustrating problem.

Just to confirm, the mistake I made looks like this?

I feel like I've made this mistake many, many times. Should I always set the full value as the hypotenuse and the components as the "x" and "y" (or "a" and "b") sides?

File size:
28 KB
Views:
32
4. Nov 4, 2013

### tiny-tim

hi aleksbooker!
the components will always be smaller than the full value, so yes, the full value has to be the hypotenuse

5. Nov 5, 2013

### aleksbooker

Gotcha. Learning something new everyday.