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1. Homework Statement
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. Homework Equations
This is a selfstudy 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 xcomponent of that tangential velocity [itex]\vec{v}_{tx}[/itex], which should be the speed at the which the beam of light is moving along the shore at that instant.
My (physicsderived) answer, 40π, is exactly half of the correct (calculusderived) 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 [itex]v_t[/itex] like so:
[itex]
\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} \\
[/itex]
I'm assuming that [itex]\vec{v_t}[/itex] is perpendicular to the beam of light, and that [itex]\vec{v_t}_x[/itex] runs along the xaxis. I'm also assuming that the angle between [itex]\vec{v_t}[/itex] and [itex]\vec{(v_t)}_x[/itex] is [itex]45^\circ[/itex], since [itex]\vec{v_t}[/itex] is always parallel to the direction of motion and the complementary angle must be 45 when [itex]\theta = 45^\circ[/itex] (the triangle totals 180, and since both the x and y sides are 2, their angles must be 45 each), then...
[itex]\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
[/itex]
But...
The calculation via related rates gives this solution...
[itex]\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[/itex]
I totally accept the calculus solution and explanation. I just don't understand why I couldn't get the same answer through physics.
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. Homework Equations
This is a selfstudy 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 xcomponent of that tangential velocity [itex]\vec{v}_{tx}[/itex], which should be the speed at the which the beam of light is moving along the shore at that instant.
My (physicsderived) answer, 40π, is exactly half of the correct (calculusderived) 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 [itex]v_t[/itex] like so:
[itex]
\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} \\
[/itex]
I'm assuming that [itex]\vec{v_t}[/itex] is perpendicular to the beam of light, and that [itex]\vec{v_t}_x[/itex] runs along the xaxis. I'm also assuming that the angle between [itex]\vec{v_t}[/itex] and [itex]\vec{(v_t)}_x[/itex] is [itex]45^\circ[/itex], since [itex]\vec{v_t}[/itex] is always parallel to the direction of motion and the complementary angle must be 45 when [itex]\theta = 45^\circ[/itex] (the triangle totals 180, and since both the x and y sides are 2, their angles must be 45 each), then...
[itex]\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
[/itex]
But...
The calculation via related rates gives this solution...
[itex]\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[/itex]
I totally accept the calculus solution and explanation. I just don't understand why I couldn't get the same answer through physics.
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