Shipnutz
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- Homework Statement
- Two square reflectors, each 1.00 cm on a side and of mass 5.00 g, are located at opposite ends of a thin, extremely light, 1.00 m rod that can rotate without friction and in a vacuum about an axle perpendicular to it through its center (Figure 1). These reflectors are small enough to be treated as point masses in moment-of-inertia calculations. Both reflectors are illuminated on one face by a sinusoidal light wave having an electric field of amplitude 1.35 N/C that falls uniformly on both surfaces and always strikes them perpendicular to the plane of their surfaces. One reflector is covered with a perfectly absorbing coating, and the other is covered with a perfectly reflecting coating.
- Relevant Equations
- 1/A dp/dt = S/c = BE/μc
F = dp/dt
F = τ = I α
ρ_rad = 2I/c (Perfect reflector)
ρ_rad = I/c (Perfectly absorbed)
(Intensity) = E_max^2/2μc
(inertia) = mL^2
Okay so I actually have the answer because my teacher basically just gave it to me, but I would really like to know why I was even wrong in the first place. Here's my steps:
1. Knowing the momentum transfer per unit area is described by: 1/A dp/dt = S/c. I can begin by relating some known classical quantities to describe the rotational motion.
2. Taking the average of the Poynting vector gives me S_av = (Intensity). Acknowledging that the reflected surface momentum is traveling from +p to -p, and the absorbed surface from +p to 0 , which I can then take the difference of 2p - p = p. Therefore, I can use 1/A dp/dt for the left side of my equation. From there I can replace S on the right side with (Intensity), then substitute (Intensity) in with E_max^2/2μc, which gives me:
1/A dp/dt = E_max^2/2μc^2
3. Realizing that dp/dt = F and F = τ = (Inertia)α, I can substitute dp/dt with (Inertia)α, giving me 1/A(Inertia)α = E_max^2/2μc^2
4. Now, according to the parallel axis theorem, I_p = I_cm + mL^2, but because the rod has a negligible mass I just need to use ∑mL^2 = mL^2 + mL^2 for both reflecting plates. This yields (Inertia) = 2mL^2.
5. So now I can substitute the inertia in and solve for α:
1/A (2mL^2)α = E_max^2/2μc^2
α = E_max^2*A/4mL^2μc^2
When I solved it this way I was off by a factor of two, and I'm not sure why. Any clarification would be greatly appreciated.
1. Knowing the momentum transfer per unit area is described by: 1/A dp/dt = S/c. I can begin by relating some known classical quantities to describe the rotational motion.
2. Taking the average of the Poynting vector gives me S_av = (Intensity). Acknowledging that the reflected surface momentum is traveling from +p to -p, and the absorbed surface from +p to 0 , which I can then take the difference of 2p - p = p. Therefore, I can use 1/A dp/dt for the left side of my equation. From there I can replace S on the right side with (Intensity), then substitute (Intensity) in with E_max^2/2μc, which gives me:
1/A dp/dt = E_max^2/2μc^2
3. Realizing that dp/dt = F and F = τ = (Inertia)α, I can substitute dp/dt with (Inertia)α, giving me 1/A(Inertia)α = E_max^2/2μc^2
4. Now, according to the parallel axis theorem, I_p = I_cm + mL^2, but because the rod has a negligible mass I just need to use ∑mL^2 = mL^2 + mL^2 for both reflecting plates. This yields (Inertia) = 2mL^2.
5. So now I can substitute the inertia in and solve for α:
1/A (2mL^2)α = E_max^2/2μc^2
α = E_max^2*A/4mL^2μc^2
When I solved it this way I was off by a factor of two, and I'm not sure why. Any clarification would be greatly appreciated.