Recent content by Ichigo449

Yes it is, so the moment arms are ##rsin(\beta)##. But the angular momentum argument is correct?

With the motor turned off and the angle to the vertical fixed at ##\beta## angular momentum should now be conserved (unsure about this). If it is conserved then the value at time 0 is: ##L = m[r_{0} + \delta]^{2}\omega##, which would be equal to the angular momentum at any subsequent time...

So how do I solve part d then? I agree it's definitely an interesting situation but can't wrap my head around how to solve it. Do I go back to the Lagrangian, find new equations of motion and then solve them subject to ω being the angular velocity at time 0?

Okay, I think I have a pretty good handle on 3/4 of the question then. But I'm still not sure how to even begin part d. Even just a hint for how to get started would be much appreciated.

Then it seems impossible for the equilibrium to ever be stable. A sinusoidal displacement from equilibrium would require that ##\beta^{2} ω^{2}## is a negative number, which can't happen since ##\beta## and ##\omega## are presumably positive real numbers.

Working with the displacement from equilibrium, $ r= r_{0} + ε(t)$, the equation of motion becomes: $\ddot{ε(t)} -gcosβ -ε(t)sin^{2}βω^{2} =-gcosβ$, which simplifies to, $\ddot{ε(t)} -ε(t)sin^{2}βω^{2} = 0$. Linearizing this expression gives: $\ddot{ε(t)} -ε(t)β^{2}ω^{2} =0$. In order for...

Homework Statement A bead of mass m slides in a frictionless hollow open-ended tube of length L which is held at an angle of β to the vertical and rotated by a motor at an angular velocity ω. The apparatus is in a vertical gravitational field. a) Find the bead's equations of motion b) Find...

Homework Statement Consider a given monochromatic component of sunlight. The electric field drives a given air molecule. Each oscillating charge of the air molecule radiates waves in all directions, some of which travel to the eye of a given observer. But, for a given molecule (call it No.1)...

So I went through the calculations and 4% is lost on first transmission, 0.0199 on each reflection and 4% is again lost on exiting. This is only a net loss in intensity of 8.0398%. So I made a mistake somewhere. For the reflections at oblique incidence I used that ##R = r_s^2 = tan^2 (\theta_t...

So for part b) am I applying the Fresnel equations correctly?

Thank you. I think I understand the process now. The Law of Reflection forces the complementary angles at both points to be equal, and these two angles are included in the triangle I mentioned so ##\beta = \pi/4## because each complementary angle is ##3\pi/8##.

I can't see a subtended angle for one reflection point, but ##\beta## is the subtended angle from both reflection points.

Could you give me a hint for what triangles to consider?

Since the light rays form a right triangle and the Law of Reflection holds 2\theta_i =\pi/4.

The question is from http://optics.byu.edu/BYUOpticsBook_2015.pdf Problem 3.5