Recent content by pelmel92

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    Solving for θ in a Parabolic Orbit

    Homework Statement A comet of mass m moves in a parabolic orbit in the ecliptic plane (the plane of Earth’s orbit), so its perihelion distance ρ (its closest distance to the Sun) is less than Ro (the orbital distance of the Earth around the Sun) and occurs when θ = 0 for the comet. (The...
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    The Path of Light Using Calculus of Variations

    Homework Statement A rectangular solid of height h increases in density as its height increases, so the index of refraction of the solid increases with height according to: n(y) = 1.20(2y + 1) where y is the distance, in meters, from the origin (see diagram). A beam of light...
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    Gravitational Potential Due to a Thin Rod of Varying Density.

    Alrighty, figured that bit out on my own as well... polar coordinates work out nicely.
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    Gravitational Potential Due to a Thin Rod of Varying Density.

    Ah, I think I may have figured it out... x isn't actually changing here, so my attempts to relate it to y were completely unnecessary? The answer then (I think) would be: φ(x) = --Gk[√(x^2+L^2) - x] . Part b has me a bit stumped though... help would be lovely.
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    Gravitational Potential Due to a Thin Rod of Varying Density.

    Homework Statement GRAVITATIONAL POTENTIAL AND FIELD DUE TO A “THIN” ROD A thin rod of length L lies along the +y-axis, with one end at the origin (see diagram). Assume: • The rod has length only- no thickness in other directions. • The density of the rod increases proportionally to...
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    Angular Momentum Conservation with a Parakeet Landing on a Turntable

    i guess that's always possible... it's a webassign homework, which has been known to fudge the grading on occasion... but i feel much better about it if you think my approach looked good, thanks for all the help :)
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    Angular Momentum Conservation with a Parakeet Landing on a Turntable

    hold on...thought i caught it but I'm still a bit confused here... .5MR^2ω1 +mR^2ω1= .5MR^2ωf +mvR so .5MR^2ω1 +mR^2ω1 -mvR = .5MR^2ωf and (.5MR^2ω1)/(.5MR^2) +(mR^2ω1)/(.5MR^2) - (mvR)/(.5MR^2) = ωf right? which means that ω1 +2(mω1)/M - 2(mv)/(MR) = ωf maybe I'm just being dense...
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    Calculating Change in Entropy: 1100 kg Cars Colliding at 75km/hr

    i imagine you're meant to use the relationship ∆S=k*log(w) where ∆s=entropy change, k is 1.380 6504(24)×10-23 J/K, and w is work... work is kinetic energy, so... got it?
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    Angular Momentum Conservation with a Parakeet Landing on a Turntable

    ωf= ω1 + 2mω1/M -2mv/(MR) do i just suck at algebra?
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    Which Ramp Requires Less Work?

    actually ignoring friction, as long as you're raising both the same height (in this case, from the ground to the truck bed) the **work will be the same regardless of the path** (i.e. the steepness of the ramp). work is force times distance, but while the force required to lift something up a...
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    Angular Momentum Conservation with a Parakeet Landing on a Turntable

    Homework Statement Consider a cylindrical turntable whose mass is M and radius is R, turning with an initial angular speed ω1. (a) A parakeet of mass m, after hovering in flight above the outer edge of the turntable, gently lands on it and stays in one place on it, as shown below. What is the...
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    Gravitational/Spring Potential Energy problem

    if it helps, here's the diagram they give http://www.webassign.net/tipler6/7-p-094.gif
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    Gravitational/Spring Potential Energy problem

    well i started off with .5kx^2=mg∆H but since the question defined h as the height the object fell when it first touches the uncompressed spring, i thought i would probably need to consider that in my ∆H. so from there P.E.=mg(h+xsin(ø))=.5kx^2 then in quad format: 0=.5kx^2 - mgsin(ø)x -mgh
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    Gravitational/Spring Potential Energy problem

    thanks for giving me a hand :) do you mean i should subtract the square root instead of add? or should i have originally set .5kx^2 equal to -mg∆h? I'm a little confused as to how to handle this problem :/
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    Gravitational/Spring Potential Energy problem

    haha whoops, i copied that over without the h's, but the answer comes up wrong even as √(2mgh/h) or √(-2mgh/k)
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