- #1
cj
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The following seemed simple enough to me . . . I'm somewhat sure about the requisite physics, but shakey on the integrals:
A particle of mass m is released from rest a distance b from a fixed origin of force that attracts a particle according to the inverse square law:
F(x) = -kx-2
Show that the time required for the particle to reach the origin is:
π(mb3/8k)1/2
And then I reviewed the hints provided by the author -- the very first of which completely stumped me. My rusty calc skills not withstanding, how is the following hint true:
Show that dx/dt = -(2k/m)1/2 · (1/x - 1/b)1/2
the negative sign results from the physical situation
the subsequent hint is also a mystery to me:
Show that t = sqrt(mb3/2k) · ∫sqrt[y/(1-y)]dy
where y = x/b (evaluated from 1 to 0)
the 3rd hint is likewise elusive to me:
Show that setting y=sin2θ results in t = sqrt(mb3/2k) · ∫2sin2θdθ (evaluated from π/2 to 0)
let alone the final result of π(mb3/8k)
A particle of mass m is released from rest a distance b from a fixed origin of force that attracts a particle according to the inverse square law:
F(x) = -kx-2
Show that the time required for the particle to reach the origin is:
π(mb3/8k)1/2
And then I reviewed the hints provided by the author -- the very first of which completely stumped me. My rusty calc skills not withstanding, how is the following hint true:
Show that dx/dt = -(2k/m)1/2 · (1/x - 1/b)1/2
the negative sign results from the physical situation
the subsequent hint is also a mystery to me:
Show that t = sqrt(mb3/2k) · ∫sqrt[y/(1-y)]dy
where y = x/b (evaluated from 1 to 0)
the 3rd hint is likewise elusive to me:
Show that setting y=sin2θ results in t = sqrt(mb3/2k) · ∫2sin2θdθ (evaluated from π/2 to 0)
let alone the final result of π(mb3/8k)
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