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Deceptively Difficult Physics Integration Problem (Restorative Forces)?

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cj
#1
Jul18-04, 05:51 PM
P: 66
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)
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AKG
#2
Jul18-04, 08:43 PM
Sci Advisor
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P: 2,586
[tex]F(x) = -kx^{-2}[/tex]

[tex]ma = -kx^{-2}[/tex]

[tex]dv/dt = -(k/m)x^{-2}[/tex]

[tex](dv/dx)(dx/dt) = -(k/m)x^{-2}[/tex]

[tex]vdv = -(k/m)x^{-2}dx[/tex]

[tex]\frac{v(t)^2}{2} = \frac{k}{m} \left (\frac{1}{x(t)} - \frac{1}{b} \right )[/tex]

I believe you can figure out the first hint from here. Note that the integration was done from time=0 to time=t. Given what you have, see if you can work with it to get the second hint.
AKG
#3
Jul18-04, 09:01 PM
Sci Advisor
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P: 2,586
Actually, it's quite simple:

[tex]dx/dt = -(2k/m)^{1/2}(1/x - 1/b)^{1/2}[/tex]

[tex](1/x - 1/b)^{-1/2}dx = -(2k/m)^{1/2}dt[/tex]

[tex]\left ( \frac{bx}{b-x} \right ) ^{1/2}dx = -(2k/m)^{1/2}dt[/tex]

[tex]\sqrt {\frac{y}{1-y}}dy = -(2k/mb^3)^{1/2}dt[/tex]

Now, integrate from [itex]t=0[/itex] to [itex]t=t_f[/itex]. You know that [itex]y(t_f) = x(t_f)/b = 0/b = 0[/itex] and [itex]y(0) = x(0)/b = b/b = 1[/itex]. So, you'll have:

[tex]\int _1 ^0 \sqrt{\frac{y}{1-y}}dy = -\sqrt{(2k/mb^3)}t_f[/tex]

The rest should be pretty simple, unless you're just rusty on the calculus. They're suggesting a substitution: y = sinē[itex]\theta[/itex].

[tex]\int _1 ^0 \sqrt{\frac{y}{1-y}}dy[/tex]

[tex]= \int _{\pi /2} ^0 \sqrt{\frac{\sin ^2 \theta}{1 - \sin ^2 \theta}}(2\sin \theta \cos \theta d\theta )[/tex]

[tex]= 2\int _{\pi /2} ^0 \sqrt{\frac{\sin ^2 \theta}{\cos ^2 \theta}} \sin \theta \cos \theta d\theta[/tex]

[tex]= 2\int _{\pi /2} ^0 \frac{\sin \theta}{\cos \theta} \sin \theta \cos \theta d\theta[/tex]

[tex]= 2\int _{\pi /2} ^0 \sin ^2 \theta d\theta[/tex]

[tex]= \int _{\pi /2} ^0 1 - \cos (2\theta ) d\theta[/tex]

[tex]= \theta - \frac{1}{2}\sin (2\theta )[/tex]

[tex]= (-\pi /2)[/tex]

EDIT: no, this looks fine. This is the integral on the left side. Now isolate [itex]t_f[/itex] and you're done.

cj
#4
Jul21-04, 06:06 AM
P: 66
Deceptively Difficult Physics Integration Problem (Restorative Forces)?

Thanks very much -- this is quite helpful.

However (with regard to the last line in your
post below), if you integrate from t = 0 to t = t
is the order not (1/b - 1/x) as opposed to (1x - 1/b)??

Thanks!


Quote Quote by AKG
[tex]F(x) = -kx^{-2}[/tex]

[tex]ma = -kx^{-2}[/tex]

[tex]dv/dt = -(k/m)x^{-2}[/tex]

[tex](dv/dx)(dx/dt) = -(k/m)x^{-2}[/tex]

[tex]vdv = -(k/m)x^{-2}dx[/tex]

[tex]\frac{v(t)^2}{2} = \frac{k}{m} \left (\frac{1}{x(t)} - \frac{1}{b} \right )[/tex]

I believe you can figure out the first hint from here. Note that the integration was done from time=0 to time=t. Given what you have, see if you can work with it to get the second hint.
AKG
#5
Jul21-04, 10:37 AM
Sci Advisor
HW Helper
P: 2,586
cj

The second last line of what you quoted was:

[tex]vdv = -(k/m)x^{-2}dx[/tex]

Now, showing all my steps:

[tex]\int _{v(0)} ^{v(t)} vdv = \int _{x(0)} ^{x(t)} -(k/m)x^{-2}dx[/tex]

[tex]\int _0 ^{v(t)} vdv = (k/m) \int _b ^{x(t)} -x^{-2}dx[/tex]

[tex]\frac{v^2}{2}|_{0}^{v(t)} = (k/m)(1/x)|_{b} ^{x(t)}[/tex]

[tex]\frac{v(t)^2}{2} = (k/m)\left (\frac{1}{x(t)} - \frac{1}{b} \right )[/tex]

And the last line was:

[tex]\frac{v(t)^2}{2} = \frac{k}{m} \left (\frac{1}{x(t)} - \frac{1}{b} \right )[/tex]

There's no difference between the last two lines here, just changed a [itex](k/m)[/itex] to a [itex]\frac{k}{m}[/itex].
cj
#6
Jul21-04, 12:43 PM
P: 66
I certainly understand your reasoning -- thanks again.

What is still perplexing me is that, since the initial conditions have the
object located at x=b, shouldn't -- technically -- the interval be taken as:

[tex]\frac{v^2}{2}|_{v(t)}^{(0)} = (k/m)(1/x)|_{x(t)} ^{(b)}[/tex]

rather than:

[tex]\frac{v^2}{2}|_{0}^{v(t)} = (k/m)(1/x)|_{b} ^{x(t)}[/tex]


??

Quote Quote by AKG
cj

The second last line of what you quoted was:

[tex]vdv = -(k/m)x^{-2}dx[/tex]

Now, showing all my steps:

[tex]\int _{v(0)} ^{v(t)} vdv = \int _{x(0)} ^{x(t)} -(k/m)x^{-2}dx[/tex]

[tex]\int _0 ^{v(t)} vdv = (k/m) \int _b ^{x(t)} -x^{-2}dx[/tex]

[tex]\frac{v^2}{2}|_{0}^{v(t)} = (k/m)(1/x)|_{b} ^{x(t)}[/tex]

[tex]\frac{v(t)^2}{2} = (k/m)\left (\frac{1}{x(t)} - \frac{1}{b} \right )[/tex]

And the last line was:

[tex]\frac{v(t)^2}{2} = \frac{k}{m} \left (\frac{1}{x(t)} - \frac{1}{b} \right )[/tex]

There's no difference between the last two lines here, just changed a [itex](k/m)[/itex] to a [itex]\frac{k}{m}[/itex].
AKG
#7
Jul21-04, 04:10 PM
Sci Advisor
HW Helper
P: 2,586
There's no difference:

[tex]\frac{v^2}{2}|_{v(t)}^{(0)} = (k/m)(1/x)|_{x(t)} ^{(b)}[/tex]

[tex]\frac{0^2}{2} - \frac{v(t)^2}{2} = (k/m)[1/b - 1/x(t)][/tex]

[tex]-\frac{v(t)^2}{2} = (k/m)[1/b - 1/x(t)][/tex]

[tex]\frac{v(t)^2}{2} = (k/m)[1/x(t) - 1/b][/tex]

Work out this one, and you'll see it's the same:

[tex]\frac{v^2}{2}|_{0}^{v(t)} = (k/m)(1/x)|_{b} ^{x(t)}[/tex]
cj
#8
Jul22-04, 10:26 AM
P: 66
Got it -- thanks for answering this, as well as all my
other questions.

I was thinking there was a convention that said
something like the integration interval should
be taken as from the final or "end" state to the
initial state.

Again, thanks for all your answers!


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