Deceptively Difficult Physics Integration Problem (Restorative Forces)?

[cj]

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)

[AKG]

F(x) = -kx^{-2}

ma = -kx^{-2}

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

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

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

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

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]

Actually, it's quite simple:

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

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

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

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

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

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

The rest should be pretty simple, unless you're just rusty on the calculus. They're suggesting a substitution: y = sin²\theta.

\int _1 ^0 \sqrt{\frac{y}{1-y}}dy

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

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

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

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

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

= \theta - \frac{1}{2}\sin (2\theta )

= (-\pi /2)

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

[cj]

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!


F(x) = -kx^{-2}

ma = -kx^{-2}

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

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

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

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

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]

cj

The second last line of what you quoted was:

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

Now, showing all my steps:

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

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

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

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

And the last line was:

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

There's no difference between the last two lines here, just changed a (k/m) to a \frac{k}{m}.

[cj]

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:

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

rather than:

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


??

cj

The second last line of what you quoted was:

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

Now, showing all my steps:

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

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

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

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

And the last line was:

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

There's no difference between the last two lines here, just changed a (k/m) to a \frac{k}{m}.

[AKG]

There's no difference:

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

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

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

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

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

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

[cj]

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!