# Deceptively Difficult Physics Integration Problem (Restorative Forces)?

by cj
Tags: deceptively, difficult, forces, integration, physics, restorative
 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)
 Sci Advisor HW Helper P: 2,589 $$F(x) = -kx^{-2}$$ $$ma = -kx^{-2}$$ $$dv/dt = -(k/m)x^{-2}$$ $$(dv/dx)(dx/dt) = -(k/m)x^{-2}$$ $$vdv = -(k/m)x^{-2}dx$$ $$\frac{v(t)^2}{2} = \frac{k}{m} \left (\frac{1}{x(t)} - \frac{1}{b} \right )$$ 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.
 Sci Advisor HW Helper P: 2,589 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.
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 by 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.
 Sci Advisor HW Helper P: 2,589 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 $\frac{k}{m}$.
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:

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

rather than:

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

??

 Quote by 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 $\frac{k}{m}$.
 Sci Advisor HW Helper P: 2,589 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: $$\frac{v^2}{2}|_{0}^{v(t)} = (k/m)(1/x)|_{b} ^{x(t)}$$
 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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