GREPhysics.NET: GR9677 #83

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GR9677 #83

Problem

GREPhysics.NET Official Solution

Alternate Solutions

This problem is still being typed.

Mechanics $\Rightarrow$ }Rippled Surface

The simple intuitive way to solve this is to note that for $d \rightarrow \infty$ , $v \rightarrow 0$ , since one gets an infinitely steep (vertical line) hill, and the only way for the particle to stay on the surface (i.e., not accelerate on it) at the vertical drop is if its velocity is 0. The onyl choice with d on the denominator is choice (D).

The more rigorous solution is due to Sara Salha.

Equating centripetal force with gravity at the top of the hill, one has $mv^2/r=mg \Rightarrow v=\sqrt{mgr}$ . The non-trivial bit comes from calculating the radius.

Recall the radius of curvature from calculus $1/r = \kappa = \frac{|\ddot{x}\dot{y}-\ddot{y}\dot{x}|}{(\dot{x}^2+\dot{y}^2)^{3/2}}$ . Defining a parameter t as the independent variable, and defining $x=t$ , $y=d\cos(kt)$ , one finds that $1/4 = \frac{k^2d \cos(kt)}{1+(kd)^2\sin^2(kt)}$ . Evaluate it at $t=0$ to find $1/r(0)=k^2d \Rightarrow r=1/(k^2d)$ , the radius of curvature at the top of the hill. Plug that into the equation for velocity above to get $v=\sqrt{\frac{mg}{kd^2}}$ , as in choice (D).

See below for user comments and alternate solutions!

Alternate Solutions

Herminso 2009-09-21 12:24:31	Physically the limit case is such that: $-mg=m\ddot{y}=m\ddot{h}(x)$ , just when the particle will stay yet on the surface al all times. The particle go out of the path if \| $\ddot{y}$ \| > \| $\ddot{h}(x)$ \|, in that case the trayectory can raising off the surface. Now, $-mg=m\ddot{h}(x)=-mdk^2\dot{x}^2 \cos ^2{(kx)}=-mdk^2v_x^2\cos ^2{(kx)}$ , so $v_x=\sqrt{g/k^2d}\frac{1}{\cos ^2{(kx)}}$ $\Rightarrow$ $v_x(x=0)=\sqrt{g/k^2d}$ , and then $v_x\le\sqrt{g/k^2d}$ . Reply to this comment
Rus Almighty 2008-04-29 08:35:33	A condition for the body to stay on the track: $\frac{Vy}{Vx} = \frac{\partial h}{\partial x}$ let's derive both sides by t and remember that on the top $kx= 2 \pi$ , Vx is V , $\dot{Vy} = -g$ and $\dot{Vx} = 0$ 'cuase there is only gravitational force on the top. we get: $\frac{\dot{Vy}}{Vx} = \frac{\partial\dot{h}}{\partial x}$ now let's calculate $\frac{\partial\dot{h}}{\partial x}$ $\frac{\partial h}{\partial x} = - kd sin kx$ $\frac{\partial\dot{h}}{\partial x} = - k^2 d sin kx dot(x) = - Vx k^2 d cos kx$ And from now on it's only algebra: $- \frac{g}{Vx} = - Vx k^2 d cos kx$ $kx= 2 \pi$ $Vx= \sqrt{\frac{g}{k^2 d}}$ QED Reply to this comment

Comments

sullx
2009-11-04 22:50:31

In the case that one does not realize the limiting scenario that Yosun provided, one could approximate the solution by doing some baby physics:

For a particle traveling over one of the given "ripples" consider the force equation:

$mg-N=\frac{mv^2}{d}$

This crudely approximates the ripple as a circle with radius d. The max speed just before the particle lifts off the ripple will be when $N$ $\Rightarrow$ $0$ . Solving for v gives $v=\sqrt{\frac{g}{d}}$ as the upper limit.

Only answer D resembles this.

cilginfizikci
2010-03-08 23:42:31

you have solved the equation wrongly... V= (gd)^0.5... and according to ur solution the answer is A... check ur calc. pls

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NEC

Herminso
2009-09-21 12:24:31

Physically the limit case is such that:

$-mg=m\ddot{y}=m\ddot{h}(x)$ ,

just when the particle will stay yet on the surface al all times. The particle go out of the path if | $\ddot{y}$ | > | $\ddot{h}(x)$ |, in that case the trayectory can raising off the surface. Now,

$-mg=m\ddot{h}(x)=-mdk^2\dot{x}^2 \cos ^2{(kx)}=-mdk^2v_x^2\cos ^2{(kx)}$ ,

so $v_x=\sqrt{g/k^2d}\frac{1}{\cos ^2{(kx)}}$ $\Rightarrow$ $v_x(x=0)=\sqrt{g/k^2d}$ , and then $v_x\le\sqrt{g/k^2d}$ .

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Alternate Solution - Unverified

jw111
2008-11-07 21:51:08

draw a free projection track on question first

the trajectory is

$y=-(1/2)g\frac{x^2}{v^2}$

the condition that the particle stay on the surface is
the trajectory ALWAYS below the suface
that is at x = 0 (as you can see from your drawing)

| $\frac{d^2y}{dx^2}$ | > | $\frac{d^2h}{dx^2}$ |

$\frac{g}{v^2}$ > $dk^2$

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NEC

ayabepaula
2008-10-23 17:24:21

${a_{max}}=g$

by definition

${a_{max}}=(amplitude)\times{(\omega)^{2}}$

Since

$y(x)=Asin(\omega.t+\phi)$

Here

Amplitude= $d$

then

${a_{max}}=g=d.{\omega}^{2}$

but

$\omega=v.k$

then

$v=\sqrt{\frac{g}{{k}^{2}d}}$

chemicalsoul
2009-10-25 18:49:16

this is the best way !

Plantis
2010-04-03 13:05:28

It is the best solution. It's so easy and represend all physics of this situation.

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NEC

dannyboytward
2008-10-17 10:36:56

Another solution:

Consider a reference frame moving at speed v. The particle only moves vertically. The surface is described by

d*cos[k(x-vt)]

The acceleration of the surface is the second time derivative

-d*(k*v)^2*cos[k(x-vt)]

which has maximum value of d*k^2*v^2. If this value is less than g, we have our answer.

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NEC

Rus Almighty
2008-04-29 08:35:33

A condition for the body to stay on the track:
$\frac{Vy}{Vx} = \frac{\partial h}{\partial x}$

let's derive both sides by t and remember that on the top $kx= 2 \pi$ , Vx is V , $\dot{Vy} = -g$ and $\dot{Vx} = 0$ 'cuase there is only gravitational force on the top.

we get:
$\frac{\dot{Vy}}{Vx} = \frac{\partial\dot{h}}{\partial x}$

now let's calculate $\frac{\partial\dot{h}}{\partial x}$
$\frac{\partial h}{\partial x} = - kd sin kx$
$\frac{\partial\dot{h}}{\partial x} = - k^2 d sin kx dot(x) = - Vx k^2 d cos kx$
And from now on it's only algebra:
$- \frac{g}{Vx} = - Vx k^2 d cos kx$
$kx= 2 \pi$
$Vx= \sqrt{\frac{g}{k^2 d}}$
QED

Rus Almighty
2008-04-29 08:44:53

fix,
instead of $\frac{\partial\dot{h}}{\partial x} = - k^2 d sin kx dot(x) = - Vx k^2 d cos kx$
$\frac{\partial\dot{h}}{\partial x} = - k^2 d cos kx \dot x = - Vx k^2 d cos kx$

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Alternate Solution - Unverified

panos85
2007-10-22 11:14:45

Am I the only one thinking that it is more natural to let d go to 0 rather than infinity? If d=0, the particle will stay on track for any v, so (D).

elisa
2008-04-02 19:35:16

Great way to see it. The site is wonderful! Thank you everyone.

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NEC

emailzac
2007-10-08 10:47:34

I'm pretty sure you can use limiting case here. If d gets very large, and k stays the same, the curve is going to be tighter, meaning the velocity's limit is going to decrease, therefore D is the only logical choice.

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NEC

georgi
2007-08-30 22:53:17

there is another great solution that is intuitive. in order for the particle to stay its velocity must be less than or equal to the phase velocity w/k of the wave. we are given k so we calculate w. w = sqrt(spring constant/m). recall F = spring constant * distance ==> mg = spring constant *d so spring constant = mg/d ==> w = sqrt(g/d) and voila your answer!

sawtooth
2007-10-29 10:40:17

I am sure this was one of the ways to solve the problem, that the guys at ETS wanted us to use. On the other hand, I am not conviced by your arguments georgi. It is not concise enough. I mean, for harmonic motion, you have to have a force that is propotional to the distance (or accelaration for that matter). $F=mg$ and we know $g$ doesn't vary that much in small distances, and it is not that small to approximate its change as linear in distance. Furthermore, I think the frictionless surface demands a force vertical to the surface and the projection to the vertical axis has to have something to do with the vertical accelaration that has to be linear in distance. Something in the ways of skozmedia. I think there's another, similar, excercise in 8677 that uses chain rules and stuff. Anyway, I don;t have the time to write a complete answer, but even if it gives the right answer, I am not quite satisfied. Any thoughts?

p3ace
2008-07-26 15:23:37

The expression for h is the solution to the harmonic oscillator problem. That is why it is valid.

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NEC

skozmedia
2006-03-27 23:14:32

Actually, there's a much more direct way. Use the fact that y(x)=-.5gt^2 and h(x)=-dcos(kx) along with x=vt. Substitute in, calculate dy/dx and dh/dx. You will need to use l'Hopital's rule for x=0, but that's easy.

Jeremy
2007-11-10 11:33:16

I just did something similar... Start by expanding $h(x)$ about $x=0$ (the first hilltop): $h(x)= d \cos (k x) \approx d-\frac{1}{2}d k^{2} x^{2}$ . Now consider the path of a particle starting at the top of the hill with tangential velocity $v$ ; its position is given by two equations: $x=v t$ , and $y=d-\frac{1}{2}g t^{2}$ . For comparison with $h(x)$ , combine the two previous equations to eliminate time: $y(x)=d-\frac{1}{2}\frac{g}{v^{2}} x^{2}$ . Thus, $d k^{2}=\frac{g}{v^{2}}$ or $v_{max}=\sqrt{\frac{g}{d k^{2}}}$ gives the maximal velocity. With this velocity, the particle follows the surface in freefall, e.g. if you were in a car going over the hill at this speed, you would experience weightlessness. At higher speeds, the particle will leave the surface. At lower speeds, the particle is kept at the surface via a normal force. The expansion about $x=0$ is justified because this is where the largest curvature is. (In fact, this is the same assumption used in the "rigorous" version of the official solution).

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NEC

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