GR9677 #83


Problem


This problem is still being typed. 
Mechanics}Rippled Surface
The simple intuitive way to solve this is to note that for , , 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 . The nontrivial bit comes from calculating the radius.
Recall the radius of curvature from calculus . Defining a parameter t as the independent variable, and defining , , one finds that . Evaluate it at to find , the radius of curvature at the top of the hill. Plug that into the equation for velocity above to get , as in choice (D).


Alternate Solutions 
ryanmes 20181024 21:32:38  The easiest way to look at so many problems on this exam is limits. If d increases, then you need a smaller velocity to stay on the track. There is only one answer where this is the case (when d is in the denominator). And this is the answer.   elxcats 20131014 20:33:49  The condition for the object to fly off is that the normal force is equal to zero. So let's set up our force equation!
But
So
using
Doing basically the same thing again to find
Plugging that into the force equation we find
Solving for
The question asks for the velocity which the above condition will always be met. This means you want to maximize the cosine term (1) which gives you answer (D).
Hope that gives some insight!   Herminso 20090921 12:24:31  Physically the limit case is such that:
,
just when the particle will stay yet on the surface al all times. The particle go out of the path if  > , in that case the trayectory can raising off the surface. Now,
,
so , and then .   Rus Almighty 20080429 08:35:33  A condition for the body to stay on the track:
let's derive both sides by t and remember that on the top , Vx is V , and 'cuase there is only gravitational force on the top.
we get:
now let's calculate
And from now on it's only algebra:
QED  

Comments 
ryanmes 20181024 21:32:38  The easiest way to look at so many problems on this exam is limits. If d increases, then you need a smaller velocity to stay on the track. There is only one answer where this is the case (when d is in the denominator). And this is the answer.   ana123 20170909 03:29:16  h\'\'(x) = g\r\n\r\n\r\nv= \r\n
ana123 20170909 03:41:20 
Excuse the formatting.\r\n\r\nCorrection\r\nh\'\'(x)= g\r\n\r\n\r\n

  elxcats 20131014 20:33:49  The condition for the object to fly off is that the normal force is equal to zero. So let's set up our force equation!
But
So
using
Doing basically the same thing again to find
Plugging that into the force equation we find
Solving for
The question asks for the velocity which the above condition will always be met. This means you want to maximize the cosine term (1) which gives you answer (D).
Hope that gives some insight!   mpdude8 20120420 18:26:28  The only choices that give you the correct units of m/s for velocity are C and D.
hoboboxer 20130411 18:44:03 
Not true. Choice (A) gives units of m/s as well.

  blackhole_taw 20101109 13:12:15 
put and .
extreme value for is , which is eventually equal to g.
thus   Chespirito 20101003 01:19:18  Take a limit as d goes to 0: the surface is flat. The particle can then go at any horizontal speed and still touch the surface. The only answer which allows for the speed to be arbitrarily large is D.
flyboy621 20101105 22:00:01 
Nice solution.

  sullx 20091104 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:
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 . Solving for v gives as the upper limit.
Only answer D resembles this.
cilginfizikci 20100308 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

  Herminso 20090921 12:24:31  Physically the limit case is such that:
,
just when the particle will stay yet on the surface al all times. The particle go out of the path if  > , in that case the trayectory can raising off the surface. Now,
,
so , and then .
wavicle 20120804 17:46:58 
tl; dr  can skip...I'm studying for the quals and using the occasional "interesting" GRE problem as a refresher. Since this "solution" is listed at the top, I had to login to clarify some things.
You will not get such a simple equation when you do
Note
I'm not trying to get the signs right, just showing that this was done incorrectly. It's much easier to work with for reasons seen above.

  jw111 20081107 21:51:08  draw a free projection track on question first
the trajectory is
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)
 > 
>
  ayabepaula 20081023 17:24:21 
by definition
Since
Here
Amplitude=
then
but
then
chemicalsoul 20091025 18:49:16 
this is the best way !

Plantis 20100403 13:05:28 
It is the best solution. It's so easy and represend all physics of this situation.

mvgnzls 20110919 18:53:13 
BEST SOLUTION!!!

walczyk 20111111 10:29:40 
This is pretty great. This problem is an uncommon example where energy is NOT the right/easy way to go about this, but dealing with force is. A particle whipping around a curve will sense a centripetal force IF it stays on the track. In this case the force of gravity is this centripetal force, at the highest point of the curve of course. g=a_c and the rest just falls right out. Great solution, this should be at the top, maybe if someone rewrites it.

  dannyboytward 20081017 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(xvt)]
The acceleration of the surface is the second time derivative
d*(k*v)^2*cos[k(xvt)]
which has maximum value of d*k^2*v^2. If this value is less than g, we have our answer.   Rus Almighty 20080429 08:35:33  A condition for the body to stay on the track:
let's derive both sides by t and remember that on the top , Vx is V , and 'cuase there is only gravitational force on the top.
we get:
now let's calculate
And from now on it's only algebra:
QED
Rus Almighty 20080429 08:44:53 
fix,
instead of

  panos85 20071022 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 20080402 19:35:16 
Great way to see it. The site is wonderful! Thank you everyone.

  emailzac 20071008 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.   georgi 20070830 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 20071029 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). and we know 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 20080726 15:23:37 
The expression for h is the solution to the harmonic oscillator problem. That is why it is valid.

  skozmedia 20060327 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 20071110 11:33:16 
I just did something similar... Start by expanding about (the first hilltop): . Now consider the path of a particle starting at the top of the hill with tangential velocity ; its position is given by two equations: , and . For comparison with , combine the two previous equations to eliminate time: . Thus, or 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 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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