GR8677 #6


Problem



Mechanics}Vectors
Since there is only one force acting, i.e., the gravitational force, one can find the tangential acceleration by projecting in the tangential direction. Equivalently, one dots gravity with the tangential unit vector, .
There's a long way to do this, wherein one writes out the full Gibbsean vector formalism, and then there's a short and elegant way. (The elegant solution is due to Teodora Popa.)
The problem gives . Thus, , where in the last step, one notes that the ratio forms the tangent of the indicated angle.
One recalls the Pythagorean identity , and the definition of in terms of and . Thus, one gets . Square both sides to get .
Solve to get .
The angle between the vectors and is , and thus the tangential acceleration is .
Beautiful problem.


Alternate Solutions 
pranav 20121102 22:23:53  (same concept as one given...but simpler math)
Differentiating y w.r.t x gives:
Draw a right triangle, with one angle .
Now label the side adjacent as "2", opposite as "x".
The hypotenuse becomes .
Now we want the tangential acceleration, which would be the "g" multiplied by the cosine of the angle opposite to i.e. , since is downward parallel to the side we labeled "x" (i.e. the dot product basically)
Therefore, answer is:
(D)   malianil 20111105 05:52:47  For those who are not inclined to solve it with trigonometric functions can take a snapshot drawing of the force diagram at arbitrary location (y,x) where rn and then use euclidean geometry to find our that the Force component of the tangent is a function of x and is rnrnIt is simpler this way for me.   nakib 20100402 11:51:51  (A) Lol!
(B) Of course not! The tangent is not pointing downwards.
(C) Wrong units.
(D) Right units, maybe correct. Also, goes to as goes to .
(E) Wrong units.
(D) is the answer.
The rigorous solutions are beautiful, but are not feasible under GRE exam conditions...
Shahbaz Ahmed Chughtai 20120609 22:51:44 
Very Nice. GRE is designed to use common sense !!!
But I salute the website owners for such a hard work !

  p3ace 20080515 06:11:52  I apologize for what I just said. It came out wrong and I feel terrible after having reread it. What I meant to say was, if you want to know how to just crank out the answer, this is how I would do it.   p3ace 20080515 06:08:15  The process of elimination is great in a pinch on a test but it doesn't demonstrate any physics, or in this case math.
To me the most straight forward way to do this is:
The tangential direction is just the direction of r vector,
r=ix+jy=ix+j(x^2)/4.
r hat or the unit vector in the tangential direction is just r/magnitude, i.e.
r hat = [ix+j(x^2)/4]/{x^2+[(x^2)/4)]^2}^(1/2)
You can factor an x out of the denominator and cancel it with the x in the numerator, leaving
r hat = (i+jx/4)/[1+(x^2)/16]^(1/2)
Now dot the acceleration vector, a=jg with the unit vector in the tangential direction to get the tangential acceleration,
a tangential = g[(x/4)/(1+x^2/16)].
Now, to get the form in the answer, multiply through in the denominator by the 4. Inside the radical, it becomes a 16 so that you have,
a tangential = gx/[16+x^2)^(1/2), Wah, LA, choice D.
Sorry I'm not a latex jockey. To me this is radically simple, more so than the other solutions, just because this is what makes sense to me. I know that others see it differently and whatever works for you is a okay, so this is offered up to those of use who think in these terms.
  kevglynn 20061031 09:10:31  I'm surprised no one noticed this one... As x > infinite, acceration must approach g (a > g), so choice (D) is the only possibility.   clmw 20051102 09:08:00  One can remove some of the trig nastiness in the above solution by just noting that the tangetial acceleration equals the normalized tangent vector times g. Using x as our simple parameter, the y component of the tangent vector equals dy/dx=x/2 and the x component equals dx/dx=1. If we normalize this vector (1,x/2) and then multiply g by the normalized y component we get the answer pretty straightforwardly (without using sin/cos identities)
Chris   maryrose 20051101 13:08:02  It can also be solved by noting the units and realizing that it cannot be zero or g. That leaves only D.   rreyes 20051031 09:47:42  nice solution! :)
let me just note that we can also answer this problem by method of elimination by observing that as x>\infty, a must > g. this eliminates all answers except D.  

Comments 
calvin_physics 20140327 15:04:37  Just ignore the math. Take limits.
x = 0, acceleration must be zero.
x = big, acceleration must be g.
We only have two choices left. (D) and (E)
E gives wrong unit, it gives you an extra x.
D has the right unit. Bingo.   pranav 20121102 22:23:53  (same concept as one given...but simpler math)
Differentiating y w.r.t x gives:
Draw a right triangle, with one angle .
Now label the side adjacent as "2", opposite as "x".
The hypotenuse becomes .
Now we want the tangential acceleration, which would be the "g" multiplied by the cosine of the angle opposite to i.e. , since is downward parallel to the side we labeled "x" (i.e. the dot product basically)
Therefore, answer is:
(D)   fkvkfdlek 20120804 21:07:40  can this be solved using lagrange multiplier?
I set up L' = L  and then was trying to rewrite tangential acceleration as
i cant quite finish the substitution to get the inside of the square root purely in terms of x   whatever 20111111 15:28:18  parametrization of the line is = (x,), then taking its derivative with respect to x and dividing by it's length to get the tangential unit vector we get
:=(1,) /
our vector for acceleration is
= (0,g)
the tangential component of acceleration is then their inner product
= * = gx/ , D
  malianil 20111105 05:52:47  For those who are not inclined to solve it with trigonometric functions can take a snapshot drawing of the force diagram at arbitrary location (y,x) where rn and then use euclidean geometry to find our that the Force component of the tangent is a function of x and is rnrnIt is simpler this way for me.   nakib 20100402 11:51:51  (A) Lol!
(B) Of course not! The tangent is not pointing downwards.
(C) Wrong units.
(D) Right units, maybe correct. Also, goes to as goes to .
(E) Wrong units.
(D) is the answer.
The rigorous solutions are beautiful, but are not feasible under GRE exam conditions...
archard 20100605 18:04:39 
The problem specifies the the coordinates are dimensionless units, so you can't eliminate C and E.

physicsworks 20100709 09:40:46 
nakib is right
when the acceleration must be . There is only one appropriate choice for this, no matter what dimensions are given.

Shahbaz Ahmed Chughtai 20120609 22:51:44 
Very Nice. GRE is designed to use common sense !!!
But I salute the website owners for such a hard work !

eighthlock 20130811 13:07:06 
Notice that the prompt states x and y are dimensionless units. Therefore you can't use dimensional analysis, because (D) and (E) have the same dimensions.

mike1999 20140716 14:15:40 
Actually, the problem specifies that x and y are dimensionless, so you can't 100% narrow it down like that (actually, perhaps if you substituted the dimensional quantity...). Anyway, you can also get it from limiting cases. As x>0, the answer should > 0. As x > inf, the answer should > g (because the tangent gets infinitely steep). Thus D is correct.

  sirius 20081105 19:37:19  Here's an easier way:
Only (A),(B),and (D) have the correct units, first of all. You know that the particle is accelerating, and that its acceleration can't be greater or even equal to g, no matter what x is. Only (D) satisfies these conditions.
Why can't the acceleration be g? The track can never be vertical since y is constrained to a onetoone function.
jmason86 20090810 19:41:52 
As Yosun pointed out below, the problem states that x and y are unitless so you can't (technically) use dimensional analysis.
Limits (x>0 and x> infinity) solves this whole problem without the need for units and it is still quick.

  p3ace 20080515 06:11:52  I apologize for what I just said. It came out wrong and I feel terrible after having reread it. What I meant to say was, if you want to know how to just crank out the answer, this is how I would do it.   p3ace 20080515 06:08:15  The process of elimination is great in a pinch on a test but it doesn't demonstrate any physics, or in this case math.
To me the most straight forward way to do this is:
The tangential direction is just the direction of r vector,
r=ix+jy=ix+j(x^2)/4.
r hat or the unit vector in the tangential direction is just r/magnitude, i.e.
r hat = [ix+j(x^2)/4]/{x^2+[(x^2)/4)]^2}^(1/2)
You can factor an x out of the denominator and cancel it with the x in the numerator, leaving
r hat = (i+jx/4)/[1+(x^2)/16]^(1/2)
Now dot the acceleration vector, a=jg with the unit vector in the tangential direction to get the tangential acceleration,
a tangential = g[(x/4)/(1+x^2/16)].
Now, to get the form in the answer, multiply through in the denominator by the 4. Inside the radical, it becomes a 16 so that you have,
a tangential = gx/[16+x^2)^(1/2), Wah, LA, choice D.
Sorry I'm not a latex jockey. To me this is radically simple, more so than the other solutions, just because this is what makes sense to me. I know that others see it differently and whatever works for you is a okay, so this is offered up to those of use who think in these terms.
neon37 20081005 11:42:48 
hey p3ace, you didnt quite get the answer though. The choice D has not .

ajkp2557 20091027 11:18:17 
Good approach, but note that the normalized tangent vector is the time derivative of the position vector (rdot) divided by the magnitude of rdot.

shen 20100811 08:25:04 
Actually u got the wrong tangential vector. It should be found using the gradient dy/dx.
r = [1, x/2]/(1+x^2/4)^1/2
You got the correct answer through a careless mistake that correctly make up for your wrong formulation. Cheers. You can check your answers.

  StrangeQuark 20070616 09:25:22  I did this problem by noting infinity conditions which in the test I am more then happy to do however in study I would like a more concrete answer, which I found on your site (thank you). However I have a question, you say that gravity is the only force acting, isn't there a constraining force from the track, i.e. a normal force that points perpendicular to the track that needs to be accounted for?
Jeremy 20071111 11:24:55 
I wondered about the official answer's omission of the normal force as well, but now I understand why it's not necessary. We only care about forces that have tangential components, and thus contribute to the tangential acceleration.

  kevglynn 20061031 09:17:19  Sorry, I'm an idiot and decided not to read before I wrote that :)
Tried get rid of it, too, but it seems that you can't edit even your own posts. oh well   kevglynn 20061031 09:10:31  I'm surprised no one noticed this one... As x > infinite, acceration must approach g (a > g), so choice (D) is the only possibility.
carlospardo 20071002 17:34:09 
Yeah, and also studying units and considering that, obviously, it is not a constant

Richard 20071031 12:28:05 
That's how I did it.
There is a similar problem on another GRE exam...
the limiting technique works there as well.

Poop Loops 20081102 16:23:51 
Limits and boundary conditions are your friends!
A math teacher of mine used to tell me: The more math you do, the more room for error there is.

testtest 20101111 18:45:16 
That does not eliminate (E)

testtest 20101111 18:46:34 
Tzzzzz what am I saying... Time to go to bed! (yes it does eliminate E)

  kevglynn 20061031 09:10:13  I'm surprised no one noticed this one... As x > infinite, acceration must approach g (a > g), so choice (D) is the only possibility.   clmw 20051102 09:08:00  One can remove some of the trig nastiness in the above solution by just noting that the tangetial acceleration equals the normalized tangent vector times g. Using x as our simple parameter, the y component of the tangent vector equals dy/dx=x/2 and the x component equals dx/dx=1. If we normalize this vector (1,x/2) and then multiply g by the normalized y component we get the answer pretty straightforwardly (without using sin/cos identities)
Chris
Jeremy 20071111 11:57:51 
I think this solution is much faster than the official one, so I thought I'd write out the equations. Let represent a unit vector in the tangential direction. We want to find the net tangential acceleration , or
,
where is the angle between and . In the end, I guess this is the same idea expressed in the official solution, but without undue trigonometric hardship.

ajkp2557 20091027 11:15:13 
Great solution!
Side note for those that (like me) have forgotten: the tangent vector is the time derivative of the position vector. (Which makes sense physically, if you think about what the velocity vector is telling us.)

  clmw 20051102 09:00:08    maryrose 20051101 13:08:02  It can also be solved by noting the units and realizing that it cannot be zero or g. That leaves only D.
yosun 20051101 16:04:57 
actually, maryrose, the problem gives "dimensionless units". thus one can't eliminate the other choices (other than 0 and g) that easily...

mrmeep 20080907 18:46:07 
The problem specifically says y and x are unitless, so does that thinking still hold?

  rreyes 20051031 09:47:42  nice solution! :)
let me just note that we can also answer this problem by method of elimination by observing that as x>\infty, a must > g. this eliminates all answers except D.  

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