GREPhysics.NET: GR8677 #6

GR8677 #6

Problem

GREPhysics.NET Official Solution

Mechanics $\Rightarrow$ }Vectors

Since there is only one force acting, i.e., the gravitational force, one can find the tangential acceleration by projecting $\vec{g}$ in the tangential direction. Equivalently, one dots gravity with the tangential unit vector, $\vec{v_t}=\vec{g} \cdot \hat{t}$ .

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 $f(x) = y = x^2/4$ . Thus, $df/dx = dy/dx = x/2 = \tan\alpha$ , where in the last step, one notes that the ratio $dy/dx$ forms the tangent of the indicated angle.

One recalls the Pythagorean identity $\sin^2\alpha + \cos^2\alpha = 1$ , and the definition of $\tan\alpha$ in terms of $\sin\alpha$ and $\cos\alpha$ . Thus, one gets $x/2=\tan\alpha = \frac{\sin\alpha}{\cos\alpha}=\frac{\sin\alpha}{\sqrt{1-\sin^2\alpha}}$ . Square both sides to get $x^2/4 = \tan^2\alpha = \frac{\sin^2\alpha}{\cos^2\alpha}=\frac{\sin^2\alpha}{1-\sin^2\alpha}$ .

Solve $x^2/4 = \frac{\sin^2\alpha}{1-\sin^2\alpha}$ to get $\sin^2 \alpha = \frac{x^2/4}{1+x^2/4} = \frac{x^2}{4+x^2}$ .

The angle between the vectors $\vec{g}$ and $\hat{t}$ is $\pi/2-\alpha$ , and thus the tangential acceleration is $g \sin \alpha = \frac{g x}{\sqrt{x^2+4}}$ .

Beautiful problem.

Alternate Solutions

p3ace 2008-05-15 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. Reply to this comment
p3ace 2008-05-15 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. Reply to this comment
kevglynn 2006-10-31 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. Reply to this comment
clmw 2005-11-02 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 Reply to this comment
maryrose 2005-11-01 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. Reply to this comment
rreyes 2005-10-31 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. Reply to this comment

Comments

sirius
2008-11-05 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 one-to-one function.

jmason86
2009-08-10 19:41:52

As Yosun pointed out below, the problem states that x and y are unit-less 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.

Reply to this comment

p3ace
2008-05-15 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.

Reply to this comment

p3ace
2008-05-15 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
2008-10-05 11:42:48

hey p3ace, you didnt quite get the answer though. The choice D has $\frac{gx}{\sqrt{x^2 + 4}}$ not $\frac{gx}{\sqrt{x^2 + 16}}$ .

ajkp2557
2009-10-27 11:18:17

Good approach, but note that the normalized tangent vector is the time derivative of the position vector (r-dot) divided by the magnitude of r-dot.

Reply to this comment

StrangeQuark
2007-06-16 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
2007-11-11 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.

Reply to this comment

kevglynn
2006-10-31 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

Reply to this comment

kevglynn
2006-10-31 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
2007-10-02 17:34:09

Yeah, and also studying units and considering that, obviously, it is not a constant

Richard
2007-10-31 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
2008-11-02 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.

Reply to this comment

kevglynn
2006-10-31 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.

Reply to this comment

clmw
2005-11-02 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
2007-11-11 11:57:51

I think this solution is much faster than the official one, so I thought I'd write out the equations. Let $\hat{t}$ represent a unit vector in the tangential direction. We want to find the net tangential acceleration $\vec{g} \cdot \hat{t}$ , or

$g \cos(\theta)=\frac{g dy}{\sqrt{(dx)^{2}+(dy)^{2}}}=\frac{g y'}{\sqrt{1+(y')^{2}}}=\frac{\frac{1}{2}g x}{\sqrt{1+\frac{1}{4}x^{2}}}=\frac{g x}{\sqrt{x^{2}+4}}$ ,

where $\theta$ is the angle between $\vec{g}$ and $\hat{t}$ . In the end, I guess this is the same idea expressed in the official solution, but without undue trigonometric hardship.

ajkp2557
2009-10-27 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.)

Reply to this comment

clmw
2005-11-02 09:00:08

Reply to this comment

maryrose
2005-11-01 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
2005-11-01 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
2008-09-07 18:46:07

The problem specifically says y and x are unitless, so does that thinking still hold?

Reply to this comment

rreyes
2005-10-31 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.

Reply to this comment

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