## PGRE 8677 #6

newguy1234
Posts: 4
Joined: Tue Dec 16, 2008 3:30 pm

### PGRE 8677 #6

Hello I am going over the old ETS exams and I am having trouble with #6 of the 8677 exam. The question is:

A particle is initially at rest at the top of a curved frictionless track. The x&y coordinates of the track are related in dimensionless units by y=x^2/4, where the positive y-axis is in the vertical downward direction. As the particle slides down the track, what is its tangential acceleration.

Now I viewed the solutions on the site grephysics.net but I feel there is a better way to do the problem like just taking some derivatives or something but I cant figure it out. So if you solved it a different way from the people on that site when you were studying please post here and let me know. Thanks for your time!

physics_auth
Posts: 163
Joined: Sat Jul 18, 2009 7:24 pm

### Re: PGRE 8677 #6

When I solved this exercise I didn't attempt an analytic solution. In fact, I thought like that: when y -> + infinity (in other words when x -> + infinity since x > 0) then the slope of the parabola tends to become perpendicular. On the particle are exerted its weight and the normal reaction (no friction). When the slope becomes perpendicular, the tangential acceleration (which is along the tangential direction) in this limit becomes perpendicular, and the only perpendicular force is mg, thus the tangential acceleration a_t -> g, if x -> + infinity. If you look all answers only (D) applies to this. All other answers give infinite or zero acceleration. Try this.
Physics_auth

newguy1234
Posts: 4
Joined: Tue Dec 16, 2008 3:30 pm

### Re: PGRE 8677 #6

yes I too came across the limiting case solution however I was am seeking an analytic solution to the problem.. Thanks for your help though

nathan12343
Posts: 249
Joined: Sat Dec 20, 2008 9:34 am

### Re: PGRE 8677 #6

Don't worry about the analytic solution. It would take me (don't know about you) at least 10 minutes to draw the free-body diagram and work through the equations to arrive at an expression for the tangential acceleration. Limiting cases are an excellent way to pare down the number of possible answers and in some cases (such as this question) get the right answer.

grae313
Posts: 2297
Joined: Tue May 29, 2007 8:46 pm

### Re: PGRE 8677 #6

It seems to me like the OP is not so interested in his ability to get the question right using the quickest method--he did that successfully--he's curious aside from any practical concerns if there is a quick, analytical solution involving derivatives...

newguy1234
Posts: 4
Joined: Tue Dec 16, 2008 3:30 pm

### Re: PGRE 8677 #6

grae313 wrote:It seems to me like the OP is not so interested in his ability to get the question right using the quickest method--he did that successfully--he's curious aside from any practical concerns if there is a quick, analytical solution involving derivatives...

Yes this is exactly what I am looking for. I tried setting up the Lagrangian and applied the Euler-Lagrange equations but I was unsuccessful in arriving at the correct answer.

quizivex
Posts: 1035
Joined: Tue Jan 09, 2007 6:13 am

### Re: PGRE 8677 #6

Yea, it's always more instructive to know the "solution", not just that the answer is correct. But beware that some of the problems on the GRE would be suicidal to attempt analytically. I recall one practice test problem asking which formula represents the modes of vibration of a string fixed at one end with a mass M at the other... Even if you knew how to derive it from DEQ's, testing M->0 and M->infinity was the only way to get the answer in a reasonable amount of time.

You know the tangential acceleration (TA) is the tangential component of m*g. If theta is the angle the curve makes with the vertical, you get:
cos(theta) = TA / mg

You also know you can relate the tangent of the angle with the slope of the track. In this case, tan(theta) = dx/dy = 1/(dy/dx)

Now relate tan() and cos() using trig, do some substitution and get the answer. (I think...)

sravanskarri
Posts: 58
Joined: Sat Jun 14, 2008 10:19 pm

### Re: PGRE 8677 #6

I think we can try this:
y= x^2/4;
dy/dx = x/2;
Considering a tangent to the curve passing through the origin(0,0) and putting any point on the tangent (1,x/2) into vector form
t bar = 1 i+ (x/2)j
ay = gj
ay(dot)[ tbar (unit)]
projecting g along the direction of the tangent = (g j) (dot) (1 i + (x/2)j)/sqrt(1+ x^2/4)
= gx/sqrt(1+x^2/4)

PS: The solution makes few assumptions and hence not a general one

sravanskarri
Posts: 58
Joined: Sat Jun 14, 2008 10:19 pm

### Re: PGRE 8677 #6

i just checked the grephysics.net..as always the case some one already had a soln of this kind.   newguy1234
Posts: 4
Joined: Tue Dec 16, 2008 3:30 pm

### Re: PGRE 8677 #6

quizivex wrote:Yea, it's always more instructive to know the "solution", not just that the answer is correct. But beware that some of the problems on the GRE would be suicidal to attempt analytically. I recall one practice test problem asking which formula represents the modes of vibration of a string fixed at one end with a mass M at the other... Even if you knew how to derive it from DEQ's, testing M->0 and M->infinity was the only way to get the answer in a reasonable amount of time.

You know the tangential acceleration (TA) is the tangential component of m*g. If theta is the angle the curve makes with the vertical, you get:
cos(theta) = TA / mg

You also know you can relate the tangent of the angle with the slope of the track. In this case, tan(theta) = dx/dy = 1/(dy/dx)

Now relate tan() and cos() using trig, do some substitution and get the answer. (I think...)
Thanks this worked out good for me!

cos(B)= TA/g

tan(B)=dx/dy= 1/(dy/dx)= 2/x

cos(B)= 1/Sqrt(1+tan(B)^2)

Now sub in

TA/g = 1/Sqrt(1+4/x^2) -------------------------------->Multiply by x/x

TA/g = x/Sqrt(x^2+4)

TA= gx/Sqrt(x^2+4)