PGRE 8677 #6

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 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 yaxis 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!
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 yaxis 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!

 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
Physics_auth

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 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

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 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 freebody 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.
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 methodhe did that successfullyhe's curious aside from any practical concerns if there is a quick, analytical solution involving derivatives...

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 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 methodhe did that successfullyhe'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 EulerLagrange equations but I was unsuccessful in arriving at the correct answer.
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.
As for your problem:
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...)
As for your problem:
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...)

 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
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

 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.

 Posts: 4
 Joined: Tue Dec 16, 2008 3:30 pm
Re: PGRE 8677 #6
Thanks this worked out good for me!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.
As for your problem:
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...)
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)