GR9277 #93



Alternate Solutions 
his dudeness 20100905 15:55:11  Yosun's method is clearly by far the best for this particular problem. However, if anyone is curious about how the relation in (E) is actually derived, here you go:
The acceleration can be split into radial and transverse components. Since gravity is the only force contributing to transverse acceleration, We have . To calculate , we note that . From conservation of energy, we have , so we get .
All together now: . A simple trig identity gets us the (E).   jburkart 20071102 00:23:15  A real way to do the problem is as follows. Kinetic energy is , where is the length of the string; potential is . Setting the Lagrangian to , Lagrange's equation is . In our case and we find , which then yields . But this is only the acceleration along the direction of motion; there is also a perpendicular centripetal acceleration given by . To find we can invoke conservation of energy: , so , so the centripetal acceleration is .
Taking the norm of the the two acceleration components, we get , answer (E).
A common mistake would be to omit the calculation of the centripetal acceleration, which would yield an incorrect answer of , which ETS kindly does not give as an option.  

Comments 
his dudeness 20100905 15:55:11  Yosun's method is clearly by far the best for this particular problem. However, if anyone is curious about how the relation in (E) is actually derived, here you go:
The acceleration can be split into radial and transverse components. Since gravity is the only force contributing to transverse acceleration, We have . To calculate , we note that . From conservation of energy, we have , so we get .
All together now: . A simple trig identity gets us the (E).
physicsres.com 20141115 12:55:37 
good answer just a typo alert.

unurbat 20160628 23:24:16 
Where did the \"r\" disappear when you try to find radial acceleration?

  elzoido238 20081106 11:02:47  Answer choice C should be 2g sin.   Poop Loops 20081025 20:22:52  Oh Jesus, people, get all this nonsense away. Calculations on the GRE? That's suicide!
Here's how you think of it:
When is , the mass wants to go straight down, so a = g.
That's it. Only E survives that.
But, if you had a different choice of answers, you can also think that acceleration is a maximum at , because it also has centripetal acceleration, which adds to the gravitational force at that point. This obviously corresponds to sin, so there you go.
ajkp2557 20091103 13:57:32 
"Calculations on the GRE? That's suicide!"
Best. Advice. Ever. The PGRE is entirely about test taking strategy. Yes, once you are in grad school it will be very important to know how to correctly work every problem, but spending four minutes on a problem on a test that only gives you 1.7 minutes per problem isn't a smart approach. You get the same credit for the answer, regardless of how you got it.
Work smarter, not harder. Best of luck!

  ayabepaula 20081022 00:36:22  time= zero
time
runcibleshaw 20150816 06:54:34 
I don\'t understand why the acceleration is 2g at the bottom. The total acceleration is the combination of g, which points down and the centripetal acceleration which points up, so it must be less than g total, right?

  wangjj0120 20081008 12:42:19  I am confused, why can't I use
force on string, so , and , where =length of string. From this, is correct, but is wrong. Obviously the concept is incorrect here, but why?
iplayterran 20100313 17:03:00 
The tension force on the string is the sum of and the centripetal force . If you use Newton's second law,
and decompose the forces into radial and angular components, you have
find velocity using conservation of energy, then find the magnitude of the above vector gives the correct answer E.

  jburkart 20071102 00:23:15  A real way to do the problem is as follows. Kinetic energy is , where is the length of the string; potential is . Setting the Lagrangian to , Lagrange's equation is . In our case and we find , which then yields . But this is only the acceleration along the direction of motion; there is also a perpendicular centripetal acceleration given by . To find we can invoke conservation of energy: , so , so the centripetal acceleration is .
Taking the norm of the the two acceleration components, we get , answer (E).
A common mistake would be to omit the calculation of the centripetal acceleration, which would yield an incorrect answer of , which ETS kindly does not give as an option.
kroner 20091006 22:29:16 
You don't need to do all that work to find the tangential acceleration. It's just the tangential component of gravity, which is .

  kb 20071028 22:49:32  is E is correct, can someone explain why
a(pi/2) = sqrt(3+1)g = 2g?
kb 20071029 16:52:27 
sorry, typo
"if E is correct, can someone explain why a(pi/2) = sqrt(3+1)g = 2g?"
and should go under "help!"

jburkart 20071102 00:25:23 
Because at that point the mass is experiencing a lot of g's! Try swinging in a swing sometimeyou feel a lot of acceleration at the bottom.

evanb 20080626 19:26:17 
At the top, the energy is , and at the bottom it is .
This leaves us with , and , but , so .

  rmyers 20061201 13:29:01  The first four words should be edited out of this solution.
lambda 20071021 22:13:29 
I agree. This test can't be finished in the time allowed if one tries to apply "rigorous physics"!

ndubuisi 20111105 01:52:04 
But one must answer a quetn well,rather than apply tricks n fail

  Andresito 20060330 12:05:12  Thank you for the advice. :)
ayabepaula 20081022 00:03:57 
1)http://dev.physicslab.org/img/db0bf4f12f9b428f81a6536588e8a7ce.gif
2)http://www.physicslab.net/applets/simplependulum
then
Gravitational acceleration:
********radial direction***************
then
(negative sign > and > opposite direction.
( rdirection)
Gravitational acceleration
********tangential direction********
(tdirection)
********total acceleration*********
***********************************************
using
********
therefore:
The magnitude of the total acceleration of the mass as a function of the angle is:
********
********
Paula Ayabe Pereira

 

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A real way to do the problem is as follows. Kinetic energy is , where is the length of the string; potential is . Setting the Lagrangian to , Lagrange's equation is . In our case and we find , which then yields . But this is only the acceleration along the direction of motion; there is also a perpendicular centripetal acceleration given by . To find we can invoke conservation of energy: , so , so the centripetal acceleration is .
Taking the norm of the the two acceleration components, we get , answer (E).
A common mistake would be to omit the calculation of the centripetal acceleration, which would yield an incorrect answer of , which ETS kindly does not give as an option.

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