GREPhysics.NET
GR | #
<< Previous | Next >>
Login | Register
   
  GR9277 #93
Problem
GREPhysics.NET Official Solution    Alternate Solutions
\prob{93}
9277_93

THe figure above shows a small mass connected to a string, which is attached to a vertical post. If the mass is released when the string is horizontal as shown, the magnitude of the total acceleration of the mass as a function of the angle $\theta$ is


  1. $g\sin\theta$
  2. $2g\cos\theta$
  3. $2g\cos\theta$
  4. $g\sqrt{3\cos^2\theta+1}$
  5. $g\sqrt{3\sin^2\theta+1}$
Mechanics}Boundary Condition

Getting low on time, one should begin scoring points based more of testing strategy than sound rigorous physics. At the initial release point, the acceleration is due to gravity and the tension is 0 (no centripetal acceleration). The only choice that gives a(\theta=0) = g is choice (E).
See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
jburkart
2007-11-02 00:23:15		A real way to do the problem is as follows. Kinetic energy is T=mR^2\dot{\theta}^2/2, where R is the length of the string; potential is U=-mgR\sin\theta. Setting the Lagrangian to L=T-U, Lagrange's equation is \frac{\partial L}{\partial q}=\frac{\partial}{\partial t}\frac{\partial L}{\partial\dot{q}}. In our case q=\theta and we find \ddot{\theta}=g\cos\theta/R, which then yields a_\parallel=\ddot{\theta}/R=g\cos\theta. But this is only the acceleration along the direction of motion; there is also a perpendicular centripetal acceleration given by v^2/R. To find v we can invoke conservation of energy: mv^2/2=mgR\sin\theta, so v^2=2gR\sin\theta, so the centripetal acceleration is a_{\perp}=v^2/R=2g\sin\theta.

Taking the norm of the the two acceleration components, we get a=g\sqrt{\cos^2\theta+4\sin^2\theta}=g\sqrt{1+3\sin^2\theta}, answer (E).

A common mistake would be to omit the calculation of the centripetal acceleration, which would yield an incorrect answer of g\cos\theta, which ETS kindly does not give as an option.		Alternate Solution - Unverified
Comments
jburkart
2007-11-02 00:23:15		A real way to do the problem is as follows. Kinetic energy is T=mR^2\dot{\theta}^2/2, where R is the length of the string; potential is U=-mgR\sin\theta. Setting the Lagrangian to L=T-U, Lagrange's equation is \frac{\partial L}{\partial q}=\frac{\partial}{\partial t}\frac{\partial L}{\partial\dot{q}}. In our case q=\theta and we find \ddot{\theta}=g\cos\theta/R, which then yields a_\parallel=\ddot{\theta}/R=g\cos\theta. But this is only the acceleration along the direction of motion; there is also a perpendicular centripetal acceleration given by v^2/R. To find v we can invoke conservation of energy: mv^2/2=mgR\sin\theta, so v^2=2gR\sin\theta, so the centripetal acceleration is a_{\perp}=v^2/R=2g\sin\theta.

Taking the norm of the the two acceleration components, we get a=g\sqrt{\cos^2\theta+4\sin^2\theta}=g\sqrt{1+3\sin^2\theta}, answer (E).

A common mistake would be to omit the calculation of the centripetal acceleration, which would yield an incorrect answer of g\cos\theta, which ETS kindly does not give as an option.		Alternate Solution - Unverified
kb
2007-10-28 22:49:32		is E is correct, can someone explain why
a(pi/2) = sqrt(3+1)g = 2g?
kb
2007-10-29 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
2007-11-02 00:25:23		 Because at that point the mass is experiencing a lot of g's! Try swinging in a swing sometime--you feel a lot of acceleration at the bottom.
evanb
2008-06-26 19:26:17		 At the top, the energy is mgh, and at the bottom it is .5mv^2.

This leaves us with v^2 = 2 g h, and a = v^2 / r, but r = h, so a = 2g.
NEC
rmyers
2006-12-01 13:29:01		The first four words should be edited out of this solution.
lambda
2007-10-21 22:13:29		 I agree. This test can't be finished in the time allowed if one tries to apply "rigorous physics"!
NEC
Andresito
2006-03-30 12:05:12		Thank you for the advice. :)NEC

Post A Comment!
Username:
Password:
Click here to register.
This comment is best classified as a: (mouseover)
 
Mouseover the respective type above for an explanation of each type.

Bare Basic LaTeX Rosetta Stone
LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces .
type this... to get...
$\int_0^\infty$
$\partial$
$\Rightarrow$
$\ddot{x},\dot{x}$
$\sqrt{z}$
$\langle my \rangle$
$\left( abacadabra \right)_{me}$
$\vec{E}$
$\frac{a}{b}$
 
The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it...
Chat Archives | Delete left banner ad | Donate
(Click to view chat.)

Anonymous:

Hate being Anonymous? Login or Register

EssayEdge.com Admissions Essay Help Poser 7 - Complete 3D Figure Design

Upgrade to Poser 7 Now

Huge Textbook Selection, Low Prices – Phat Campus.

Google
 

All-Solutions List | Latest Comments | Unanswered Questions (Cries for Help!) |::| About

© 2005-2007 by Yosun Chang, All rights reserved.

This site is not affiliated with ETS, nor is it officially endorsed (yet) by ETS. This site is the work of an individual named Yosun Chang. The solutions, sans user comments and ETS questions, are © 2005 by Yosun Chang except where noted. All rights reserved. / The comments appending particular solutions are due to the respective users. / Educational Testing Service, ETS, the ETS logo, Graduate Record Examinations, and GRE are registered trademarks of Educational Testing Service. The examination questions are © 1997 and © 2001 by ETS.