GR9677 #68



Alternate Solutions 
jmason86 20091004 15:04:26  MOE:
As Yosun said: PE = mgscos. Eliminate (B) and (C)
Also as Yosun said: Translational KE = 1/2m. Eliminate (D)
Only (A) and (E) remain.
Rotational KE = something... another term. (A) is only the Translational KE and the PE; it lacks a Rotational KE term. Eliminate (A).
(E) Remains.  

Comments 
ngendler 20151023 05:06:16  Does no one else see a DOT over the m in choice (E)???
camcam 20190912 16:14:14 
It looks like a printing error, there are similar dots all over the page if you look closely. That one\'s just in an unfortunate spot, I guess.

  nc 20141016 15:48:23  Without actually calculating the Lagrangian, we know that kinetic energy term will depend on omega, so that eliminates A and D. We also know that L = T  U and that U = mgscos(theta). The only choice that makes sense, then, is E.   Sagan_fan 20130523 15:08:36  The noeffort MOE: L = T  U
Here T has two components, rotational and translational, so we want an answer with 3 terms; Eliminate (A) and (D).
Since the potential should be subtracted, eliminate (C).
By inspection, U has an angle term, eliminating (B)   jmason86 20091004 15:04:26  MOE:
As Yosun said: PE = mgscos. Eliminate (B) and (C)
Also as Yosun said: Translational KE = 1/2m. Eliminate (D)
Only (A) and (E) remain.
Rotational KE = something... another term. (A) is only the Translational KE and the PE; it lacks a Rotational KE term. Eliminate (A).
(E) Remains.   f4hy 20090403 18:28:36  LIMITS!
I first eliminated C since the wrong sign and then B because no theta dependance.
From there I thought what would happen if . Well then it should be like a free falling particle so D is out. The spinning must give some kinetic energy so there goes A. E is the only one left.   r10101 20071030 19:13:43  After eliminating (B) and (C) by the potential term, use variable dependency: T = T for sure, so (A) and (D) are both out as well.
p3ace 20080411 10:25:59 
This is a great observation, but if there were other choices with dependence in the middle term, one would have to solve more exactly.
I would use the expression for in spherical coordinates which should be memorized, and in cylindrical coord.'s as well.

p3ace 20080411 10:44:23 
I hit post prematurely. Sorry.rnAs I was saying, the expressions for velocity in the other coordinates are very handy for problems like this one because one can simply dot r vector with itself to get v^2 for kinetic energy. rnIn this case, The spherical symmetry lends itself perfectly to the problem because s is just r, and is '. Also, is fixed, so ' is zero. Then dot the velocity with itself and multiply by (1/2)m and WahLah. You have it.rnI would type in the origianl expression for velocity in spherical coordinates but I'm not used to Latex and would rather be studying. I got it from Mechanics by Symon, and I've used it to solve numersous problems with cylindrical or spherical symmetry.

 

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