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Mechanics}Lagrangians


The potential energy of the mass is obviously U=mgs\cos\theta, and thus one eliminates choices (B) and (C). (To wit: L=T-U, (C) has the wrong sign).

The translational part of the kinetic energy is easily just 1/2 m \dot{s}^2. The rotational part requires the calculation of the moment of inertia for a point particle, which is just I=mr^2, where r=s\sin\theta, in this case. Thus, the rotational kinetic energy is 1/2 m (s\sin\theta)^2 \omega^2. The only choice that has the right rotational kinetic energy term is choice (E).

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
jmason86
2009-10-04 15:04:26
MOE:
As Yosun said: PE = mgscos\theta. Eliminate (B) and (C)
Also as Yosun said: Translational KE = 1/2m\dot{s}^2. 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.
Alternate Solution - Unverified
Comments
ngendler
2015-10-23 05:06:16
Does no one else see a DOT over the m in choice (E)???NEC
nc
2014-10-16 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. NEC
Sagan_fan
2013-05-23 15:08:36
The no-effort 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)
NEC
jmason86
2009-10-04 15:04:26
MOE:
As Yosun said: PE = mgscos\theta. Eliminate (B) and (C)
Also as Yosun said: Translational KE = 1/2m\dot{s}^2. 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.
Alternate Solution - Unverified
f4hy
2009-04-03 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 \theta =0. 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.
NEC
r10101
2007-10-30 19:13:43
After eliminating (B) and (C) by the potential term, use variable dependency: T = T(\omega) for sure, so (A) and (D) are both out as well.
p3ace
2008-04-11 10:25:59
This is a great observation, but if there were other choices with \omega dependence in the middle term, one would have to solve more exactly.
I would use the expression for \dot{r}} in spherical coordinates which should be memorized, and in cylindrical coord.'s as well.


p3ace
2008-04-11 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 \omega is \phi'. Also, \theta is fixed, so \theta' is zero. Then dot the velocity with itself and multiply by (1/2)m and Wah-Lah. 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.
NEC

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