GREPhysics.NET: GR9677 #68

GR9677 #68

Problem

GREPhysics.NET Official Solution

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Mechanics $\Rightarrow$ }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).

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.

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Comments

ngendler
2015-10-23 05:06:16

Does no one else see a DOT over the m in choice (E)???

camcam
2019-09-12 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.

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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.

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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)

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jmason86
2009-10-04 15:04:26

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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.

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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.

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