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

Comments

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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