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  GR8677 #76
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Verbatim question for GR8677 #76
Mechanics}Conservation of Energy

Recall the formula for inertia I=\int dm r^2. A hoop has constant r=R, thus the integral is trivial and I_{hoop}=MR^2. One can break up the kinetic energy into the pure rolling (about center of mass) and pure translation bit,

where one recalls that v=R\omega.

Plug the value of angular velocity \omega into the equation for angular momentum L=I\omega, to get MR^2 \sqrt{\frac{gh}{R^2}}=MR\sqrt{gh}, as in choice (A).

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2011-09-26 20:24:02
energy before = energy after
Mgh= \frac{1}{2}M\{v^2} + \frac{1}{2}I\{w^2} w=v/r
Mgh= \frac{1}{2}M\{v^2} + \frac{1}{2}(M\{R^2})\frac{v^2}{R^2}
Mgh= \frac{1}{2}M\{v^2} + \frac{1}{2}M\{v^2}
Mgh= M\{v^2}
v= \sqrt{gh}
now, L= R x Mv = RMv = RM\sqrt{gh}
2010-08-02 14:25:18
Another POE technique here is to realize that the hoop's angular momentum must depend on its radius (think about a hoop with really huge radius versus one with a tiny radius), so only answers (A) and (B) remain. Of course, something more is needed to get the exact answer, but in a pinch you'll be down to a 50-50 random guess at the worst.
2011-06-30 09:44:29
And if you, in general, are unlucky, like me, then it's really useless.
2012-04-19 16:05:57
POE is never useless, and if you can narrow it down to a 1 in 2 guess, guessing is a better option than skipping.
2006-07-19 16:23:40
also, units rule out choices (C), (D), and (E). only (A) and (B) have units of angular momentum
2009-08-10 20:34:14
Totally agreed. I did this.. but they were really mean in giving you a factor of 2 to distinguish the last choice. Damn ETS. Isn't that the first rule of real physics: Factors of 2 don't matter
2006-03-19 10:39:52
This solution is confusing. Please specify why the factor of (1/2) disappears when you equate the translational energy with the potential energy.

Thank you
2006-07-20 17:16:01
v=R*omega so the translational kinetic energy is equal to the rotational kinetic energy, and that value is (1/2)*M*R^2*omega^2. so when you add 1/2 plus 1/2 you get 1 making the total kinetic energy M*R^2*omega^2. nothing "disappears" here.
2011-09-26 20:23:25
energy before = energy after
Mgh= \frac{1}{2}M\{v^2} + \frac{1}{2}I\{w^2} w=v/r
Mgh= \frac{1}{2}M\{v^2} + \frac{1}{2}(M\{R^2})\frac{v^2}{R^2}
Mgh= \frac{1}{2}M\{v^2} + \frac{1}{2}M\{v^2}
Mgh= M\{v^2}
v= \sqrt{gh}
now, L= R x Mv = RMv = RM\sqrt{gh}


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