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GR8677 #76
Problem
 GREPhysics.NET Official Solution Alternate Solutions

Mechanics$\Rightarrow$}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,

$\begin{eqnarray} Mgh&=&\frac{1}{2}I\omega^2 + \frac{1}{2}Mv^2\\ &=&\frac{1}{2}MR^2\omega^2 + \frac{1}{2}Mv^2\\ Mgh&=&MR^2\omega^2\\ gh&=&R^2\omega^2\\ &\Rightarrow& \omega = \sqrt{\frac{gh}{R^2}} ,
\end{eqnarray}$

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).

Alternate Solutions
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mvgnzls
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$\sqrt{gh}$
signminus
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.
 ali82011-06-30 09:44:29 And if you, in general, are unlucky, like me, then it's really useless.
 mpdude82012-04-19 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.
 danielsw986672019-09-13 12:56:02 The answer is A. cash loans south africa no credit check
ben
2006-07-19 16:23:40
also, units rule out choices (C), (D), and (E). only (A) and (B) have units of angular momentum
 jmason862009-08-10 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 haha
Andresito
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
 ben2006-07-20 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.
 mvgnzls2011-09-26 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$\sqrt{gh}$
 danielsw986672019-10-21 05:55:38 Because the answer is A. semenax review

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

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