GREPhysics.NET: GR8677 #76

GR8677 #76

Problem

GREPhysics.NET Official Solution

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,

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

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).

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Comments

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 $\sqrt{gh}$

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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.

ali8
2011-06-30 09:44:29

And if you, in general, are unlucky, like me, then it's really useless.

mpdude8
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.

danielsw98667
2019-09-13 12:56:02

The answer is A. cash loans south africa no credit check

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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

jmason86
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
haha

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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

ben
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.

mvgnzls
2011-09-26 20:23:25

danielsw98667
2019-10-21 05:55:38

Because the answer is A. semenax review

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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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