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

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

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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.

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