GREPhysics.NET
GR | # Login | Register
   
  GR8677 #76
Problem
GREPhysics.NET Official Solution    Alternate Solutions
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).

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
There are no Alternate Solutions for this problem. Be the first to post one!
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}
NEC
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
NEC
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
NEC
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
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}


danielsw98667
2019-10-21 05:55:38
Because the answer is A. semenax review
NEC

Post A Comment!
You are replying to:
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.

Username:
Password:
Click here to register.
This comment is best classified as a: (mouseover)
 
Mouseover the respective type above for an explanation of each type.

Bare Basic LaTeX Rosetta Stone

LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces .
type this... to get...
$\int_0^\infty$
$\partial$
$\Rightarrow$
$\ddot{x},\dot{x}$
$\sqrt{z}$
$\langle my \rangle$
$\left( abacadabra \right)_{me}$
$\vec{E}$
$\frac{a}{b}$
 
The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it...