GR8677 #76



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mvgnzls 20110926 20:24:02  energy before = energy after
Mgh= M + I w=v/r
Mgh= M + (M)
Mgh= M + M
Mgh= M
v=
now, L= R x Mv = RMv = RM
  signminus 20100802 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 5050 random guess at the worst.
ali8 20110630 09:44:29 
And if you, in general, are unlucky, like me, then it's really useless.

mpdude8 20120419 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.

  ben 20060719 16:23:40  also, units rule out choices (C), (D), and (E). only (A) and (B) have units of angular momentum
jmason86 20090810 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 20060319 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 20060720 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 20110926 20:23:25 
energy before = energy after
Mgh= M + I w=v/r
Mgh= M + (M)
Mgh= M + M
Mgh= M
v=
now, L= R x Mv = RMv = RM

 

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