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GR0177 #25
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Mechanics}Moment of Inertia

The moment of inertia of the center penny about the center is just

The moment of inertia of any one of the other pennis about the center is given by the parallel axis theorem, , where d is the distance from the new point from the center of mass. for each penny, and thus one has , since the distance from the center of each penny to the center of the configuration is 2r.

Since there are 6 pennies on the outside, one has the total inertia , as in choice (E).  Alternate Solutions
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gmlaghari85
2014-08-13 00:49:34
Make it simple. Take any diagonal..

I = Icm + md^2= 1/2 mr^2 + Md^2 where M=3m on diagonal and d = 3r
then I = 1/2 mr^2 + 72mr^2 = 55/2 mr^2
 gmlaghari852014-08-13 00:51:15 sorry its 27mr^2
 Ac314152017-08-31 16:08:17 I understand this method. But why does it work?? physik
2011-04-07 11:22:50
whats wrong with just doing this:rnrnIp:=I for the 6 penniesrnrnIp=6*(Mr^2)/2rnwe conclude that r=3R because thats how far each disk's boundery is from the center of the systemrn=> Ip= 6 *(3R)^2rn=> Ip= 6*(9) /2rnso Ip=(54/2)MR^2rnrnNow the center penny is just one disk Icp= (1/2)MR^2rnrnTotal I = Ip + Icp = (55/2)MR^2 Choice Ern
 physik2011-04-07 16:24:34 Im sorry about the look of this thing. What I wanted to write was moment of inertia for the 6 disks is Ip=6*(MR^2)/2 here R =3r => R^2 = 9r^2 Ip=((6*9)/2) Mr^2 and the moment of inertia of the central disk is Icp=MR^2/2 Total moment I = (MR^2)/2 + (54/2)(MR^2) anmuhich
2009-03-14 10:35:10
A faster way to do this is to treat the whole configuration as a disk and approximate the moment of intertia:

I=(1/2)(M)(R*R)

M= 7m and R= 3r

This gives I=(56/2)m *r*r which makes sense because this answer should be a little over the actual answer which is now obviously E
 f4hy2009-04-02 15:08:03 I wish I thought of that. Thanks.
 mr_eggs2009-08-16 21:23:26 but 9*7 is 63...
 bcomnes2011-09-22 16:11:53 It is an interesting way to get a ballpark number, but the number you actually get will not make sense if the answer choices include numbers between what this method gets and the actual correct answer.      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}\$