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Mechanics}Moment of Inertia

The moment of inertia of the center penny about the center is just 1/2 m r^2

The moment of inertia of any one of the other pennis about the center is given by the parallel axis theorem, I=I_{CM} + md^2, where d is the distance from the new point from the center of mass. I_{CM}=mr^2/2 for each penny, and thus one has I=mr^2/2+md^2=mr^2/2+m(2r)^2=9/2mr^2, 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 I=1/2mr^2+54/2mr^2=55/2mr^2, as in choice (E).

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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
2014-08-13 00:51:15
sorry its 27mr^2
2017-08-31 16:08:17
I understand this method. But why does it work??
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 \frac{M}{2}*(3R)^2rn=> Ip= 6*(9)\frac{M}{R^2} /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
2011-04-07 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

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)
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:


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
2009-04-02 15:08:03
I wish I thought of that. Thanks.
2009-08-16 21:23:26
but 9*7 is 63...
2011-09-22 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.

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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

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