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![[Post New]](/templates/default/images/icon_minipost_new.gif) 12 Feb 2008 20:53:07 IST
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okay, marks edited
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Will nip in at times to solve problems :)
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12 Feb 2008 20:56:09 IST
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Methink the qn and ans given to problem ( 6 ) are also self- cotradictory !! It is said that the mass of the wheels are negligble . Then how in the total energy expression of the system can a rotational energy term possibly occur ? On the other hand consider a wheel of mass M be attached to the end of the spring . If u carry out the analysis u will find that  is indeed greater than sqrt( k/m ) . Plz explain it also .
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12 Feb 2008 21:05:09 IST
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well feynmann, you are a genius but the last statement you wrote is not in accordance with what i got . it was greater than rootk/m. i used a shortcut but you can conserve energy, take it as constant,diff it,you must be knowing all this and then find out
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Impossible To be Impossible is Impossible |
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12 Feb 2008 21:07:55 IST
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feynmann... time period is asked here right? So if w is greater, T will be lesser.
Just for your information - he has considered the mass of the wheels as well, though your point is absolutely correct.
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12 Feb 2008 21:08:55 IST
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edited - lol, thanks for correcting me anchit. I am forgetting my own questions.
@feynmann - w comes out to be sqrt(k/(m+m'))
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12 Feb 2008 21:09:13 IST
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karthik don't now create a confusion here. frequency is asked and omega comes out to be less
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12 Feb 2008 21:12:14 IST
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1/2mvsquare+1/2Iwsquare+1/2kxsquare=const
this would give effective m greater and frequency is rootk/m which would be lesser
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Impossible To be Impossible is Impossible |
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12 Feb 2008 21:13:08 IST
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posted the ans in my previous post
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12 Feb 2008 21:21:42 IST
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What is this happening ? U first told that total m is same for both cases , now u are including mass of the wheels . And deriving answers on this basis !! It would be fair to stick to ur qn . Again anchitsaini drop the 1/2 I w^2 term in the energy expression . According to the original qn it is not allowed . Instead apply no slipping condition and see the result .
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12 Feb 2008 21:42:58 IST
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Agree with Feynmann.
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http://14-69-8.blogspot.com
My blog |
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12 Feb 2008 22:20:48 IST
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I am very confused here.
1st of all..ques 5 seems to have been cancelled because of the typo. And tarinbansal tried ques 1,2,3,4,6. each of which are valid. So, he cant get 40. Either he gets 50 or less than 40.
2nd ly....answer given to ques 6 is not agreed upon.
It is clearly given that mass is negligible and therefore, rotational energy will not be considered. So,
1/2 mv2 + 1/2 kx2= c
v = dx/dt
Differentiating 1st eq with respect to x,
=> 1/2m x 2v dv/dx + kx=0
=> mvdv/dx = -kx
=> mdv/dt= -kx
=> md2x/dt2= -kx
=> d2x/dt2= -(k/m)x
Comparing -(k/m)x = -  2x,
= root(k/m). Hence, (a)
Karthik...jaldi clarify kar yaar. Please. Mera dimag kharab ho raha hai.
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13 Feb 2008 00:05:27 IST
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Hmm, that qn 6 sure has caused lots of comedy and confusion. Since its stated that the mass is negligible, yeah, then option a) is correct. I blindly typed the qns, and then assumed that the solutions are correct, I should have cross checked properly. I agree that its option a) according to the wordings of the question. However, feynmann seems to have a different opinion.
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13 Feb 2008 00:07:44 IST
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anyway I'll confirm with ma juniors who go to this class about this qn.
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13 Feb 2008 00:17:34 IST
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btw, feynmann, can you provide the complete working of what you have done for qn 6, so that we all can come to a conclusion.
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13 Feb 2008 01:28:39 IST
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