DOUBT IN ROTATION.

JEE Main 2015 Tutorials › Mechanics questions › DOUBT IN ROTATION.

#Physics #Mechanics

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5 Answers

1

Anirudh Kumar ·2010-02-24 03:32:27

is the asnwer

= Ï‰²2Î±*2Ï€ where Î±= 5gk(k+1)2R(k²+1)

please verify

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1

Manmay kumar Mohanty ·2010-02-24 04:56:26

ANS. IS (1+K²)w_o²R8pik(k+1)g

w is actually omega..

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29

govind ·2010-02-24 09:09:13

So by balancing the forces we get
N₁ + kN₂ = Mg
N₂ = kN₁

N₁ + k²N₁ = Mg
(k² + 1) N₁ = Mg ..........................1

by balancing the torque abt the centre of cylinder we get..
(kN₁ + kN₂ ) = MR² Î± / 2
k(N₁ + kN₁) = MRÎ±/2
k(k+1)Mg / (k² + 1) = MRÎ±/2 from 1

So Î± = 2k(k+1)g / (k² + 1)R
Now 0 = Ï‰ + Î±t
so t = -Ï‰/Î±

Now Î¸ = 1/2Î±t²
= 1/2 Î± * Ï‰² / Î±²
= 1/2 Ï‰² / Î±

= Ï‰²(k² +1)R / 4k(k+1)g..
Now number of turns = Î¸ / 2Ï€R....
: )

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Manmay kumar Mohanty ·2010-02-24 20:03:55

GOVIND THE GENIUS HAS DONE IT AGAIN. THNKS GOVIND.

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pranav ·2010-02-24 21:09:19

well done govind!!!

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