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Scorching goIITian

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29 Oct 2008 18:21:53 IST

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Rotational Motion

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Rotational Motion

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Bhimsen Padalkar

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29 Oct 2008 18:24:10 IST

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A rod of length L is fixed vertically in the sand. If the rod falls to the ground in vertical plane. Prove that the speed of its tip when it reaches the ground is under root 3gL.

akshay A NEW BEGINNING...

Blazing goIITian

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29 Oct 2008 20:21:59 IST

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simple apply energy conservation

mgL = 1/2 I w ^2 ( w = angular velocity )

I = 2ML^2/3...

and hence w = underroot of 3g/ L

hence v = underroot of 3g/L * L = underoot 3gL...

Karthik M

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29 Oct 2008 21:10:39 IST

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Akshay has made a grave mistake. His energy conservation equation is wrong, and his MI formula is also wrong.

Here's the solution :

The center of mass of the rod falls down by a distance of L/2, and not L. Hence,

$\frac{mgL}{2} = \frac{1}{2} \frac{ml^2}{3} \omega ^2$

which gives $\omega = \sqrt{3gR}$

Decoder

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29 Oct 2008 21:12:50 IST

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i think akshay's anser is correct...can u explain.why u r saying so ?? :)

Karthik M

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29 Oct 2008 21:13:54 IST

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Posted in a haste... see my previous post.

akshay A NEW BEGINNING...

Blazing goIITian

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29 Oct 2008 21:16:05 IST

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@ karthik2007 i am taking the rotation abt the lower most point
and u are taking abt COM
u are free to take both.....

Decoder

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29 Oct 2008 21:23:37 IST

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actually .. mv^2 /2 + I(cm)w^2 /2 = mgl/2.. by com..

& mgl/2 = 1/2 ml^2 / 3 w^2... (by frame of hinge pt)..

i think its over..

akshay A NEW BEGINNING...

Blazing goIITian

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29 Oct 2008 21:24:27 IST

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ya ...right one by decoder..

Bhimsen Padalkar

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29 Oct 2008 21:26:17 IST

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But we should consider velocity of tip of rod. Why centre of mass ?

Karthik M

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29 Oct 2008 21:26:44 IST

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akshay, even I've taken the rotation about the bottom most point. The MI of a rod about one of its ends is ml^2/3, and not 2ml^2/3.

akshay A NEW BEGINNING...

Blazing goIITian

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29 Oct 2008 21:29:49 IST

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it is ML ^3 when it is parallel to axis passing through COM
but here it is perpendicular...

Decoder

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29 Oct 2008 21:30:32 IST

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@bhim ..while talking in hinge frame we r having pure rotational motion..

while in com frame..we both have tranlational velocity as well as rotational velocity..

Karthik M

Blazing goIITian

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29 Oct 2008 21:36:57 IST

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There is nothing like parallel/perpendicular when you consider a thin rod. Most importantly, the center of mass comes down by a distance of L/2 as it is situated at a height L/2 above the ground. So either way your method has a flaw.

ramyani chakrabarty

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29 Oct 2008 21:38:05 IST

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I m complicating da prob :

Q. A rod of length L, mass m , is standing on a smooth horizontal surface. A slight disturbance causes da lower end to slip on da smooth surface and da rod starts falling. Find the vel of COM of the rod at da instant when it makes an angle @ with horizontal.

* edited

Decoder

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29 Oct 2008 21:43:04 IST

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three step problem..

1)find constriant relation of bottom most pt velocity and angular velocity of rod..
2) apply com mom conservation ..

3) get ur answer by widening ur eyes. :)

akshay A NEW BEGINNING...

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29 Oct 2008 21:43:32 IST

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in ur Q u have written rough horizontal surface then u have written
is slip on smooth surface??

Decoder

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29 Oct 2008 21:44:36 IST

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:D :D :D :D

ramyani chakrabarty

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29 Oct 2008 21:47:41 IST

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now it is smooth.

ramyani chakrabarty

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29 Oct 2008 21:51:19 IST

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I widened my eyes !

but gimme da soln.

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